Related
I have a 20 x 4000 dataframe in Python using pandas. Two of these columns are named Year and quarter. I'd like to create a variable called period that makes Year = 2000 and quarter= q2 into 2000q2.
Can anyone help with that?
If both columns are strings, you can concatenate them directly:
df["period"] = df["Year"] + df["quarter"]
If one (or both) of the columns are not string typed, you should convert it (them) first,
df["period"] = df["Year"].astype(str) + df["quarter"]
Beware of NaNs when doing this!
If you need to join multiple string columns, you can use agg:
df['period'] = df[['Year', 'quarter', ...]].agg('-'.join, axis=1)
Where "-" is the separator.
Small data-sets (< 150rows)
[''.join(i) for i in zip(df["Year"].map(str),df["quarter"])]
or slightly slower but more compact:
df.Year.str.cat(df.quarter)
Larger data sets (> 150rows)
df['Year'].astype(str) + df['quarter']
UPDATE: Timing graph Pandas 0.23.4
Let's test it on 200K rows DF:
In [250]: df
Out[250]:
Year quarter
0 2014 q1
1 2015 q2
In [251]: df = pd.concat([df] * 10**5)
In [252]: df.shape
Out[252]: (200000, 2)
UPDATE: new timings using Pandas 0.19.0
Timing without CPU/GPU optimization (sorted from fastest to slowest):
In [107]: %timeit df['Year'].astype(str) + df['quarter']
10 loops, best of 3: 131 ms per loop
In [106]: %timeit df['Year'].map(str) + df['quarter']
10 loops, best of 3: 161 ms per loop
In [108]: %timeit df.Year.str.cat(df.quarter)
10 loops, best of 3: 189 ms per loop
In [109]: %timeit df.loc[:, ['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 567 ms per loop
In [110]: %timeit df[['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 584 ms per loop
In [111]: %timeit df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)
1 loop, best of 3: 24.7 s per loop
Timing using CPU/GPU optimization:
In [113]: %timeit df['Year'].astype(str) + df['quarter']
10 loops, best of 3: 53.3 ms per loop
In [114]: %timeit df['Year'].map(str) + df['quarter']
10 loops, best of 3: 65.5 ms per loop
In [115]: %timeit df.Year.str.cat(df.quarter)
10 loops, best of 3: 79.9 ms per loop
In [116]: %timeit df.loc[:, ['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 230 ms per loop
In [117]: %timeit df[['Year','quarter']].astype(str).sum(axis=1)
1 loop, best of 3: 230 ms per loop
In [118]: %timeit df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)
1 loop, best of 3: 9.38 s per loop
Answer contribution by #anton-vbr
df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})
df['period'] = df[['Year', 'quarter']].apply(lambda x: ''.join(x), axis=1)
Yields this dataframe
Year quarter period
0 2014 q1 2014q1
1 2015 q2 2015q2
This method generalizes to an arbitrary number of string columns by replacing df[['Year', 'quarter']] with any column slice of your dataframe, e.g. df.iloc[:,0:2].apply(lambda x: ''.join(x), axis=1).
You can check more information about apply() method here
The method cat() of the .str accessor works really well for this:
>>> import pandas as pd
>>> df = pd.DataFrame([["2014", "q1"],
... ["2015", "q3"]],
... columns=('Year', 'Quarter'))
>>> print(df)
Year Quarter
0 2014 q1
1 2015 q3
>>> df['Period'] = df.Year.str.cat(df.Quarter)
>>> print(df)
Year Quarter Period
0 2014 q1 2014q1
1 2015 q3 2015q3
cat() even allows you to add a separator so, for example, suppose you only have integers for year and period, you can do this:
>>> import pandas as pd
>>> df = pd.DataFrame([[2014, 1],
... [2015, 3]],
... columns=('Year', 'Quarter'))
>>> print(df)
Year Quarter
0 2014 1
1 2015 3
>>> df['Period'] = df.Year.astype(str).str.cat(df.Quarter.astype(str), sep='q')
>>> print(df)
Year Quarter Period
0 2014 1 2014q1
1 2015 3 2015q3
Joining multiple columns is just a matter of passing either a list of series or a dataframe containing all but the first column as a parameter to str.cat() invoked on the first column (Series):
>>> df = pd.DataFrame(
... [['USA', 'Nevada', 'Las Vegas'],
... ['Brazil', 'Pernambuco', 'Recife']],
... columns=['Country', 'State', 'City'],
... )
>>> df['AllTogether'] = df['Country'].str.cat(df[['State', 'City']], sep=' - ')
>>> print(df)
Country State City AllTogether
0 USA Nevada Las Vegas USA - Nevada - Las Vegas
1 Brazil Pernambuco Recife Brazil - Pernambuco - Recife
Do note that if your pandas dataframe/series has null values, you need to include the parameter na_rep to replace the NaN values with a string, otherwise the combined column will default to NaN.
