Related
I have a pandas data frame df like:
a b
A 1
A 2
B 5
B 5
B 4
C 6
I want to group by the first column and get second column as lists in rows:
A [1,2]
B [5,5,4]
C [6]
Is it possible to do something like this using pandas groupby?
You can do this using groupby to group on the column of interest and then apply list to every group:
In [1]: df = pd.DataFrame( {'a':['A','A','B','B','B','C'], 'b':[1,2,5,5,4,6]})
df
Out[1]:
a b
0 A 1
1 A 2
2 B 5
3 B 5
4 B 4
5 C 6
In [2]: df.groupby('a')['b'].apply(list)
Out[2]:
a
A [1, 2]
B [5, 5, 4]
C [6]
Name: b, dtype: object
In [3]: df1 = df.groupby('a')['b'].apply(list).reset_index(name='new')
df1
Out[3]:
a new
0 A [1, 2]
1 B [5, 5, 4]
2 C [6]
A handy way to achieve this would be:
df.groupby('a').agg({'b':lambda x: list(x)})
Look into writing Custom Aggregations: https://www.kaggle.com/akshaysehgal/how-to-group-by-aggregate-using-py
If performance is important go down to numpy level:
import numpy as np
df = pd.DataFrame({'a': np.random.randint(0, 60, 600), 'b': [1, 2, 5, 5, 4, 6]*100})
def f(df):
keys, values = df.sort_values('a').values.T
ukeys, index = np.unique(keys, True)
arrays = np.split(values, index[1:])
df2 = pd.DataFrame({'a':ukeys, 'b':[list(a) for a in arrays]})
return df2
Tests:
In [301]: %timeit f(df)
1000 loops, best of 3: 1.64 ms per loop
In [302]: %timeit df.groupby('a')['b'].apply(list)
100 loops, best of 3: 5.26 ms per loop
To solve this for several columns of a dataframe:
In [5]: df = pd.DataFrame( {'a':['A','A','B','B','B','C'], 'b':[1,2,5,5,4,6],'c'
...: :[3,3,3,4,4,4]})
In [6]: df
Out[6]:
a b c
0 A 1 3
1 A 2 3
2 B 5 3
3 B 5 4
4 B 4 4
5 C 6 4
In [7]: df.groupby('a').agg(lambda x: list(x))
Out[7]:
b c
a
A [1, 2] [3, 3]
B [5, 5, 4] [3, 4, 4]
C [6] [4]
This answer was inspired from Anamika Modi's answer. Thank you!
Use any of the following groupby and agg recipes.
# Setup
df = pd.DataFrame({
'a': ['A', 'A', 'B', 'B', 'B', 'C'],
'b': [1, 2, 5, 5, 4, 6],
'c': ['x', 'y', 'z', 'x', 'y', 'z']
})
df
a b c
0 A 1 x
1 A 2 y
2 B 5 z
3 B 5 x
4 B 4 y
5 C 6 z
To aggregate multiple columns as lists, use any of the following:
df.groupby('a').agg(list)
df.groupby('a').agg(pd.Series.tolist)
b c
a
A [1, 2] [x, y]
B [5, 5, 4] [z, x, y]
C [6] [z]
To group-listify a single column only, convert the groupby to a SeriesGroupBy object, then call SeriesGroupBy.agg. Use,
df.groupby('a').agg({'b': list}) # 4.42 ms
df.groupby('a')['b'].agg(list) # 2.76 ms - faster
a
A [1, 2]
B [5, 5, 4]
C [6]
Name: b, dtype: object
As you were saying the groupby method of a pd.DataFrame object can do the job.
Example
L = ['A','A','B','B','B','C']
N = [1,2,5,5,4,6]
import pandas as pd
df = pd.DataFrame(zip(L,N),columns = list('LN'))
groups = df.groupby(df.L)
groups.groups
{'A': [0, 1], 'B': [2, 3, 4], 'C': [5]}
which gives and index-wise description of the groups.
To get elements of single groups, you can do, for instance
groups.get_group('A')
L N
0 A 1
1 A 2
groups.get_group('B')
L N
2 B 5
3 B 5
4 B 4
It is time to use agg instead of apply .
When
df = pd.DataFrame( {'a':['A','A','B','B','B','C'], 'b':[1,2,5,5,4,6], 'c': [1,2,5,5,4,6]})
If you want multiple columns stack into list , result in pd.DataFrame
df.groupby('a')[['b', 'c']].agg(list)
# or
df.groupby('a').agg(list)
If you want single column in list, result in ps.Series
df.groupby('a')['b'].agg(list)
#or
df.groupby('a')['b'].apply(list)
Note, result in pd.DataFrame is about 10x slower than result in ps.Series when you only aggregate single column, use it in multicolumns case .
Just a suplement. pandas.pivot_table is much more universal and seems more convenient：
"""data"""
df = pd.DataFrame( {'a':['A','A','B','B','B','C'],
'b':[1,2,5,5,4,6],
'c':[1,2,1,1,1,6]})
print(df)
a b c
0 A 1 1
1 A 2 2
2 B 5 1
3 B 5 1
4 B 4 1
5 C 6 6
"""pivot_table"""
pt = pd.pivot_table(df,
values=['b', 'c'],
index='a',
aggfunc={'b': list,
'c': set})
print(pt)
b c
a
A [1, 2] {1, 2}
B [5, 5, 4] {1}
C [6] {6}
If looking for a unique list while grouping multiple columns this could probably help:
df.groupby('a').agg(lambda x: list(set(x))).reset_index()
Building upon #B.M answer, here is a more general version and updated to work with newer library version: (numpy version 1.19.2, pandas version 1.2.1)
And this solution can also deal with multi-indices:
However this is not heavily tested, use with caution.
If performance is important go down to numpy level:
import pandas as pd
import numpy as np
np.random.seed(0)
df = pd.DataFrame({'a': np.random.randint(0, 10, 90), 'b': [1,2,3]*30, 'c':list('abcefghij')*10, 'd': list('hij')*30})
def f_multi(df,col_names):
if not isinstance(col_names,list):
col_names = [col_names]
values = df.sort_values(col_names).values.T
col_idcs = [df.columns.get_loc(cn) for cn in col_names]
other_col_names = [name for idx, name in enumerate(df.columns) if idx not in col_idcs]
other_col_idcs = [df.columns.get_loc(cn) for cn in other_col_names]
# split df into indexing colums(=keys) and data colums(=vals)
keys = values[col_idcs,:]
vals = values[other_col_idcs,:]
# list of tuple of key pairs
multikeys = list(zip(*keys))
# remember unique key pairs and ther indices
ukeys, index = np.unique(multikeys, return_index=True, axis=0)
# split data columns according to those indices
arrays = np.split(vals, index[1:], axis=1)
# resulting list of subarrays has same number of subarrays as unique key pairs
# each subarray has the following shape:
# rows = number of non-grouped data columns
# cols = number of data points grouped into that unique key pair
# prepare multi index
idx = pd.MultiIndex.from_arrays(ukeys.T, names=col_names)
list_agg_vals = dict()
for tup in zip(*arrays, other_col_names):
col_vals = tup[:-1] # first entries are the subarrays from above
col_name = tup[-1] # last entry is data-column name
list_agg_vals[col_name] = col_vals
df2 = pd.DataFrame(data=list_agg_vals, index=idx)
return df2
Tests:
In [227]: %timeit f_multi(df, ['a','d'])
2.54 ms ± 64.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [228]: %timeit df.groupby(['a','d']).agg(list)
4.56 ms ± 61.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Results:
for the random seed 0 one would get:
The easiest way I have found to achieve the same thing, at least for one column, which is similar to Anamika's answer, just with the tuple syntax for the aggregate function.
df.groupby('a').agg(b=('b','unique'), c=('c','unique'))
Let us using df.groupby with list and Series constructor
pd.Series({x : y.b.tolist() for x , y in df.groupby('a')})
Out[664]:
A [1, 2]
B [5, 5, 4]
C [6]
dtype: object
Here I have grouped elements with "|" as a separator
import pandas as pd
df = pd.read_csv('input.csv')
df
Out[1]:
Area Keywords
0 A 1
1 A 2
2 B 5
3 B 5
4 B 4
5 C 6
df.dropna(inplace = True)
df['Area']=df['Area'].apply(lambda x:x.lower().strip())
print df.columns
df_op = df.groupby('Area').agg({"Keywords":lambda x : "|".join(x)})
df_op.to_csv('output.csv')
Out[2]:
df_op
Area Keywords
A [1| 2]
B [5| 5| 4]
C [6]
Answer based on #EdChum's comment on his answer. Comment is this -
groupby is notoriously slow and memory hungry, what you could do is sort by column A, then find the idxmin and idxmax (probably store this in a dict) and use this to slice your dataframe would be faster I think
Let's first create a dataframe with 500k categories in first column and total df shape 20 million as mentioned in question.
df = pd.DataFrame(columns=['a', 'b'])
df['a'] = (np.random.randint(low=0, high=500000, size=(20000000,))).astype(str)
df['b'] = list(range(20000000))
print(df.shape)
df.head()
# Sort data by first column
df.sort_values(by=['a'], ascending=True, inplace=True)
df.reset_index(drop=True, inplace=True)
# Create a temp column
df['temp_idx'] = list(range(df.shape[0]))
# Take all values of b in a separate list
all_values_b = list(df.b.values)
print(len(all_values_b))
# For each category in column a, find min and max indexes
gp_df = df.groupby(['a']).agg({'temp_idx': [np.min, np.max]})
gp_df.reset_index(inplace=True)
gp_df.columns = ['a', 'temp_idx_min', 'temp_idx_max']
# Now create final list_b column, using min and max indexes for each category of a and filtering list of b.
gp_df['list_b'] = gp_df[['temp_idx_min', 'temp_idx_max']].apply(lambda x: all_values_b[x[0]:x[1]+1], axis=1)
print(gp_df.shape)
gp_df.head()
This above code takes 2 minutes for 20 million rows and 500k categories in first column.
Sorting consumes O(nlog(n)) time which is the most time consuming operation in the solutions suggested above
For a simple solution (containing single column) pd.Series.to_list would work and can be considered more efficient unless considering other frameworks
e.g.
import pandas as pd
from string import ascii_lowercase
import random
def generate_string(case=4):
return ''.join([random.choice(ascii_lowercase) for _ in range(case)])
df = pd.DataFrame({'num_val':[random.randint(0,100) for _ in range(20000000)],'string_val':[generate_string() for _ in range(20000000)]})
%timeit df.groupby('string_val').agg({'num_val':pd.Series.to_list})
For 20 million records it takes about 17.2 seconds. compared to apply(list) which takes about 19.2 and lambda function which takes about 20.6s
Just to add up to previous answers, In my case, I want the list and other functions like min and max. The way to do that is:
df = pd.DataFrame({
'a':['A','A','B','B','B','C'],
'b':[1,2,5,5,4,6]
})
df=df.groupby('a').agg({
'b':['min', 'max',lambda x: list(x)]
})
#then flattening and renaming if necessary
df.columns = df.columns.to_flat_index()
df.rename(columns={('b', 'min'): 'b_min', ('b', 'max'): 'b_max', ('b', '<lambda_0>'): 'b_list'},inplace=True)
It's a bit old but I was directed here. Is there anyway to group it by multiple different columns?
"column1", "column2", "column3"
"foo", "val1", 3
"foo", "val2", 0
"foo", "val2", 3
"bar", "other", 99
to this:
"column1", "column2", "column3"
"foo", "val1", [ 3 ]
"foo", "val2", [ 0, 3 ]
"bar", "other", [ 99 ]
I have a pandas data frame df like:
a b
A 1
A 2
B 5
B 5
B 4
C 6
I want to group by the first column and get second column as lists in rows:
A [1,2]
B [5,5,4]
C [6]
Is it possible to do something like this using pandas groupby?
You can do this using groupby to group on the column of interest and then apply list to every group:
In [1]: df = pd.DataFrame( {'a':['A','A','B','B','B','C'], 'b':[1,2,5,5,4,6]})
df
Out[1]:
a b
0 A 1
1 A 2
2 B 5
3 B 5
4 B 4
5 C 6
In [2]: df.groupby('a')['b'].apply(list)
Out[2]:
a
A [1, 2]
B [5, 5, 4]
C [6]
Name: b, dtype: object
In [3]: df1 = df.groupby('a')['b'].apply(list).reset_index(name='new')
df1
Out[3]:
a new
0 A [1, 2]
1 B [5, 5, 4]
2 C [6]
A handy way to achieve this would be:
df.groupby('a').agg({'b':lambda x: list(x)})
Look into writing Custom Aggregations: https://www.kaggle.com/akshaysehgal/how-to-group-by-aggregate-using-py
If performance is important go down to numpy level:
import numpy as np
df = pd.DataFrame({'a': np.random.randint(0, 60, 600), 'b': [1, 2, 5, 5, 4, 6]*100})
def f(df):
keys, values = df.sort_values('a').values.T
ukeys, index = np.unique(keys, True)
arrays = np.split(values, index[1:])
df2 = pd.DataFrame({'a':ukeys, 'b':[list(a) for a in arrays]})
return df2
Tests:
In [301]: %timeit f(df)
1000 loops, best of 3: 1.64 ms per loop
In [302]: %timeit df.groupby('a')['b'].apply(list)
100 loops, best of 3: 5.26 ms per loop
To solve this for several columns of a dataframe:
In [5]: df = pd.DataFrame( {'a':['A','A','B','B','B','C'], 'b':[1,2,5,5,4,6],'c'
...: :[3,3,3,4,4,4]})
In [6]: df
Out[6]:
a b c
0 A 1 3
1 A 2 3
2 B 5 3
3 B 5 4
4 B 4 4
5 C 6 4
In [7]: df.groupby('a').agg(lambda x: list(x))
Out[7]:
b c
a
A [1, 2] [3, 3]
B [5, 5, 4] [3, 4, 4]
C [6] [4]
This answer was inspired from Anamika Modi's answer. Thank you!
Use any of the following groupby and agg recipes.
# Setup
df = pd.DataFrame({
'a': ['A', 'A', 'B', 'B', 'B', 'C'],
'b': [1, 2, 5, 5, 4, 6],
'c': ['x', 'y', 'z', 'x', 'y', 'z']
})
df
a b c
0 A 1 x
1 A 2 y
2 B 5 z
3 B 5 x
4 B 4 y
5 C 6 z
To aggregate multiple columns as lists, use any of the following:
df.groupby('a').agg(list)
df.groupby('a').agg(pd.Series.tolist)
b c
a
A [1, 2] [x, y]
B [5, 5, 4] [z, x, y]
C [6] [z]
To group-listify a single column only, convert the groupby to a SeriesGroupBy object, then call SeriesGroupBy.agg. Use,
df.groupby('a').agg({'b': list}) # 4.42 ms
df.groupby('a')['b'].agg(list) # 2.76 ms - faster
a
A [1, 2]
B [5, 5, 4]
C [6]
Name: b, dtype: object
As you were saying the groupby method of a pd.DataFrame object can do the job.
Example
L = ['A','A','B','B','B','C']
N = [1,2,5,5,4,6]
import pandas as pd
df = pd.DataFrame(zip(L,N),columns = list('LN'))
groups = df.groupby(df.L)
groups.groups
{'A': [0, 1], 'B': [2, 3, 4], 'C': [5]}
which gives and index-wise description of the groups.
To get elements of single groups, you can do, for instance
groups.get_group('A')
L N
0 A 1
1 A 2
groups.get_group('B')
L N
2 B 5
3 B 5
4 B 4
It is time to use agg instead of apply .
When
df = pd.DataFrame( {'a':['A','A','B','B','B','C'], 'b':[1,2,5,5,4,6], 'c': [1,2,5,5,4,6]})
If you want multiple columns stack into list , result in pd.DataFrame
df.groupby('a')[['b', 'c']].agg(list)
# or
df.groupby('a').agg(list)
If you want single column in list, result in ps.Series
df.groupby('a')['b'].agg(list)
#or
df.groupby('a')['b'].apply(list)
Note, result in pd.DataFrame is about 10x slower than result in ps.Series when you only aggregate single column, use it in multicolumns case .
Just a suplement. pandas.pivot_table is much more universal and seems more convenient：
"""data"""
df = pd.DataFrame( {'a':['A','A','B','B','B','C'],
'b':[1,2,5,5,4,6],
'c':[1,2,1,1,1,6]})
print(df)
a b c
0 A 1 1
1 A 2 2
2 B 5 1
3 B 5 1
4 B 4 1
5 C 6 6
"""pivot_table"""
pt = pd.pivot_table(df,
values=['b', 'c'],
index='a',
aggfunc={'b': list,
'c': set})
print(pt)
b c
a
A [1, 2] {1, 2}
B [5, 5, 4] {1}
C [6] {6}
If looking for a unique list while grouping multiple columns this could probably help:
df.groupby('a').agg(lambda x: list(set(x))).reset_index()
Building upon #B.M answer, here is a more general version and updated to work with newer library version: (numpy version 1.19.2, pandas version 1.2.1)
And this solution can also deal with multi-indices:
However this is not heavily tested, use with caution.
If performance is important go down to numpy level:
import pandas as pd
import numpy as np
np.random.seed(0)
df = pd.DataFrame({'a': np.random.randint(0, 10, 90), 'b': [1,2,3]*30, 'c':list('abcefghij')*10, 'd': list('hij')*30})
def f_multi(df,col_names):
if not isinstance(col_names,list):
col_names = [col_names]
values = df.sort_values(col_names).values.T
col_idcs = [df.columns.get_loc(cn) for cn in col_names]
other_col_names = [name for idx, name in enumerate(df.columns) if idx not in col_idcs]
other_col_idcs = [df.columns.get_loc(cn) for cn in other_col_names]
# split df into indexing colums(=keys) and data colums(=vals)
keys = values[col_idcs,:]
vals = values[other_col_idcs,:]
# list of tuple of key pairs
multikeys = list(zip(*keys))
# remember unique key pairs and ther indices
ukeys, index = np.unique(multikeys, return_index=True, axis=0)
# split data columns according to those indices
arrays = np.split(vals, index[1:], axis=1)
# resulting list of subarrays has same number of subarrays as unique key pairs
# each subarray has the following shape:
# rows = number of non-grouped data columns
# cols = number of data points grouped into that unique key pair
# prepare multi index
idx = pd.MultiIndex.from_arrays(ukeys.T, names=col_names)
list_agg_vals = dict()
for tup in zip(*arrays, other_col_names):
col_vals = tup[:-1] # first entries are the subarrays from above
col_name = tup[-1] # last entry is data-column name
list_agg_vals[col_name] = col_vals
df2 = pd.DataFrame(data=list_agg_vals, index=idx)
return df2
Tests:
In [227]: %timeit f_multi(df, ['a','d'])
2.54 ms ± 64.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [228]: %timeit df.groupby(['a','d']).agg(list)
4.56 ms ± 61.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Results:
for the random seed 0 one would get:
The easiest way I have found to achieve the same thing, at least for one column, which is similar to Anamika's answer, just with the tuple syntax for the aggregate function.
df.groupby('a').agg(b=('b','unique'), c=('c','unique'))
Let us using df.groupby with list and Series constructor
pd.Series({x : y.b.tolist() for x , y in df.groupby('a')})
Out[664]:
A [1, 2]
B [5, 5, 4]
C [6]
dtype: object
Here I have grouped elements with "|" as a separator
import pandas as pd
df = pd.read_csv('input.csv')
df
Out[1]:
Area Keywords
0 A 1
1 A 2
2 B 5
3 B 5
4 B 4
5 C 6
df.dropna(inplace = True)
df['Area']=df['Area'].apply(lambda x:x.lower().strip())
print df.columns
df_op = df.groupby('Area').agg({"Keywords":lambda x : "|".join(x)})
df_op.to_csv('output.csv')
Out[2]:
df_op
Area Keywords
A [1| 2]
B [5| 5| 4]
C [6]
Answer based on #EdChum's comment on his answer. Comment is this -
groupby is notoriously slow and memory hungry, what you could do is sort by column A, then find the idxmin and idxmax (probably store this in a dict) and use this to slice your dataframe would be faster I think
Let's first create a dataframe with 500k categories in first column and total df shape 20 million as mentioned in question.
df = pd.DataFrame(columns=['a', 'b'])
df['a'] = (np.random.randint(low=0, high=500000, size=(20000000,))).astype(str)
df['b'] = list(range(20000000))
print(df.shape)
df.head()
# Sort data by first column
df.sort_values(by=['a'], ascending=True, inplace=True)
df.reset_index(drop=True, inplace=True)
# Create a temp column
df['temp_idx'] = list(range(df.shape[0]))
# Take all values of b in a separate list
all_values_b = list(df.b.values)
print(len(all_values_b))
# For each category in column a, find min and max indexes
gp_df = df.groupby(['a']).agg({'temp_idx': [np.min, np.max]})
gp_df.reset_index(inplace=True)
gp_df.columns = ['a', 'temp_idx_min', 'temp_idx_max']
# Now create final list_b column, using min and max indexes for each category of a and filtering list of b.
gp_df['list_b'] = gp_df[['temp_idx_min', 'temp_idx_max']].apply(lambda x: all_values_b[x[0]:x[1]+1], axis=1)
print(gp_df.shape)
gp_df.head()
This above code takes 2 minutes for 20 million rows and 500k categories in first column.