Use of a lamba function this time with string.format().
import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'Quarter': ['q1', 'q2']})
print df
df['YearQuarter'] = df[['Year','Quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)
print df
Quarter Year
0 q1 2014
1 q2 2015
Quarter Year YearQuarter
0 q1 2014 2014q1
1 q2 2015 2015q2
This allows you to work with non-strings and reformat values as needed.
import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'Quarter': [1, 2]})
print df.dtypes
print df
df['YearQuarter'] = df[['Year','Quarter']].apply(lambda x : '{}q{}'.format(x[0],x[1]), axis=1)
print df
Quarter int64
Year object
dtype: object
Quarter Year
0 1 2014
1 2 2015
Quarter Year YearQuarter
0 1 2014 2014q1
1 2 2015 2015q2
generalising to multiple columns, why not:
columns = ['whatever', 'columns', 'you', 'choose']
df['period'] = df[columns].astype(str).sum(axis=1)
You can use lambda:
combine_lambda = lambda x: '{}{}'.format(x.Year, x.quarter)
And then use it with creating the new column:
df['period'] = df.apply(combine_lambda, axis = 1)
Let us suppose your dataframe is df with columns Year and Quarter.
import pandas as pd
df = pd.DataFrame({'Quarter':'q1 q2 q3 q4'.split(), 'Year':'2000'})
Suppose we want to see the dataframe;
df
>>> Quarter Year
0 q1 2000
1 q2 2000
2 q3 2000
3 q4 2000
Finally, concatenate the Year and the Quarter as follows.
df['Period'] = df['Year'] + ' ' + df['Quarter']
You can now print df to see the resulting dataframe.
df
>>> Quarter Year Period
0 q1 2000 2000 q1
1 q2 2000 2000 q2
2 q3 2000 2000 q3
3 q4 2000 2000 q4
If you do not want the space between the year and quarter, simply remove it by doing;
df['Period'] = df['Year'] + df['Quarter']
Although the #silvado answer is good if you change df.map(str) to df.astype(str) it will be faster:
import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})
In [131]: %timeit df["Year"].map(str)
10000 loops, best of 3: 132 us per loop
In [132]: %timeit df["Year"].astype(str)
10000 loops, best of 3: 82.2 us per loop
Here is an implementation that I find very versatile:
In [1]: import pandas as pd
In [2]: df = pd.DataFrame([[0, 'the', 'quick', 'brown'],
...: [1, 'fox', 'jumps', 'over'],
...: [2, 'the', 'lazy', 'dog']],
...: columns=['c0', 'c1', 'c2', 'c3'])
In [3]: def str_join(df, sep, *cols):
...: from functools import reduce
...: return reduce(lambda x, y: x.astype(str).str.cat(y.astype(str), sep=sep),
...: [df[col] for col in cols])
...:
In [4]: df['cat'] = str_join(df, '-', 'c0', 'c1', 'c2', 'c3')
In [5]: df
Out[5]:
c0 c1 c2 c3 cat
0 0 the quick brown 0-the-quick-brown
1 1 fox jumps over 1-fox-jumps-over
2 2 the lazy dog 2-the-lazy-dog
more efficient is
def concat_df_str1(df):
""" run time: 1.3416s """
return pd.Series([''.join(row.astype(str)) for row in df.values], index=df.index)
and here is a time test:
import numpy as np
import pandas as pd
from time import time
def concat_df_str1(df):
""" run time: 1.3416s """
return pd.Series([''.join(row.astype(str)) for row in df.values], index=df.index)
def concat_df_str2(df):
""" run time: 5.2758s """
return df.astype(str).sum(axis=1)
def concat_df_str3(df):
""" run time: 5.0076s """
df = df.astype(str)
return df[0] + df[1] + df[2] + df[3] + df[4] + \
df[5] + df[6] + df[7] + df[8] + df[9]
def concat_df_str4(df):
""" run time: 7.8624s """
return df.astype(str).apply(lambda x: ''.join(x), axis=1)
def main():
df = pd.DataFrame(np.zeros(1000000).reshape(100000, 10))
df = df.astype(int)
time1 = time()
df_en = concat_df_str4(df)
print('run time: %.4fs' % (time() - time1))
print(df_en.head(10))
if __name__ == '__main__':
main()
final, when sum(concat_df_str2) is used, the result is not simply concat, it will trans to integer.