Sorting consumes O(nlog(n)) time which is the most time consuming operation in the solutions suggested above
For a simple solution (containing single column) pd.Series.to_list would work and can be considered more efficient unless considering other frameworks
e.g.
import pandas as pd
from string import ascii_lowercase
import random
def generate_string(case=4):
return ''.join([random.choice(ascii_lowercase) for _ in range(case)])
df = pd.DataFrame({'num_val':[random.randint(0,100) for _ in range(20000000)],'string_val':[generate_string() for _ in range(20000000)]})
%timeit df.groupby('string_val').agg({'num_val':pd.Series.to_list})
For 20 million records it takes about 17.2 seconds. compared to apply(list) which takes about 19.2 and lambda function which takes about 20.6s
Just to add up to previous answers, In my case, I want the list and other functions like min and max. The way to do that is:
df = pd.DataFrame({
'a':['A','A','B','B','B','C'],
'b':[1,2,5,5,4,6]
})
df=df.groupby('a').agg({
'b':['min', 'max',lambda x: list(x)]
})
#then flattening and renaming if necessary
df.columns = df.columns.to_flat_index()
df.rename(columns={('b', 'min'): 'b_min', ('b', 'max'): 'b_max', ('b', '<lambda_0>'): 'b_list'},inplace=True)
It's a bit old but I was directed here. Is there anyway to group it by multiple different columns?
"column1", "column2", "column3"
"foo", "val1", 3
"foo", "val2", 0
"foo", "val2", 3
"bar", "other", 99
to this:
"column1", "column2", "column3"
"foo", "val1", [ 3 ]
"foo", "val2", [ 0, 3 ]
"bar", "other", [ 99 ]
I have a pandas dataframe in which one column of text strings contains comma-separated values. I want to split each CSV field and create a new row per entry (assume that CSV are clean and need only be split on ','). For example, a should become b:
In [7]: a
Out[7]:
var1 var2
0 a,b,c 1
1 d,e,f 2
In [8]: b
Out[8]:
var1 var2
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
So far, I have tried various simple functions, but the .apply method seems to only accept one row as return value when it is used on an axis, and I can't get .transform to work. Any suggestions would be much appreciated!
Example data:
from pandas import DataFrame
import numpy as np
a = DataFrame([{'var1': 'a,b,c', 'var2': 1},
{'var1': 'd,e,f', 'var2': 2}])
b = DataFrame([{'var1': 'a', 'var2': 1},
{'var1': 'b', 'var2': 1},
{'var1': 'c', 'var2': 1},
{'var1': 'd', 'var2': 2},
{'var1': 'e', 'var2': 2},
{'var1': 'f', 'var2': 2}])
I know this won't work because we lose DataFrame meta-data by going through numpy, but it should give you a sense of what I tried to do:
def fun(row):
letters = row['var1']
letters = letters.split(',')
out = np.array([row] * len(letters))
out['var1'] = letters
a['idx'] = range(a.shape[0])
z = a.groupby('idx')
z.transform(fun)
UPDATE 3: it makes more sense to use Series.explode() / DataFrame.explode() methods (implemented in Pandas 0.25.0 and extended in Pandas 1.3.0 to support multi-column explode) as is shown in the usage example:
for a single column:
In [1]: df = pd.DataFrame({'A': [[0, 1, 2], 'foo', [], [3, 4]],
...: 'B': 1,
...: 'C': [['a', 'b', 'c'], np.nan, [], ['d', 'e']]})
In [2]: df
Out[2]:
A B C
0 [0, 1, 2] 1 [a, b, c]
1 foo 1 NaN
2 [] 1 []
3 [3, 4] 1 [d, e]
In [3]: df.explode('A')
Out[3]:
A B C
0 0 1 [a, b, c]
0 1 1 [a, b, c]
0 2 1 [a, b, c]
1 foo 1 NaN
2 NaN 1 []
3 3 1 [d, e]
3 4 1 [d, e]
for multiple columns (for Pandas 1.3.0+):
In [4]: df.explode(['A', 'C'])
Out[4]:
A B C
0 0 1 a
0 1 1 b
0 2 1 c
1 foo 1 NaN
2 NaN 1 NaN
3 3 1 d
3 4 1 e
UPDATE 2: more generic vectorized function, which will work for multiple normal and multiple list columns
def explode(df, lst_cols, fill_value='', preserve_index=False):
# make sure `lst_cols` is list-alike
if (lst_cols is not None
and len(lst_cols) > 0
and not isinstance(lst_cols, (list, tuple, np.ndarray, pd.Series))):
lst_cols = [lst_cols]
# all columns except `lst_cols`
idx_cols = df.columns.difference(lst_cols)
# calculate lengths of lists
lens = df[lst_cols[0]].str.len()
# preserve original index values
idx = np.repeat(df.index.values, lens)
# create "exploded" DF
res = (pd.DataFrame({
col:np.repeat(df[col].values, lens)
for col in idx_cols},
index=idx)
.assign(**{col:np.concatenate(df.loc[lens>0, col].values)
for col in lst_cols}))
# append those rows that have empty lists
if (lens == 0).any():
# at least one list in cells is empty
res = (res.append(df.loc[lens==0, idx_cols], sort=False)
.fillna(fill_value))
# revert the original index order
res = res.sort_index()
# reset index if requested
if not preserve_index:
res = res.reset_index(drop=True)
return res
Demo:
Multiple list columns - all list columns must have the same # of elements in each row:
In [134]: df
Out[134]:
aaa myid num text
0 10 1 [1, 2, 3] [aa, bb, cc]
1 11 2 [] []
2 12 3 [1, 2] [cc, dd]
3 13 4 [] []
In [135]: explode(df, ['num','text'], fill_value='')
Out[135]:
aaa myid num text
0 10 1 1 aa
1 10 1 2 bb
2 10 1 3 cc
3 11 2
4 12 3 1 cc
5 12 3 2 dd
6 13 4
preserving original index values:
In [136]: explode(df, ['num','text'], fill_value='', preserve_index=True)
Out[136]:
aaa myid num text
0 10 1 1 aa
0 10 1 2 bb
0 10 1 3 cc
1 11 2
2 12 3 1 cc
2 12 3 2 dd
3 13 4
Setup:
df = pd.DataFrame({
'aaa': {0: 10, 1: 11, 2: 12, 3: 13},
'myid': {0: 1, 1: 2, 2: 3, 3: 4},
'num': {0: [1, 2, 3], 1: [], 2: [1, 2], 3: []},
'text': {0: ['aa', 'bb', 'cc'], 1: [], 2: ['cc', 'dd'], 3: []}
})
CSV column:
In [46]: df
Out[46]:
var1 var2 var3
0 a,b,c 1 XX
1 d,e,f,x,y 2 ZZ
In [47]: explode(df.assign(var1=df.var1.str.split(',')), 'var1')
Out[47]:
var1 var2 var3
0 a 1 XX
1 b 1 XX
2 c 1 XX
3 d 2 ZZ
4 e 2 ZZ
5 f 2 ZZ
6 x 2 ZZ
7 y 2 ZZ
using this little trick we can convert CSV-like column to list column:
In [48]: df.assign(var1=df.var1.str.split(','))
Out[48]:
var1 var2 var3
0 [a, b, c] 1 XX
1 [d, e, f, x, y] 2 ZZ
UPDATE: generic vectorized approach (will work also for multiple columns):
Original DF:
In [177]: df
Out[177]:
var1 var2 var3
0 a,b,c 1 XX
1 d,e,f,x,y 2 ZZ
Solution:
first let's convert CSV strings to lists:
In [178]: lst_col = 'var1'
In [179]: x = df.assign(**{lst_col:df[lst_col].str.split(',')})
In [180]: x
Out[180]:
var1 var2 var3
0 [a, b, c] 1 XX
1 [d, e, f, x, y] 2 ZZ
Now we can do this:
In [181]: pd.DataFrame({
...: col:np.repeat(x[col].values, x[lst_col].str.len())
...: for col in x.columns.difference([lst_col])
...: }).assign(**{lst_col:np.concatenate(x[lst_col].values)})[x.columns.tolist()]
...:
Out[181]:
var1 var2 var3
0 a 1 XX
1 b 1 XX
2 c 1 XX
3 d 2 ZZ
4 e 2 ZZ
5 f 2 ZZ
6 x 2 ZZ
7 y 2 ZZ
OLD answer:
Inspired by #AFinkelstein solution, i wanted to make it bit more generalized which could be applied to DF with more than two columns and as fast, well almost, as fast as AFinkelstein's solution):
In [2]: df = pd.DataFrame(
...: [{'var1': 'a,b,c', 'var2': 1, 'var3': 'XX'},
...: {'var1': 'd,e,f,x,y', 'var2': 2, 'var3': 'ZZ'}]
...: )
In [3]: df
Out[3]:
var1 var2 var3
0 a,b,c 1 XX
1 d,e,f,x,y 2 ZZ
In [4]: (df.set_index(df.columns.drop('var1',1).tolist())
...: .var1.str.split(',', expand=True)
...: .stack()
...: .reset_index()
...: .rename(columns={0:'var1'})
...: .loc[:, df.columns]
...: )
Out[4]:
var1 var2 var3
0 a 1 XX
1 b 1 XX
2 c 1 XX
3 d 2 ZZ
4 e 2 ZZ
5 f 2 ZZ
6 x 2 ZZ
7 y 2 ZZ
After painful experimentation to find something faster than the accepted answer, I got this to work. It ran around 100x faster on the dataset I tried it on.
If someone knows a way to make this more elegant, by all means please modify my code. I couldn't find a way that works without setting the other columns you want to keep as the index and then resetting the index and re-naming the columns, but I'd imagine there's something else that works.
b = DataFrame(a.var1.str.split(',').tolist(), index=a.var2).stack()
b = b.reset_index()[[0, 'var2']] # var1 variable is currently labeled 0
b.columns = ['var1', 'var2'] # renaming var1
Pandas >= 0.25
Series and DataFrame methods define a .explode() method that explodes lists into separate rows. See the docs section on Exploding a list-like column.
Since you have a list of comma separated strings, split the string on comma to get a list of elements, then call explode on that column.
df = pd.DataFrame({'var1': ['a,b,c', 'd,e,f'], 'var2': [1, 2]})
df
var1 var2
0 a,b,c 1
1 d,e,f 2
df.assign(var1=df['var1'].str.split(',')).explode('var1')
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
Note that explode only works on a single column (for now). To explode multiple columns at once, see below.
NaNs and empty lists get the treatment they deserve without you having to jump through hoops to get it right.
df = pd.DataFrame({'var1': ['d,e,f', '', np.nan], 'var2': [1, 2, 3]})
df
var1 var2
0 d,e,f 1
1 2
2 NaN 3
df['var1'].str.split(',')
0 [d, e, f]
1 []
2 NaN
df.assign(var1=df['var1'].str.split(',')).explode('var1')
var1 var2
0 d 1
0 e 1
0 f 1
1 2 # empty list entry becomes empty string after exploding
2 NaN 3 # NaN left un-touched
This is a serious advantage over ravel/repeat -based solutions (which ignore empty lists completely, and choke on NaNs).
Exploding Multiple Columns
pandas 1.3 update
df.explode works on multiple columns starting from pandas 1.3:
df = pd.DataFrame({'var1': ['a,b,c', 'd,e,f'],
'var2': ['i,j,k', 'l,m,n'],
'var3': [1, 2]})
df
var1 var2 var3
0 a,b,c i,j,k 1
1 d,e,f l,m,n 2
(df.set_index(['var3'])
.apply(lambda col: col.str.split(','))
.explode(['var1', 'var2'])
.reset_index()
.reindex(df.columns, axis=1))
var1 var2 var3
0 a i 1
1 b j 1
2 c k 1
3 d l 2
4 e m 2
5 f n 2
On older versions, you would move the explode column inside the apply which is a lot less performant:
(df.set_index(['var3'])
.apply(lambda col: col.str.split(',').explode())
.reset_index()
.reindex(df.columns, axis=1))
The idea is to set as the index, all the columns that should NOT be exploded, then explode the remaining columns via apply. This works well when the lists are equally sized.
How about something like this:
In [55]: pd.concat([Series(row['var2'], row['var1'].split(','))
for _, row in a.iterrows()]).reset_index()
Out[55]:
index 0
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
Then you just have to rename the columns
Here's a function I wrote for this common task. It's more efficient than the Series/stack methods. Column order and names are retained.
def tidy_split(df, column, sep='|', keep=False):
"""
Split the values of a column and expand so the new DataFrame has one split
value per row. Filters rows where the column is missing.
Params
------
df : pandas.DataFrame
dataframe with the column to split and expand
column : str
the column to split and expand
sep : str
the string used to split the column's values
keep : bool
whether to retain the presplit value as it's own row
Returns
-------
pandas.DataFrame
Returns a dataframe with the same columns as `df`.
"""
indexes = list()
new_values = list()
df = df.dropna(subset=[column])
for i, presplit in enumerate(df[column].astype(str)):
values = presplit.split(sep)
if keep and len(values) > 1:
indexes.append(i)
new_values.append(presplit)
for value in values:
indexes.append(i)
new_values.append(value)
new_df = df.iloc[indexes, :].copy()
new_df[column] = new_values
return new_df
With this function, the original question is as simple as:
tidy_split(a, 'var1', sep=',')
Similar question as: pandas: How do I split text in a column into multiple rows?
You could do:
>> a=pd.DataFrame({"var1":"a,b,c d,e,f".split(),"var2":[1,2]})
>> s = a.var1.str.split(",").apply(pd.Series, 1).stack()
>> s.index = s.index.droplevel(-1)
>> del a['var1']
>> a.join(s)
var2 var1
0 1 a
0 1 b
0 1 c
1 2 d
1 2 e
1 2 f
There is a possibility to split and explode the dataframe without changing the structure of dataframe
Split and expand data of specific columns
Input:
var1 var2
0 a,b,c 1
1 d,e,f 2
#Get the indexes which are repetative with the split
df['var1'] = df['var1'].str.split(',')
df = df.explode('var1')
Out:
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
Edit-1
Split and Expand of rows for Multiple columns
Filename RGB RGB_type
0 A [[0, 1650, 6, 39], [0, 1691, 1, 59], [50, 1402... [r, g, b]
1 B [[0, 1423, 16, 38], [0, 1445, 16, 46], [0, 141... [r, g, b]
Re indexing based on the reference column and aligning the column value information with stack
df = df.reindex(df.index.repeat(df['RGB_type'].apply(len)))
df = df.groupby('Filename').apply(lambda x:x.apply(lambda y: pd.Series(y.iloc[0])))
df.reset_index(drop=True).ffill()
Out:
Filename RGB_type Top 1 colour Top 1 frequency Top 2 colour Top 2 frequency
Filename
A 0 A r 0 1650 6 39
1 A g 0 1691 1 59
2 A b 50 1402 49 187
B 0 B r 0 1423 16 38
1 B g 0 1445 16 46
2 B b 0 1419 16 39
TL;DR
import pandas as pd
import numpy as np
def explode_str(df, col, sep):
s = df[col]
i = np.arange(len(s)).repeat(s.str.count(sep) + 1)
return df.iloc[i].assign(**{col: sep.join(s).split(sep)})
def explode_list(df, col):
s = df[col]
i = np.arange(len(s)).repeat(s.str.len())
return df.iloc[i].assign(**{col: np.concatenate(s)})
Demonstration
explode_str(a, 'var1', ',')
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
Let's create a new dataframe d that has lists
d = a.assign(var1=lambda d: d.var1.str.split(','))
explode_list(d, 'var1')
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
General Comments
I'll use np.arange with repeat to produce dataframe index positions that I can use with iloc.
FAQ
Why don't I use loc?
Because the index may not be unique and using loc will return every row that matches a queried index.
Why don't you use the values attribute and slice that?
When calling values, if the entirety of the the dataframe is in one cohesive "block", Pandas will return a view of the array that is the "block". Otherwise Pandas will have to cobble together a new array. When cobbling, that array must be of a uniform dtype. Often that means returning an array with dtype that is object. By using iloc instead of slicing the values attribute, I alleviate myself from having to deal with that.
Why do you use assign?
When I use assign using the same column name that I'm exploding, I overwrite the existing column and maintain its position in the dataframe.
Why are the index values repeat?
By virtue of using iloc on repeated positions, the resulting index shows the same repeated pattern. One repeat for each element the list or string.
This can be reset with reset_index(drop=True)
For Strings
I don't want to have to split the strings prematurely. So instead I count the occurrences of the sep argument assuming that if I were to split, the length of the resulting list would be one more than the number of separators.
I then use that sep to join the strings then split.
def explode_str(df, col, sep):
s = df[col]
i = np.arange(len(s)).repeat(s.str.count(sep) + 1)
return df.iloc[i].assign(**{col: sep.join(s).split(sep)})
For Lists
Similar as for strings except I don't need to count occurrences of sep because its already split.
I use Numpy's concatenate to jam the lists together.
import pandas as pd
import numpy as np
def explode_list(df, col):
s = df[col]
i = np.arange(len(s)).repeat(s.str.len())
return df.iloc[i].assign(**{col: np.concatenate(s)})
I came up with a solution for dataframes with arbitrary numbers of columns (while still only separating one column's entries at a time).
def splitDataFrameList(df,target_column,separator):
''' df = dataframe to split,
target_column = the column containing the values to split
separator = the symbol used to perform the split
returns: a dataframe with each entry for the target column separated, with each element moved into a new row.
The values in the other columns are duplicated across the newly divided rows.
'''
def splitListToRows(row,row_accumulator,target_column,separator):
split_row = row[target_column].split(separator)
for s in split_row:
new_row = row.to_dict()
new_row[target_column] = s
row_accumulator.append(new_row)
new_rows = []
df.apply(splitListToRows,axis=1,args = (new_rows,target_column,separator))
new_df = pandas.DataFrame(new_rows)
return new_df
Here is a fairly straightforward message that uses the split method from pandas str accessor and then uses NumPy to flatten each row into a single array.
The corresponding values are retrieved by repeating the non-split column the correct number of times with np.repeat.
var1 = df.var1.str.split(',', expand=True).values.ravel()
var2 = np.repeat(df.var2.values, len(var1) / len(df))
pd.DataFrame({'var1': var1,
'var2': var2})
var1 var2
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
I have been struggling with out-of-memory experience using various way to explode my lists so I prepared some benchmarks to help me decide which answers to upvote. I tested five scenarios with varying proportions of the list length to the number of lists. Sharing the results below:
Time: (less is better, click to view large version)
Peak memory usage: (less is better)
Conclusions:
#MaxU's answer (update 2), codename concatenate offers the best speed in almost every case, while keeping the peek memory usage low,
see #DMulligan's answer (codename stack) if you need to process lots of rows with relatively small lists and can afford increased peak memory,
the accepted #Chang's answer works well for data frames that have a few rows but very large lists.
Full details (functions and benchmarking code) are in this GitHub gist. Please note that the benchmark problem was simplified and did not include splitting of strings into the list - which most solutions performed in a similar fashion.