Using zip could be even quicker:
df["period"] = [''.join(i) for i in zip(df["Year"].map(str),df["quarter"])]
Graph:
import pandas as pd
import numpy as np
import timeit
import matplotlib.pyplot as plt
from collections import defaultdict
df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})
myfuncs = {
"df['Year'].astype(str) + df['quarter']":
lambda: df['Year'].astype(str) + df['quarter'],
"df['Year'].map(str) + df['quarter']":
lambda: df['Year'].map(str) + df['quarter'],
"df.Year.str.cat(df.quarter)":
lambda: df.Year.str.cat(df.quarter),
"df.loc[:, ['Year','quarter']].astype(str).sum(axis=1)":
lambda: df.loc[:, ['Year','quarter']].astype(str).sum(axis=1),
"df[['Year','quarter']].astype(str).sum(axis=1)":
lambda: df[['Year','quarter']].astype(str).sum(axis=1),
"df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1)":
lambda: df[['Year','quarter']].apply(lambda x : '{}{}'.format(x[0],x[1]), axis=1),
"[''.join(i) for i in zip(dataframe['Year'].map(str),dataframe['quarter'])]":
lambda: [''.join(i) for i in zip(df["Year"].map(str),df["quarter"])]
}
d = defaultdict(dict)
step = 10
cont = True
while cont:
lendf = len(df); print(lendf)
for k,v in myfuncs.items():
iters = 1
t = 0
while t < 0.2:
ts = timeit.repeat(v, number=iters, repeat=3)
t = min(ts)
iters *= 10
d[k][lendf] = t/iters
if t > 2: cont = False
df = pd.concat([df]*step)
pd.DataFrame(d).plot().legend(loc='upper center', bbox_to_anchor=(0.5, -0.15))
plt.yscale('log'); plt.xscale('log'); plt.ylabel('seconds'); plt.xlabel('df rows')
plt.show()
This solution uses an intermediate step compressing two columns of the DataFrame to a single column containing a list of the values.
This works not only for strings but for all kind of column-dtypes
import pandas as pd
df = pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']})
df['list']=df[['Year','quarter']].values.tolist()
df['period']=df['list'].apply(''.join)
print(df)
Result:
Year quarter list period
0 2014 q1 [2014, q1] 2014q1
1 2015 q2 [2015, q2] 2015q2
Here is my summary of the above solutions to concatenate / combine two columns with int and str value into a new column, using a separator between the values of columns. Three solutions work for this purpose.
# be cautious about the separator, some symbols may cause "SyntaxError: EOL while scanning string literal".
# e.g. ";;" as separator would raise the SyntaxError
separator = "&&"
# pd.Series.str.cat() method does not work to concatenate / combine two columns with int value and str value. This would raise "AttributeError: Can only use .cat accessor with a 'category' dtype"
df["period"] = df["Year"].map(str) + separator + df["quarter"]
df["period"] = df[['Year','quarter']].apply(lambda x : '{} && {}'.format(x[0],x[1]), axis=1)
df["period"] = df.apply(lambda x: f'{x["Year"]} && {x["quarter"]}', axis=1)
my take....
listofcols = ['col1','col2','col3']
df['combined_cols'] = ''
for column in listofcols:
df['combined_cols'] = df['combined_cols'] + ' ' + df[column]
'''
As many have mentioned previously, you must convert each column to string and then use the plus operator to combine two string columns. You can get a large performance improvement by using NumPy.
%timeit df['Year'].values.astype(str) + df.quarter
71.1 ms ± 3.76 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit df['Year'].astype(str) + df['quarter']
565 ms ± 22.3 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
One can use assign method of DataFrame:
df= (pd.DataFrame({'Year': ['2014', '2015'], 'quarter': ['q1', 'q2']}).
assign(period=lambda x: x.Year+x.quarter ))
Similar to #geher answer but with any separator you like:
SEP = " "
INPUT_COLUMNS_WITH_SEP = ",sep,".join(INPUT_COLUMNS).split(",")
df.assign(sep=SEP)[INPUT_COLUMNS_WITH_SEP].sum(axis=1)
def madd(x):
"""Performs element-wise string concatenation with multiple input arrays.
Args:
x: iterable of np.array.
Returns: np.array.