One-liner using split(___, expand=True) and the level and name arguments to reset_index():
>>> b = a.var1.str.split(',', expand=True).set_index(a.var2).stack().reset_index(level=0, name='var1')
>>> b
var2 var1
0 1 a
1 1 b
2 1 c
0 2 d
1 2 e
2 2 f
If you need b to look exactly like in the question, you can additionally do:
>>> b = b.reset_index(drop=True)[['var1', 'var2']]
>>> b
var1 var2
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
Based on the excellent #DMulligan's solution, here is a generic vectorized (no loops) function which splits a column of a dataframe into multiple rows, and merges it back to the original dataframe. It also uses a great generic change_column_order function from this answer.
def change_column_order(df, col_name, index):
cols = df.columns.tolist()
cols.remove(col_name)
cols.insert(index, col_name)
return df[cols]
def split_df(dataframe, col_name, sep):
orig_col_index = dataframe.columns.tolist().index(col_name)
orig_index_name = dataframe.index.name
orig_columns = dataframe.columns
dataframe = dataframe.reset_index() # we need a natural 0-based index for proper merge
index_col_name = (set(dataframe.columns) - set(orig_columns)).pop()
df_split = pd.DataFrame(
pd.DataFrame(dataframe[col_name].str.split(sep).tolist())
.stack().reset_index(level=1, drop=1), columns=[col_name])
df = dataframe.drop(col_name, axis=1)
df = pd.merge(df, df_split, left_index=True, right_index=True, how='inner')
df = df.set_index(index_col_name)
df.index.name = orig_index_name
# merge adds the column to the last place, so we need to move it back
return change_column_order(df, col_name, orig_col_index)
Example:
df = pd.DataFrame([['a:b', 1, 4], ['c:d', 2, 5], ['e:f:g:h', 3, 6]],
columns=['Name', 'A', 'B'], index=[10, 12, 13])
df
Name A B
10 a:b 1 4
12 c:d 2 5
13 e:f:g:h 3 6
split_df(df, 'Name', ':')
Name A B
10 a 1 4
10 b 1 4
12 c 2 5
12 d 2 5
13 e 3 6
13 f 3 6
13 g 3 6
13 h 3 6
Note that it preserves the original index and order of the columns. It also works with dataframes which have non-sequential index.
The string function split can take an option boolean argument 'expand'.
Here is a solution using this argument:
(a.var1
.str.split(",",expand=True)
.set_index(a.var2)
.stack()
.reset_index(level=1, drop=True)
.reset_index()
.rename(columns={0:"var1"}))
I do appreciate the answer of "Chang She", really, but the iterrows() function takes long time on large dataset. I faced that issue and I came to this.
# First, reset_index to make the index a column
a = a.reset_index().rename(columns={'index':'duplicated_idx'})
# Get a longer series with exploded cells to rows
series = pd.DataFrame(a['var1'].str.split('/')
.tolist(), index=a.duplicated_idx).stack()
# New df from series and merge with the old one
b = series.reset_index([0, 'duplicated_idx'])
b = b.rename(columns={0:'var1'})
# Optional & Advanced: In case, there are other columns apart from var1 & var2
b.merge(
a[a.columns.difference(['var1'])],
on='duplicated_idx')
# Optional: Delete the "duplicated_index"'s column, and reorder columns
b = b[a.columns.difference(['duplicated_idx'])]
One-liner using assign and explode:
col1 col2
0 a,b,c 1
1 d,e,f 2
df.assign(col1 = df.col1.str.split(',')).explode('col1', ignore_index=True)
Output:
col1 col2
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
Just used jiln's excellent answer from above, but needed to expand to split multiple columns. Thought I would share.
def splitDataFrameList(df,target_column,separator):
''' df = dataframe to split,
target_column = the column containing the values to split
separator = the symbol used to perform the split
returns: a dataframe with each entry for the target column separated, with each element moved into a new row.
The values in the other columns are duplicated across the newly divided rows.
'''
def splitListToRows(row, row_accumulator, target_columns, separator):
split_rows = []
for target_column in target_columns:
split_rows.append(row[target_column].split(separator))
# Seperate for multiple columns
for i in range(len(split_rows[0])):
new_row = row.to_dict()
for j in range(len(split_rows)):
new_row[target_columns[j]] = split_rows[j][i]
row_accumulator.append(new_row)
new_rows = []
df.apply(splitListToRows,axis=1,args = (new_rows,target_column,separator))
new_df = pd.DataFrame(new_rows)
return new_df
upgraded MaxU's answer with MultiIndex support
def explode(df, lst_cols, fill_value='', preserve_index=False):
"""
usage:
In [134]: df
Out[134]:
aaa myid num text
0 10 1 [1, 2, 3] [aa, bb, cc]
1 11 2 [] []
2 12 3 [1, 2] [cc, dd]
3 13 4 [] []
In [135]: explode(df, ['num','text'], fill_value='')
Out[135]:
aaa myid num text
0 10 1 1 aa
1 10 1 2 bb
2 10 1 3 cc
3 11 2
4 12 3 1 cc
5 12 3 2 dd
6 13 4
"""
# make sure `lst_cols` is list-alike
if (lst_cols is not None
and len(lst_cols) > 0
and not isinstance(lst_cols, (list, tuple, np.ndarray, pd.Series))):
lst_cols = [lst_cols]
# all columns except `lst_cols`
idx_cols = df.columns.difference(lst_cols)
# calculate lengths of lists
lens = df[lst_cols[0]].str.len()
# preserve original index values
idx = np.repeat(df.index.values, lens)
res = (pd.DataFrame({
col:np.repeat(df[col].values, lens)
for col in idx_cols},
index=idx)
.assign(**{col:np.concatenate(df.loc[lens>0, col].values)
for col in lst_cols}))
# append those rows that have empty lists
if (lens == 0).any():
# at least one list in cells is empty
res = (res.append(df.loc[lens==0, idx_cols], sort=False)
.fillna(fill_value))
# revert the original index order
res = res.sort_index()
# reset index if requested
if not preserve_index:
res = res.reset_index(drop=True)
# if original index is MultiIndex build the dataframe from the multiindex
# create "exploded" DF
if isinstance(df.index, pd.MultiIndex):
res = res.reindex(
index=pd.MultiIndex.from_tuples(
res.index,
names=['number', 'color']
)
)
return res
My version of the solution to add to this collection! :-)
# Original problem
from pandas import DataFrame
import numpy as np
a = DataFrame([{'var1': 'a,b,c', 'var2': 1},
{'var1': 'd,e,f', 'var2': 2}])
b = DataFrame([{'var1': 'a', 'var2': 1},
{'var1': 'b', 'var2': 1},
{'var1': 'c', 'var2': 1},
{'var1': 'd', 'var2': 2},
{'var1': 'e', 'var2': 2},
{'var1': 'f', 'var2': 2}])
### My solution
import pandas as pd
import functools
def expand_on_cols(df, fuse_cols, delim=","):
def expand_on_col(df, fuse_col):
col_order = df.columns
df_expanded = pd.DataFrame(
df.set_index([x for x in df.columns if x != fuse_col])[fuse_col]
.apply(lambda x: x.split(delim))
.explode()
).reset_index()
return df_expanded[col_order]
all_expanded = functools.reduce(expand_on_col, fuse_cols, df)
return all_expanded
assert(b.equals(expand_on_cols(a, ["var1"], delim=",")))
I have come up with the following solution to this problem:
def iter_var1(d):
for _, row in d.iterrows():
for v in row["var1"].split(","):
yield (v, row["var2"])
new_a = DataFrame.from_records([i for i in iter_var1(a)],
columns=["var1", "var2"])
Another solution that uses python copy package
import copy
new_observations = list()
def pandas_explode(df, column_to_explode):
new_observations = list()
for row in df.to_dict(orient='records'):
explode_values = row[column_to_explode]
del row[column_to_explode]
if type(explode_values) is list or type(explode_values) is tuple:
for explode_value in explode_values:
new_observation = copy.deepcopy(row)
new_observation[column_to_explode] = explode_value
new_observations.append(new_observation)
else:
new_observation = copy.deepcopy(row)
new_observation[column_to_explode] = explode_values
new_observations.append(new_observation)
return_df = pd.DataFrame(new_observations)
return return_df
df = pandas_explode(df, column_name)
There are a lot of answers here but I'm surprised no one has mentioned the built in pandas explode function. Check out the link below:
https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.explode.html#pandas.DataFrame.explode
For some reason I was unable to access that function, so I used the below code:
import pandas_explode
pandas_explode.patch()
df_zlp_people_cnt3 = df_zlp_people_cnt2.explode('people')
Above is a sample of my data. As you can see the people column had series of people, and I was trying to explode it. The code I have given works for list type data. So try to get your comma separated text data into list format. Also since my code uses built in functions, it is much faster than custom/apply functions.
Note: You may need to install pandas_explode with pip.
I had a similar problem, my solution was converting the dataframe to a list of dictionaries first, then do the transition. Here is the function:
import re
import pandas as pd
def separate_row(df, column_name):
ls = []
for row_dict in df.to_dict('records'):
for word in re.split(',', row_dict[column_name]):
row = row_dict.copy()
row[column_name]=word
ls.append(row)
return pd.DataFrame(ls)
Example:
>>> from pandas import DataFrame
>>> import numpy as np
>>> a = DataFrame([{'var1': 'a,b,c', 'var2': 1},
{'var1': 'd,e,f', 'var2': 2}])
>>> a
var1 var2
0 a,b,c 1
1 d,e,f 2
>>> separate_row(a, "var1")
var1 var2
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
You can also change the function a bit to support separating list type rows.
Upon adding few bits and pieces from all the solutions on this page, I was able to get something like this(for someone who need to use it right away).
parameters to the function are df(input dataframe) and key(column that has delimiter separated string). Just replace with your delimiter if that is different to semicolon ";".
def split_df_rows_for_semicolon_separated_key(key, df):
df=df.set_index(df.columns.drop(key,1).tolist())[key].str.split(';', expand=True).stack().reset_index().rename(columns={0:key}).loc[:, df.columns]
df=df[df[key] != '']
return df
Try:
vals = np.array(a.var1.str.split(",").values.tolist())
var = np.repeat(a.var2, vals.shape[1])
out = pd.DataFrame(np.column_stack((var, vals.ravel())), columns=a.columns)
display(out)
var1 var2
0 1 a
1 1 b
2 1 c
3 2 d
4 2 e
5 2 f
In recent version of pandas you can use split followed by explode
a.assign(var1=a['var1'].str.split(',')).explode('var1')
a
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
A short and simple way to change the format of the column using .apply() so that it can be used by .explod():
import string
import pandas as pd
from io import StringIO
file = StringIO(""" var1 var2
0 a,b,c 1
1 d,e,f 2""")
df = pd.read_csv(file, sep=r'\s\s+')
df['var1'] = df['var1'].apply(lambda x : str(x).split(','))
df.explode('var1')
Output:
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
I have a pandas dataframe in which one column of text strings contains comma-separated values. I want to split each CSV field and create a new row per entry (assume that CSV are clean and need only be split on ','). For example, a should become b:
In [7]: a
Out[7]:
var1 var2
0 a,b,c 1
1 d,e,f 2
In [8]: b
Out[8]:
var1 var2
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
So far, I have tried various simple functions, but the .apply method seems to only accept one row as return value when it is used on an axis, and I can't get .transform to work. Any suggestions would be much appreciated!