"""
for i, arr in enumerate(x):
if type(arr.item(0)) is not str:
x[i] = x[i].astype(str)
return reduce(np.core.defchararray.add, x)
For example:
data = list(zip([2000]*4, ['q1', 'q2', 'q3', 'q4']))
df = pd.DataFrame(data=data, columns=['Year', 'quarter'])
df['period'] = madd([df[col].values for col in ['Year', 'quarter']])
df
Year quarter period
0 2000 q1 2000q1
1 2000 q2 2000q2
2 2000 q3 2000q3
3 2000 q4 2000q4
Use .combine_first.
df['Period'] = df['Year'].combine_first(df['Quarter'])
When combining columns with strings by concatenating them using the addition operator + if any is NaN then entire output will be NaN so use fillna()
df["join"] = "some" + df["col"].fillna(df["val_if_nan"])
How can I select rows from a DataFrame based on values in some column in Pandas?
In SQL, I would use:
SELECT *
FROM table
WHERE column_name = some_value
To select rows whose column value equals a scalar, some_value, use ==:
df.loc[df['column_name'] == some_value]
To select rows whose column value is in an iterable, some_values, use isin:
df.loc[df['column_name'].isin(some_values)]
Combine multiple conditions with &:
df.loc[(df['column_name'] >= A) & (df['column_name'] <= B)]
Note the parentheses. Due to Python's operator precedence rules, & binds more tightly than <= and >=. Thus, the parentheses in the last example are necessary. Without the parentheses
df['column_name'] >= A & df['column_name'] <= B
is parsed as
df['column_name'] >= (A & df['column_name']) <= B
which results in a Truth value of a Series is ambiguous error.
To select rows whose column value does not equal some_value, use !=:
df.loc[df['column_name'] != some_value]
isin returns a boolean Series, so to select rows whose value is not in some_values, negate the boolean Series using ~:
df.loc[~df['column_name'].isin(some_values)]
For example,
import pandas as pd
import numpy as np
df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(),
'B': 'one one two three two two one three'.split(),
'C': np.arange(8), 'D': np.arange(8) * 2})
print(df)
# A B C D
# 0 foo one 0 0
# 1 bar one 1 2
# 2 foo two 2 4
# 3 bar three 3 6
# 4 foo two 4 8
# 5 bar two 5 10
# 6 foo one 6 12
# 7 foo three 7 14
print(df.loc[df['A'] == 'foo'])
yields
A B C D
0 foo one 0 0
2 foo two 2 4
4 foo two 4 8
6 foo one 6 12
7 foo three 7 14
If you have multiple values you want to include, put them in a
list (or more generally, any iterable) and use isin:
print(df.loc[df['B'].isin(['one','three'])])
yields
A B C D
0 foo one 0 0
1 bar one 1 2
3 bar three 3 6
6 foo one 6 12
7 foo three 7 14
Note, however, that if you wish to do this many times, it is more efficient to
make an index first, and then use df.loc:
df = df.set_index(['B'])
print(df.loc['one'])
yields
A C D
B
one foo 0 0
one bar 1 2
one foo 6 12
or, to include multiple values from the index use df.index.isin:
df.loc[df.index.isin(['one','two'])]
yields
A C D
B
one foo 0 0
one bar 1 2
two foo 2 4
two foo 4 8
two bar 5 10
one foo 6 12
There are several ways to select rows from a Pandas dataframe:
Boolean indexing (df[df['col'] == value] )
Positional indexing (df.iloc[...])
Label indexing (df.xs(...))
df.query(...) API
Below I show you examples of each, with advice when to use certain techniques. Assume our criterion is column 'A' == 'foo'
(Note on performance: For each base type, we can keep things simple by using the Pandas API or we can venture outside the API, usually into NumPy, and speed things up.)
Setup
The first thing we'll need is to identify a condition that will act as our criterion for selecting rows. We'll start with the OP's case column_name == some_value, and include some other common use cases.
Borrowing from #unutbu:
import pandas as pd, numpy as np
df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(),
'B': 'one one two three two two one three'.split(),
'C': np.arange(8), 'D': np.arange(8) * 2})
1. Boolean indexing
... Boolean indexing requires finding the true value of each row's 'A' column being equal to 'foo', then using those truth values to identify which rows to keep. Typically, we'd name this series, an array of truth values, mask. We'll do so here as well.
mask = df['A'] == 'foo'
We can then use this mask to slice or index the data frame
df[mask]
A B C D
0 foo one 0 0
2 foo two 2 4
4 foo two 4 8
6 foo one 6 12
7 foo three 7 14
This is one of the simplest ways to accomplish this task and if performance or intuitiveness isn't an issue, this should be your chosen method. However, if performance is a concern, then you might want to consider an alternative way of creating the mask.