Example data:
from pandas import DataFrame
import numpy as np
a = DataFrame([{'var1': 'a,b,c', 'var2': 1},
{'var1': 'd,e,f', 'var2': 2}])
b = DataFrame([{'var1': 'a', 'var2': 1},
{'var1': 'b', 'var2': 1},
{'var1': 'c', 'var2': 1},
{'var1': 'd', 'var2': 2},
{'var1': 'e', 'var2': 2},
{'var1': 'f', 'var2': 2}])
I know this won't work because we lose DataFrame meta-data by going through numpy, but it should give you a sense of what I tried to do:
def fun(row):
letters = row['var1']
letters = letters.split(',')
out = np.array([row] * len(letters))
out['var1'] = letters
a['idx'] = range(a.shape[0])
z = a.groupby('idx')
z.transform(fun)
UPDATE 3: it makes more sense to use Series.explode() / DataFrame.explode() methods (implemented in Pandas 0.25.0 and extended in Pandas 1.3.0 to support multi-column explode) as is shown in the usage example:
for a single column:
In [1]: df = pd.DataFrame({'A': [[0, 1, 2], 'foo', [], [3, 4]],
...: 'B': 1,
...: 'C': [['a', 'b', 'c'], np.nan, [], ['d', 'e']]})
In [2]: df
Out[2]:
A B C
0 [0, 1, 2] 1 [a, b, c]
1 foo 1 NaN
2 [] 1 []
3 [3, 4] 1 [d, e]
In [3]: df.explode('A')
Out[3]:
A B C
0 0 1 [a, b, c]
0 1 1 [a, b, c]
0 2 1 [a, b, c]
1 foo 1 NaN
2 NaN 1 []
3 3 1 [d, e]
3 4 1 [d, e]
for multiple columns (for Pandas 1.3.0+):
In [4]: df.explode(['A', 'C'])
Out[4]:
A B C
0 0 1 a
0 1 1 b
0 2 1 c
1 foo 1 NaN
2 NaN 1 NaN
3 3 1 d
3 4 1 e
UPDATE 2: more generic vectorized function, which will work for multiple normal and multiple list columns
def explode(df, lst_cols, fill_value='', preserve_index=False):
# make sure `lst_cols` is list-alike
if (lst_cols is not None
and len(lst_cols) > 0
and not isinstance(lst_cols, (list, tuple, np.ndarray, pd.Series))):
lst_cols = [lst_cols]
# all columns except `lst_cols`
idx_cols = df.columns.difference(lst_cols)
# calculate lengths of lists
lens = df[lst_cols[0]].str.len()
# preserve original index values
idx = np.repeat(df.index.values, lens)
# create "exploded" DF
res = (pd.DataFrame({
col:np.repeat(df[col].values, lens)
for col in idx_cols},
index=idx)
.assign(**{col:np.concatenate(df.loc[lens>0, col].values)
for col in lst_cols}))
# append those rows that have empty lists
if (lens == 0).any():
# at least one list in cells is empty
res = (res.append(df.loc[lens==0, idx_cols], sort=False)
.fillna(fill_value))
# revert the original index order
res = res.sort_index()
# reset index if requested
if not preserve_index:
res = res.reset_index(drop=True)
return res
Demo:
Multiple list columns - all list columns must have the same # of elements in each row:
In [134]: df
Out[134]:
aaa myid num text
0 10 1 [1, 2, 3] [aa, bb, cc]
1 11 2 [] []
2 12 3 [1, 2] [cc, dd]
3 13 4 [] []
In [135]: explode(df, ['num','text'], fill_value='')
Out[135]:
aaa myid num text
0 10 1 1 aa
1 10 1 2 bb
2 10 1 3 cc
3 11 2
4 12 3 1 cc
5 12 3 2 dd
6 13 4
preserving original index values:
In [136]: explode(df, ['num','text'], fill_value='', preserve_index=True)
Out[136]:
aaa myid num text
0 10 1 1 aa
0 10 1 2 bb
0 10 1 3 cc
1 11 2
2 12 3 1 cc
2 12 3 2 dd
3 13 4
Setup:
df = pd.DataFrame({
'aaa': {0: 10, 1: 11, 2: 12, 3: 13},
'myid': {0: 1, 1: 2, 2: 3, 3: 4},
'num': {0: [1, 2, 3], 1: [], 2: [1, 2], 3: []},
'text': {0: ['aa', 'bb', 'cc'], 1: [], 2: ['cc', 'dd'], 3: []}
})
CSV column:
In [46]: df
Out[46]:
var1 var2 var3
0 a,b,c 1 XX
1 d,e,f,x,y 2 ZZ
In [47]: explode(df.assign(var1=df.var1.str.split(',')), 'var1')
Out[47]:
var1 var2 var3
0 a 1 XX
1 b 1 XX
2 c 1 XX
3 d 2 ZZ
4 e 2 ZZ
5 f 2 ZZ
6 x 2 ZZ
7 y 2 ZZ
using this little trick we can convert CSV-like column to list column:
In [48]: df.assign(var1=df.var1.str.split(','))
Out[48]:
var1 var2 var3
0 [a, b, c] 1 XX
1 [d, e, f, x, y] 2 ZZ
UPDATE: generic vectorized approach (will work also for multiple columns):
Original DF:
In [177]: df
Out[177]:
var1 var2 var3
0 a,b,c 1 XX
1 d,e,f,x,y 2 ZZ
Solution:
first let's convert CSV strings to lists:
In [178]: lst_col = 'var1'
In [179]: x = df.assign(**{lst_col:df[lst_col].str.split(',')})
In [180]: x
Out[180]:
var1 var2 var3
0 [a, b, c] 1 XX
1 [d, e, f, x, y] 2 ZZ
Now we can do this:
In [181]: pd.DataFrame({
...: col:np.repeat(x[col].values, x[lst_col].str.len())
...: for col in x.columns.difference([lst_col])
...: }).assign(**{lst_col:np.concatenate(x[lst_col].values)})[x.columns.tolist()]
...:
Out[181]:
var1 var2 var3
0 a 1 XX
1 b 1 XX
2 c 1 XX
3 d 2 ZZ
4 e 2 ZZ
5 f 2 ZZ
6 x 2 ZZ
7 y 2 ZZ
OLD answer:
Inspired by #AFinkelstein solution, i wanted to make it bit more generalized which could be applied to DF with more than two columns and as fast, well almost, as fast as AFinkelstein's solution):
In [2]: df = pd.DataFrame(
...: [{'var1': 'a,b,c', 'var2': 1, 'var3': 'XX'},
...: {'var1': 'd,e,f,x,y', 'var2': 2, 'var3': 'ZZ'}]
...: )
In [3]: df
Out[3]:
var1 var2 var3
0 a,b,c 1 XX
1 d,e,f,x,y 2 ZZ
In [4]: (df.set_index(df.columns.drop('var1',1).tolist())
...: .var1.str.split(',', expand=True)
...: .stack()
...: .reset_index()
...: .rename(columns={0:'var1'})
...: .loc[:, df.columns]
...: )
Out[4]:
var1 var2 var3
0 a 1 XX
1 b 1 XX
2 c 1 XX
3 d 2 ZZ
4 e 2 ZZ
5 f 2 ZZ
6 x 2 ZZ
7 y 2 ZZ
After painful experimentation to find something faster than the accepted answer, I got this to work. It ran around 100x faster on the dataset I tried it on.
If someone knows a way to make this more elegant, by all means please modify my code. I couldn't find a way that works without setting the other columns you want to keep as the index and then resetting the index and re-naming the columns, but I'd imagine there's something else that works.
b = DataFrame(a.var1.str.split(',').tolist(), index=a.var2).stack()
b = b.reset_index()[[0, 'var2']] # var1 variable is currently labeled 0
b.columns = ['var1', 'var2'] # renaming var1
Pandas >= 0.25
Series and DataFrame methods define a .explode() method that explodes lists into separate rows. See the docs section on Exploding a list-like column.
Since you have a list of comma separated strings, split the string on comma to get a list of elements, then call explode on that column.
df = pd.DataFrame({'var1': ['a,b,c', 'd,e,f'], 'var2': [1, 2]})
df
var1 var2
0 a,b,c 1
1 d,e,f 2
df.assign(var1=df['var1'].str.split(',')).explode('var1')
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
Note that explode only works on a single column (for now). To explode multiple columns at once, see below.
NaNs and empty lists get the treatment they deserve without you having to jump through hoops to get it right.
df = pd.DataFrame({'var1': ['d,e,f', '', np.nan], 'var2': [1, 2, 3]})
df
var1 var2
0 d,e,f 1
1 2
2 NaN 3
df['var1'].str.split(',')
0 [d, e, f]
1 []
2 NaN
df.assign(var1=df['var1'].str.split(',')).explode('var1')
var1 var2
0 d 1
0 e 1
0 f 1
1 2 # empty list entry becomes empty string after exploding
2 NaN 3 # NaN left un-touched
This is a serious advantage over ravel/repeat -based solutions (which ignore empty lists completely, and choke on NaNs).
Exploding Multiple Columns
pandas 1.3 update
df.explode works on multiple columns starting from pandas 1.3:
df = pd.DataFrame({'var1': ['a,b,c', 'd,e,f'],
'var2': ['i,j,k', 'l,m,n'],
'var3': [1, 2]})
df
var1 var2 var3
0 a,b,c i,j,k 1
1 d,e,f l,m,n 2
(df.set_index(['var3'])
.apply(lambda col: col.str.split(','))
.explode(['var1', 'var2'])
.reset_index()
.reindex(df.columns, axis=1))
var1 var2 var3
0 a i 1
1 b j 1
2 c k 1
3 d l 2
4 e m 2
5 f n 2
On older versions, you would move the explode column inside the apply which is a lot less performant:
(df.set_index(['var3'])
.apply(lambda col: col.str.split(',').explode())
.reset_index()
.reindex(df.columns, axis=1))
The idea is to set as the index, all the columns that should NOT be exploded, then explode the remaining columns via apply. This works well when the lists are equally sized.
How about something like this:
In [55]: pd.concat([Series(row['var2'], row['var1'].split(','))
for _, row in a.iterrows()]).reset_index()
Out[55]:
index 0
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
Then you just have to rename the columns
Here's a function I wrote for this common task. It's more efficient than the Series/stack methods. Column order and names are retained.
def tidy_split(df, column, sep='|', keep=False):
"""
Split the values of a column and expand so the new DataFrame has one split
value per row. Filters rows where the column is missing.
Params
------
df : pandas.DataFrame
dataframe with the column to split and expand
column : str
the column to split and expand
sep : str
the string used to split the column's values
keep : bool
whether to retain the presplit value as it's own row
Returns
-------
pandas.DataFrame
Returns a dataframe with the same columns as `df`.
"""
indexes = list()
new_values = list()
df = df.dropna(subset=[column])
for i, presplit in enumerate(df[column].astype(str)):
values = presplit.split(sep)
if keep and len(values) > 1:
indexes.append(i)
new_values.append(presplit)
for value in values:
indexes.append(i)
new_values.append(value)
new_df = df.iloc[indexes, :].copy()
new_df[column] = new_values
return new_df
With this function, the original question is as simple as:
tidy_split(a, 'var1', sep=',')
Similar question as: pandas: How do I split text in a column into multiple rows?
You could do:
>> a=pd.DataFrame({"var1":"a,b,c d,e,f".split(),"var2":[1,2]})
>> s = a.var1.str.split(",").apply(pd.Series, 1).stack()
>> s.index = s.index.droplevel(-1)
>> del a['var1']
>> a.join(s)
var2 var1
0 1 a
0 1 b
0 1 c
1 2 d
1 2 e
1 2 f
There is a possibility to split and explode the dataframe without changing the structure of dataframe
Split and expand data of specific columns
Input:
var1 var2
0 a,b,c 1
1 d,e,f 2
#Get the indexes which are repetative with the split
df['var1'] = df['var1'].str.split(',')
df = df.explode('var1')
Out:
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
Edit-1
Split and Expand of rows for Multiple columns
Filename RGB RGB_type
0 A [[0, 1650, 6, 39], [0, 1691, 1, 59], [50, 1402... [r, g, b]
1 B [[0, 1423, 16, 38], [0, 1445, 16, 46], [0, 141... [r, g, b]
Re indexing based on the reference column and aligning the column value information with stack
df = df.reindex(df.index.repeat(df['RGB_type'].apply(len)))
df = df.groupby('Filename').apply(lambda x:x.apply(lambda y: pd.Series(y.iloc[0])))
df.reset_index(drop=True).ffill()
Out:
Filename RGB_type Top 1 colour Top 1 frequency Top 2 colour Top 2 frequency
Filename
A 0 A r 0 1650 6 39
1 A g 0 1691 1 59
2 A b 50 1402 49 187
B 0 B r 0 1423 16 38
1 B g 0 1445 16 46
2 B b 0 1419 16 39
TL;DR
import pandas as pd
import numpy as np
def explode_str(df, col, sep):
s = df[col]
i = np.arange(len(s)).repeat(s.str.count(sep) + 1)
return df.iloc[i].assign(**{col: sep.join(s).split(sep)})
def explode_list(df, col):
s = df[col]
i = np.arange(len(s)).repeat(s.str.len())
return df.iloc[i].assign(**{col: np.concatenate(s)})
Demonstration
explode_str(a, 'var1', ',')
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
Let's create a new dataframe d that has lists
d = a.assign(var1=lambda d: d.var1.str.split(','))
explode_list(d, 'var1')
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
General Comments
I'll use np.arange with repeat to produce dataframe index positions that I can use with iloc.
FAQ
Why don't I use loc?
Because the index may not be unique and using loc will return every row that matches a queried index.
Why don't you use the values attribute and slice that?
When calling values, if the entirety of the the dataframe is in one cohesive "block", Pandas will return a view of the array that is the "block". Otherwise Pandas will have to cobble together a new array. When cobbling, that array must be of a uniform dtype. Often that means returning an array with dtype that is object. By using iloc instead of slicing the values attribute, I alleviate myself from having to deal with that.
Why do you use assign?
When I use assign using the same column name that I'm exploding, I overwrite the existing column and maintain its position in the dataframe.
Why are the index values repeat?
By virtue of using iloc on repeated positions, the resulting index shows the same repeated pattern. One repeat for each element the list or string.
This can be reset with reset_index(drop=True)
For Strings
I don't want to have to split the strings prematurely. So instead I count the occurrences of the sep argument assuming that if I were to split, the length of the resulting list would be one more than the number of separators.
I then use that sep to join the strings then split.
def explode_str(df, col, sep):
s = df[col]
i = np.arange(len(s)).repeat(s.str.count(sep) + 1)
return df.iloc[i].assign(**{col: sep.join(s).split(sep)})
For Lists
Similar as for strings except I don't need to count occurrences of sep because its already split.
I use Numpy's concatenate to jam the lists together.
import pandas as pd
import numpy as np
def explode_list(df, col):
s = df[col]
i = np.arange(len(s)).repeat(s.str.len())
return df.iloc[i].assign(**{col: np.concatenate(s)})
I came up with a solution for dataframes with arbitrary numbers of columns (while still only separating one column's entries at a time).
def splitDataFrameList(df,target_column,separator):
''' df = dataframe to split,
target_column = the column containing the values to split
separator = the symbol used to perform the split
returns: a dataframe with each entry for the target column separated, with each element moved into a new row.
The values in the other columns are duplicated across the newly divided rows.
'''
def splitListToRows(row,row_accumulator,target_column,separator):
split_row = row[target_column].split(separator)
for s in split_row:
new_row = row.to_dict()
new_row[target_column] = s
row_accumulator.append(new_row)
new_rows = []
df.apply(splitListToRows,axis=1,args = (new_rows,target_column,separator))
new_df = pandas.DataFrame(new_rows)
return new_df
Here is a fairly straightforward message that uses the split method from pandas str accessor and then uses NumPy to flatten each row into a single array.
The corresponding values are retrieved by repeating the non-split column the correct number of times with np.repeat.
var1 = df.var1.str.split(',', expand=True).values.ravel()
var2 = np.repeat(df.var2.values, len(var1) / len(df))
pd.DataFrame({'var1': var1,
'var2': var2})
var1 var2
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
I have been struggling with out-of-memory experience using various way to explode my lists so I prepared some benchmarks to help me decide which answers to upvote. I tested five scenarios with varying proportions of the list length to the number of lists. Sharing the results below:
Time: (less is better, click to view large version)
Peak memory usage: (less is better)
Conclusions:
#MaxU's answer (update 2), codename concatenate offers the best speed in almost every case, while keeping the peek memory usage low,
see #DMulligan's answer (codename stack) if you need to process lots of rows with relatively small lists and can afford increased peak memory,
the accepted #Chang's answer works well for data frames that have a few rows but very large lists.
Full details (functions and benchmarking code) are in this GitHub gist. Please note that the benchmark problem was simplified and did not include splitting of strings into the list - which most solutions performed in a similar fashion.
One-liner using split(___, expand=True) and the level and name arguments to reset_index():
>>> b = a.var1.str.split(',', expand=True).set_index(a.var2).stack().reset_index(level=0, name='var1')
>>> b
var2 var1
0 1 a
1 1 b
2 1 c
0 2 d
1 2 e
2 2 f
If you need b to look exactly like in the question, you can additionally do:
>>> b = b.reset_index(drop=True)[['var1', 'var2']]
>>> b
var1 var2
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
Based on the excellent #DMulligan's solution, here is a generic vectorized (no loops) function which splits a column of a dataframe into multiple rows, and merges it back to the original dataframe. It also uses a great generic change_column_order function from this answer.
def change_column_order(df, col_name, index):
cols = df.columns.tolist()
cols.remove(col_name)
cols.insert(index, col_name)
return df[cols]
def split_df(dataframe, col_name, sep):
orig_col_index = dataframe.columns.tolist().index(col_name)
orig_index_name = dataframe.index.name
orig_columns = dataframe.columns
dataframe = dataframe.reset_index() # we need a natural 0-based index for proper merge
index_col_name = (set(dataframe.columns) - set(orig_columns)).pop()
df_split = pd.DataFrame(
pd.DataFrame(dataframe[col_name].str.split(sep).tolist())
.stack().reset_index(level=1, drop=1), columns=[col_name])
df = dataframe.drop(col_name, axis=1)
df = pd.merge(df, df_split, left_index=True, right_index=True, how='inner')
df = df.set_index(index_col_name)
df.index.name = orig_index_name
# merge adds the column to the last place, so we need to move it back
return change_column_order(df, col_name, orig_col_index)
Example:
df = pd.DataFrame([['a:b', 1, 4], ['c:d', 2, 5], ['e:f:g:h', 3, 6]],
columns=['Name', 'A', 'B'], index=[10, 12, 13])
df
Name A B
10 a:b 1 4
12 c:d 2 5
13 e:f:g:h 3 6
split_df(df, 'Name', ':')
Name A B
10 a 1 4
10 b 1 4
12 c 2 5
12 d 2 5
13 e 3 6
13 f 3 6
13 g 3 6
13 h 3 6
Note that it preserves the original index and order of the columns. It also works with dataframes which have non-sequential index.
The string function split can take an option boolean argument 'expand'.
Here is a solution using this argument:
(a.var1
.str.split(",",expand=True)
.set_index(a.var2)
.stack()
.reset_index(level=1, drop=True)
.reset_index()
.rename(columns={0:"var1"}))
I do appreciate the answer of "Chang She", really, but the iterrows() function takes long time on large dataset. I faced that issue and I came to this.