2. Positional indexing
Positional indexing (df.iloc[...]) has its use cases, but this isn't one of them. In order to identify where to slice, we first need to perform the same boolean analysis we did above. This leaves us performing one extra step to accomplish the same task.
mask = df['A'] == 'foo'
pos = np.flatnonzero(mask)
df.iloc[pos]
A B C D
0 foo one 0 0
2 foo two 2 4
4 foo two 4 8
6 foo one 6 12
7 foo three 7 14
3. Label indexing
Label indexing can be very handy, but in this case, we are again doing more work for no benefit
df.set_index('A', append=True, drop=False).xs('foo', level=1)
A B C D
0 foo one 0 0
2 foo two 2 4
4 foo two 4 8
6 foo one 6 12
7 foo three 7 14
4. df.query() API
pd.DataFrame.query is a very elegant/intuitive way to perform this task, but is often slower. However, if you pay attention to the timings below, for large data, the query is very efficient. More so than the standard approach and of similar magnitude as my best suggestion.
df.query('A == "foo"')
A B C D
0 foo one 0 0
2 foo two 2 4
4 foo two 4 8
6 foo one 6 12
7 foo three 7 14
My preference is to use the Boolean mask
Actual improvements can be made by modifying how we create our Boolean mask.
mask alternative 1
Use the underlying NumPy array and forgo the overhead of creating another pd.Series
mask = df['A'].values == 'foo'
I'll show more complete time tests at the end, but just take a look at the performance gains we get using the sample data frame. First, we look at the difference in creating the mask
%timeit mask = df['A'].values == 'foo'
%timeit mask = df['A'] == 'foo'
5.84 µs ± 195 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
166 µs ± 4.45 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Evaluating the mask with the NumPy array is ~ 30 times faster. This is partly due to NumPy evaluation often being faster. It is also partly due to the lack of overhead necessary to build an index and a corresponding pd.Series object.
Next, we'll look at the timing for slicing with one mask versus the other.
mask = df['A'].values == 'foo'
%timeit df[mask]
mask = df['A'] == 'foo'
%timeit df[mask]
219 µs ± 12.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
239 µs ± 7.03 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
The performance gains aren't as pronounced. We'll see if this holds up over more robust testing.
mask alternative 2
We could have reconstructed the data frame as well. There is a big caveat when reconstructing a dataframe—you must take care of the dtypes when doing so!
Instead of df[mask] we will do this
pd.DataFrame(df.values[mask], df.index[mask], df.columns).astype(df.dtypes)
If the data frame is of mixed type, which our example is, then when we get df.values the resulting array is of dtype object and consequently, all columns of the new data frame will be of dtype object. Thus requiring the astype(df.dtypes) and killing any potential performance gains.
%timeit df[m]
%timeit pd.DataFrame(df.values[mask], df.index[mask], df.columns).astype(df.dtypes)
216 µs ± 10.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
1.43 ms ± 39.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
However, if the data frame is not of mixed type, this is a very useful way to do it.
Given
np.random.seed([3,1415])
d1 = pd.DataFrame(np.random.randint(10, size=(10, 5)), columns=list('ABCDE'))
d1
A B C D E
0 0 2 7 3 8
1 7 0 6 8 6
2 0 2 0 4 9
3 7 3 2 4 3
4 3 6 7 7 4
5 5 3 7 5 9
6 8 7 6 4 7
7 6 2 6 6 5
8 2 8 7 5 8
9 4 7 6 1 5
%%timeit
mask = d1['A'].values == 7
d1[mask]
179 µs ± 8.73 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
Versus
%%timeit
mask = d1['A'].values == 7
pd.DataFrame(d1.values[mask], d1.index[mask], d1.columns)
87 µs ± 5.12 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
We cut the time in half.
mask alternative 3
#unutbu also shows us how to use pd.Series.isin to account for each element of df['A'] being in a set of values. This evaluates to the same thing if our set of values is a set of one value, namely 'foo'. But it also generalizes to include larger sets of values if needed. Turns out, this is still pretty fast even though it is a more general solution. The only real loss is in intuitiveness for those not familiar with the concept.
mask = df['A'].isin(['foo'])
df[mask]
A B C D
0 foo one 0 0
2 foo two 2 4
4 foo two 4 8
6 foo one 6 12
7 foo three 7 14
However, as before, we can utilize NumPy to improve performance while sacrificing virtually nothing. We'll use np.in1d
mask = np.in1d(df['A'].values, ['foo'])
df[mask]
A B C D
0 foo one 0 0
2 foo two 2 4
4 foo two 4 8
6 foo one 6 12
7 foo three 7 14
Timing
I'll include other concepts mentioned in other posts as well for reference.
Code Below
Each column in this table represents a different length data frame over which we test each function. Each column shows relative time taken, with the fastest function given a base index of 1.0.
res.div(res.min())
10 30 100 300 1000 3000 10000 30000
mask_standard 2.156872 1.850663 2.034149 2.166312 2.164541 3.090372 2.981326 3.131151
mask_standard_loc 1.879035 1.782366 1.988823 2.338112 2.361391 3.036131 2.998112 2.990103
mask_with_values 1.010166 1.000000 1.005113 1.026363 1.028698 1.293741 1.007824 1.016919
mask_with_values_loc 1.196843 1.300228 1.000000 1.000000 1.038989 1.219233 1.037020 1.000000
query 4.997304 4.765554 5.934096 4.500559 2.997924 2.397013 1.680447 1.398190
xs_label 4.124597 4.272363 5.596152 4.295331 4.676591 5.710680 6.032809 8.950255
mask_with_isin 1.674055 1.679935 1.847972 1.724183 1.345111 1.405231 1.253554 1.264760
mask_with_in1d 1.000000 1.083807 1.220493 1.101929 1.000000 1.000000 1.000000 1.144175
You'll notice that the fastest times seem to be shared between mask_with_values and mask_with_in1d.
res.T.plot(loglog=True)
Functions
def mask_standard(df):
mask = df['A'] == 'foo'
return df[mask]
def mask_standard_loc(df):
mask = df['A'] == 'foo'
return df.loc[mask]
def mask_with_values(df):
mask = df['A'].values == 'foo'
return df[mask]
def mask_with_values_loc(df):
mask = df['A'].values == 'foo'
return df.loc[mask]
def query(df):
return df.query('A == "foo"')
def xs_label(df):
return df.set_index('A', append=True, drop=False).xs('foo', level=-1)
def mask_with_isin(df):
mask = df['A'].isin(['foo'])
return df[mask]
def mask_with_in1d(df):
mask = np.in1d(df['A'].values, ['foo'])
return df[mask]
Testing
res = pd.DataFrame(
index=[
'mask_standard', 'mask_standard_loc', 'mask_with_values', 'mask_with_values_loc',
'query', 'xs_label', 'mask_with_isin', 'mask_with_in1d'
],
columns=[10, 30, 100, 300, 1000, 3000, 10000, 30000],
dtype=float
)
for j in res.columns:
d = pd.concat([df] * j, ignore_index=True)
for i in res.index:a
stmt = '{}(d)'.format(i)
setp = 'from __main__ import d, {}'.format(i)
res.at[i, j] = timeit(stmt, setp, number=50)
Special Timing
Looking at the special case when we have a single non-object dtype for the entire data frame.
Code Below
spec.div(spec.min())
10 30 100 300 1000 3000 10000 30000
mask_with_values 1.009030 1.000000 1.194276 1.000000 1.236892 1.095343 1.000000 1.000000
mask_with_in1d 1.104638 1.094524 1.156930 1.072094 1.000000 1.000000 1.040043 1.027100
reconstruct 1.000000 1.142838 1.000000 1.355440 1.650270 2.222181 2.294913 3.406735
Turns out, reconstruction isn't worth it past a few hundred rows.
spec.T.plot(loglog=True)
Functions
np.random.seed([3,1415])
d1 = pd.DataFrame(np.random.randint(10, size=(10, 5)), columns=list('ABCDE'))
def mask_with_values(df):
mask = df['A'].values == 'foo'
return df[mask]
def mask_with_in1d(df):
mask = np.in1d(df['A'].values, ['foo'])
return df[mask]
def reconstruct(df):
v = df.values
mask = np.in1d(df['A'].values, ['foo'])
return pd.DataFrame(v[mask], df.index[mask], df.columns)
spec = pd.DataFrame(
index=['mask_with_values', 'mask_with_in1d', 'reconstruct'],
columns=[10, 30, 100, 300, 1000, 3000, 10000, 30000],
dtype=float
)
Testing
for j in spec.columns:
d = pd.concat([df] * j, ignore_index=True)
for i in spec.index:
stmt = '{}(d)'.format(i)
setp = 'from __main__ import d, {}'.format(i)
spec.at[i, j] = timeit(stmt, setp, number=50)
tl;dr
The Pandas equivalent to
select * from table where column_name = some_value
is
table[table.column_name == some_value]
Multiple conditions:
table[(table.column_name == some_value) | (table.column_name2 == some_value2)]
or
table.query('column_name == some_value | column_name2 == some_value2')
Code example
import pandas as pd
# Create data set
d = {'foo':[100, 111, 222],
'bar':[333, 444, 555]}
df = pd.DataFrame(d)
# Full dataframe:
df
# Shows:
# bar foo
# 0 333 100
# 1 444 111
# 2 555 222
# Output only the row(s) in df where foo is 222:
df[df.foo == 222]
# Shows:
# bar foo
# 2 555 222
In the above code it is the line df[df.foo == 222] that gives the rows based on the column value, 222 in this case.