# First, reset_index to make the index a column
a = a.reset_index().rename(columns={'index':'duplicated_idx'})
# Get a longer series with exploded cells to rows
series = pd.DataFrame(a['var1'].str.split('/')
.tolist(), index=a.duplicated_idx).stack()
# New df from series and merge with the old one
b = series.reset_index([0, 'duplicated_idx'])
b = b.rename(columns={0:'var1'})
# Optional & Advanced: In case, there are other columns apart from var1 & var2
b.merge(
a[a.columns.difference(['var1'])],
on='duplicated_idx')
# Optional: Delete the "duplicated_index"'s column, and reorder columns
b = b[a.columns.difference(['duplicated_idx'])]
One-liner using assign and explode:
col1 col2
0 a,b,c 1
1 d,e,f 2
df.assign(col1 = df.col1.str.split(',')).explode('col1', ignore_index=True)
Output:
col1 col2
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
Just used jiln's excellent answer from above, but needed to expand to split multiple columns. Thought I would share.
def splitDataFrameList(df,target_column,separator):
''' df = dataframe to split,
target_column = the column containing the values to split
separator = the symbol used to perform the split
returns: a dataframe with each entry for the target column separated, with each element moved into a new row.
The values in the other columns are duplicated across the newly divided rows.
'''
def splitListToRows(row, row_accumulator, target_columns, separator):
split_rows = []
for target_column in target_columns:
split_rows.append(row[target_column].split(separator))
# Seperate for multiple columns
for i in range(len(split_rows[0])):
new_row = row.to_dict()
for j in range(len(split_rows)):
new_row[target_columns[j]] = split_rows[j][i]
row_accumulator.append(new_row)
new_rows = []
df.apply(splitListToRows,axis=1,args = (new_rows,target_column,separator))
new_df = pd.DataFrame(new_rows)
return new_df
upgraded MaxU's answer with MultiIndex support
def explode(df, lst_cols, fill_value='', preserve_index=False):
"""
usage:
In [134]: df
Out[134]:
aaa myid num text
0 10 1 [1, 2, 3] [aa, bb, cc]
1 11 2 [] []
2 12 3 [1, 2] [cc, dd]
3 13 4 [] []
In [135]: explode(df, ['num','text'], fill_value='')
Out[135]:
aaa myid num text
0 10 1 1 aa
1 10 1 2 bb
2 10 1 3 cc
3 11 2
4 12 3 1 cc
5 12 3 2 dd
6 13 4
"""
# make sure `lst_cols` is list-alike
if (lst_cols is not None
and len(lst_cols) > 0
and not isinstance(lst_cols, (list, tuple, np.ndarray, pd.Series))):
lst_cols = [lst_cols]
# all columns except `lst_cols`
idx_cols = df.columns.difference(lst_cols)
# calculate lengths of lists
lens = df[lst_cols[0]].str.len()
# preserve original index values
idx = np.repeat(df.index.values, lens)
res = (pd.DataFrame({
col:np.repeat(df[col].values, lens)
for col in idx_cols},
index=idx)
.assign(**{col:np.concatenate(df.loc[lens>0, col].values)
for col in lst_cols}))
# append those rows that have empty lists
if (lens == 0).any():
# at least one list in cells is empty
res = (res.append(df.loc[lens==0, idx_cols], sort=False)
.fillna(fill_value))
# revert the original index order
res = res.sort_index()
# reset index if requested
if not preserve_index:
res = res.reset_index(drop=True)
# if original index is MultiIndex build the dataframe from the multiindex
# create "exploded" DF
if isinstance(df.index, pd.MultiIndex):
res = res.reindex(
index=pd.MultiIndex.from_tuples(
res.index,
names=['number', 'color']
)
)
return res
My version of the solution to add to this collection! :-)
# Original problem
from pandas import DataFrame
import numpy as np
a = DataFrame([{'var1': 'a,b,c', 'var2': 1},
{'var1': 'd,e,f', 'var2': 2}])
b = DataFrame([{'var1': 'a', 'var2': 1},
{'var1': 'b', 'var2': 1},
{'var1': 'c', 'var2': 1},
{'var1': 'd', 'var2': 2},
{'var1': 'e', 'var2': 2},
{'var1': 'f', 'var2': 2}])
### My solution
import pandas as pd
import functools
def expand_on_cols(df, fuse_cols, delim=","):
def expand_on_col(df, fuse_col):
col_order = df.columns
df_expanded = pd.DataFrame(
df.set_index([x for x in df.columns if x != fuse_col])[fuse_col]
.apply(lambda x: x.split(delim))
.explode()
).reset_index()
return df_expanded[col_order]
all_expanded = functools.reduce(expand_on_col, fuse_cols, df)
return all_expanded
assert(b.equals(expand_on_cols(a, ["var1"], delim=",")))
I have come up with the following solution to this problem:
def iter_var1(d):
for _, row in d.iterrows():
for v in row["var1"].split(","):
yield (v, row["var2"])
new_a = DataFrame.from_records([i for i in iter_var1(a)],
columns=["var1", "var2"])
Another solution that uses python copy package
import copy
new_observations = list()
def pandas_explode(df, column_to_explode):
new_observations = list()
for row in df.to_dict(orient='records'):
explode_values = row[column_to_explode]
del row[column_to_explode]
if type(explode_values) is list or type(explode_values) is tuple:
for explode_value in explode_values:
new_observation = copy.deepcopy(row)
new_observation[column_to_explode] = explode_value
new_observations.append(new_observation)
else:
new_observation = copy.deepcopy(row)
new_observation[column_to_explode] = explode_values
new_observations.append(new_observation)
return_df = pd.DataFrame(new_observations)
return return_df
df = pandas_explode(df, column_name)
There are a lot of answers here but I'm surprised no one has mentioned the built in pandas explode function. Check out the link below:
https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.explode.html#pandas.DataFrame.explode
For some reason I was unable to access that function, so I used the below code:
import pandas_explode
pandas_explode.patch()
df_zlp_people_cnt3 = df_zlp_people_cnt2.explode('people')
Above is a sample of my data. As you can see the people column had series of people, and I was trying to explode it. The code I have given works for list type data. So try to get your comma separated text data into list format. Also since my code uses built in functions, it is much faster than custom/apply functions.
Note: You may need to install pandas_explode with pip.
I had a similar problem, my solution was converting the dataframe to a list of dictionaries first, then do the transition. Here is the function:
import re
import pandas as pd
def separate_row(df, column_name):
ls = []
for row_dict in df.to_dict('records'):
for word in re.split(',', row_dict[column_name]):
row = row_dict.copy()
row[column_name]=word
ls.append(row)
return pd.DataFrame(ls)
Example:
>>> from pandas import DataFrame
>>> import numpy as np
>>> a = DataFrame([{'var1': 'a,b,c', 'var2': 1},
{'var1': 'd,e,f', 'var2': 2}])
>>> a
var1 var2
0 a,b,c 1
1 d,e,f 2
>>> separate_row(a, "var1")
var1 var2
0 a 1
1 b 1
2 c 1
3 d 2
4 e 2
5 f 2
You can also change the function a bit to support separating list type rows.
Upon adding few bits and pieces from all the solutions on this page, I was able to get something like this(for someone who need to use it right away).
parameters to the function are df(input dataframe) and key(column that has delimiter separated string). Just replace with your delimiter if that is different to semicolon ";".
def split_df_rows_for_semicolon_separated_key(key, df):
df=df.set_index(df.columns.drop(key,1).tolist())[key].str.split(';', expand=True).stack().reset_index().rename(columns={0:key}).loc[:, df.columns]
df=df[df[key] != '']
return df
Try:
vals = np.array(a.var1.str.split(",").values.tolist())
var = np.repeat(a.var2, vals.shape[1])
out = pd.DataFrame(np.column_stack((var, vals.ravel())), columns=a.columns)
display(out)
var1 var2
0 1 a
1 1 b
2 1 c
3 2 d
4 2 e
5 2 f
In recent version of pandas you can use split followed by explode
a.assign(var1=a['var1'].str.split(',')).explode('var1')
a
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
A short and simple way to change the format of the column using .apply() so that it can be used by .explod():
import string
import pandas as pd
from io import StringIO
file = StringIO(""" var1 var2
0 a,b,c 1
1 d,e,f 2""")
df = pd.read_csv(file, sep=r'\s\s+')
df['var1'] = df['var1'].apply(lambda x : str(x).split(','))
df.explode('var1')
Output:
var1 var2
0 a 1
0 b 1
0 c 1
1 d 2
1 e 2
1 f 2
I'm wondering about existing pandas functionalities, that I might not been able to find so far.
Bascially, I have a data frame with various columns. I'd like to select specific rows depending on the values of certain colums (FYI: i was interested in the value of column D, that had several parameters described in A-C).
E.g. I want to know which row(s) have A==1 & B==2 & C==5?
df
A B C D
0 1 2 4 a
1 1 2 5 b
2 1 3 4 c
df_result
1 1 2 5 b
So far I have been able to basically reduce this:
import pandas as pd
df = pd.DataFrame({'A': [1,1,1],
'B': [2,2,3],
'C': [4,5,4],
'D': ['a', 'b', 'c']})
df_A = df[df['A'] == 1]
df_B = df_A[df_A['B'] == 2]
df_C = df_B[df_B['C'] == 5]
To this:
parameter = [['A', 1],
['B', 2],
['C', 5]]
df_filtered = df
for x, y in parameter:
df_filtered = df_filtered[df_filtered[x] == y]
which yielded the same results. But I wonder if there's another way? Maybe without loop in one line?
You could use query() method to filter data, and construct filter expression from parameters like
In [288]: df.query(' and '.join(['{0}=={1}'.format(x[0], x[1]) for x in parameter]))
Out[288]:
A B C D
1 1 2 5 b
Details
In [296]: df
Out[296]:
A B C D
0 1 2 4 a
1 1 2 5 b
2 1 3 4 c
In [297]: query = ' and '.join(['{0}=={1}'.format(x[0], x[1]) for x in parameter])
In [298]: query
Out[298]: 'A==1 and B==2 and C==5'
In [299]: df.query(query)
Out[299]:
A B C D
1 1 2 5 b
Just for the information if others are interested, I would have done it this way:
import numpy as np
matched = np.all([df[vn] == vv for vn, vv in parameters], axis=0)
df_filtered = df[matched]
But I like the query function better, now that I have seen it #John Galt.