Multiple conditions are also possible:
df[(df.foo == 222) | (df.bar == 444)]
# bar foo
# 1 444 111
# 2 555 222
But at that point I would recommend using the query function, since it's less verbose and yields the same result:
df.query('foo == 222 | bar == 444')
I find the syntax of the previous answers to be redundant and difficult to remember. Pandas introduced the query() method in v0.13 and I much prefer it. For your question, you could do df.query('col == val').
Reproduced from The query() Method (Experimental):
In [167]: n = 10
In [168]: df = pd.DataFrame(np.random.rand(n, 3), columns=list('abc'))
In [169]: df
Out[169]:
a b c
0 0.687704 0.582314 0.281645
1 0.250846 0.610021 0.420121
2 0.624328 0.401816 0.932146
3 0.011763 0.022921 0.244186
4 0.590198 0.325680 0.890392
5 0.598892 0.296424 0.007312
6 0.634625 0.803069 0.123872
7 0.924168 0.325076 0.303746
8 0.116822 0.364564 0.454607
9 0.986142 0.751953 0.561512
# pure python
In [170]: df[(df.a < df.b) & (df.b < df.c)]
Out[170]:
a b c
3 0.011763 0.022921 0.244186
8 0.116822 0.364564 0.454607
# query
In [171]: df.query('(a < b) & (b < c)')
Out[171]:
a b c
3 0.011763 0.022921 0.244186
8 0.116822 0.364564 0.454607
You can also access variables in the environment by prepending an #.
exclude = ('red', 'orange')
df.query('color not in #exclude')
More flexibility using .query with pandas >= 0.25.0:
Since pandas >= 0.25.0 we can use the query method to filter dataframes with pandas methods and even column names which have spaces. Normally the spaces in column names would give an error, but now we can solve that using a backtick (`) - see GitHub:
# Example dataframe
df = pd.DataFrame({'Sender email':['ex#example.com', "reply#shop.com", "buy#shop.com"]})
Sender email
0 ex#example.com
1 reply#shop.com
2 buy#shop.com
Using .query with method str.endswith:
df.query('`Sender email`.str.endswith("#shop.com")')
Output
Sender email
1 reply#shop.com
2 buy#shop.com
Also we can use local variables by prefixing it with an # in our query:
domain = 'shop.com'
df.query('`Sender email`.str.endswith(#domain)')
Output
Sender email
1 reply#shop.com
2 buy#shop.com
For selecting only specific columns out of multiple columns for a given value in Pandas:
select col_name1, col_name2 from table where column_name = some_value.
Options loc:
df.loc[df['column_name'] == some_value, [col_name1, col_name2]]
or query:
df.query('column_name == some_value')[[col_name1, col_name2]]
In newer versions of Pandas, inspired by the documentation (Viewing data):
df[df["colume_name"] == some_value] #Scalar, True/False..
df[df["colume_name"] == "some_value"] #String
Combine multiple conditions by putting the clause in parentheses, (), and combining them with & and | (and/or). Like this:
df[(df["colume_name"] == "some_value1") & (pd[pd["colume_name"] == "some_value2"])]
Other filters
pandas.notna(df["colume_name"]) == True # Not NaN
df['colume_name'].str.contains("text") # Search for "text"
df['colume_name'].str.lower().str.contains("text") # Search for "text", after converting to lowercase
Faster results can be achieved using numpy.where.
For example, with unubtu's setup -
In [76]: df.iloc[np.where(df.A.values=='foo')]
Out[76]:
A B C D
0 foo one 0 0
2 foo two 2 4
4 foo two 4 8
6 foo one 6 12
7 foo three 7 14
Timing comparisons:
In [68]: %timeit df.iloc[np.where(df.A.values=='foo')] # fastest
1000 loops, best of 3: 380 µs per loop
In [69]: %timeit df.loc[df['A'] == 'foo']
1000 loops, best of 3: 745 µs per loop
In [71]: %timeit df.loc[df['A'].isin(['foo'])]
1000 loops, best of 3: 562 µs per loop
In [72]: %timeit df[df.A=='foo']
1000 loops, best of 3: 796 µs per loop
In [74]: %timeit df.query('(A=="foo")') # slowest
1000 loops, best of 3: 1.71 ms per loop
Here is a simple example
from pandas import DataFrame
# Create data set
d = {'Revenue':[100,111,222],
'Cost':[333,444,555]}
df = DataFrame(d)
# mask = Return True when the value in column "Revenue" is equal to 111
mask = df['Revenue'] == 111
print mask
# Result:
# 0 False
# 1 True
# 2 False
# Name: Revenue, dtype: bool
# Select * FROM df WHERE Revenue = 111
df[mask]
# Result:
# Cost Revenue
# 1 444 111
To add: You can also do df.groupby('column_name').get_group('column_desired_value').reset_index() to make a new data frame with specified column having a particular value. E.g.,
import pandas as pd
df = pd.DataFrame({'A': 'foo bar foo bar foo bar foo foo'.split(),
'B': 'one one two three two two one three'.split()})
print("Original dataframe:")
print(df)
b_is_two_dataframe = pd.DataFrame(df.groupby('B').get_group('two').reset_index()).drop('index', axis = 1)
#NOTE: the final drop is to remove the extra index column returned by groupby object
print('Sub dataframe where B is two:')
print(b_is_two_dataframe)
Running this gives:
Original dataframe:
A B
0 foo one
1 bar one
2 foo two
3 bar three
4 foo two
5 bar two
6 foo one
7 foo three
Sub dataframe where B is two:
A B
0 foo two
1 foo two
2 bar two
You can also use .apply:
df.apply(lambda row: row[df['B'].isin(['one','three'])])
It actually works row-wise (i.e., applies the function to each row).
The output is
A B C D
0 foo one 0 0
1 bar one 1 2
3 bar three 3 6
6 foo one 6 12
7 foo three 7 14
The results is the same as using as mentioned by #unutbu
df[[df['B'].isin(['one','three'])]]
1. Use f-strings inside query() calls
If the column name used to filter your dataframe comes from a local variable, f-strings may be useful. For example,
col = 'A'
df.query(f"{col} == 'foo'")
In fact, f-strings can be used for the query variable as well (except for datetime):
col = 'A'
my_var = 'foo'
df.query(f"{col} == '{my_var}'") # if my_var is a string
my_num = 1
df.query(f"{col} == {my_num}") # if my_var is a number
my_date = '2022-12-10'
df.query(f"{col} == #my_date") # must use # for datetime though
2. Install numexpr to speed up query() calls
The pandas documentation recommends installing numexpr to speed up numeric calculation when using query(). Use pip install numexpr (or conda, sudo etc. depending on your environment) to install it.
For larger dataframes (where performance actually matters), df.query() with numexpr engine performs much faster than df[mask]. In particular, it performs better for the following cases.
Logical and/or comparison operators on columns of strings
If a column of strings are compared to some other string(s) and matching rows are to be selected, even for a single comparison operation, query() performs faster than df[mask]. For example, for a dataframe with 80k rows, it's 30% faster1 and for a dataframe with 800k rows, it's 60% faster.2
df[df.A == 'foo']
df.query("A == 'foo'") # <--- performs 30%-60% faster
This gap increases as the number of operations increases (if 4 comparisons are chained df.query() is 2-2.3 times faster than df[mask])1,2 and/or the dataframe length increases.2
Multiple operations on numeric columns
If multiple arithmetic, logical or comparison operations need to be computed to create a boolean mask to filter df, query() performs faster. For example, for a frame with 80k rows, it's 20% faster1 and for a frame with 800k rows, it's 2 times faster.2
df[(df.B % 5) **2 < 0.1]
df.query("(B % 5) **2 < 0.1") # <--- performs 20%-100% faster.
This gap in performance increases as the number of operations increases and/or the dataframe length increases.2
The following plot shows how the methods perform as the dataframe length increases.3
3. Call pandas methods inside query()
Numexpr currently supports only logical (&, |, ~), comparison (==, >, <, >=, <=, !=) and basic arithmetic operators (+, -, *, /, **, %).
For example, it doesn't support integer division (//). However, calling the equivalent pandas method (floordiv()) works.
df.query('B.floordiv(2) <= 3') # or
df.query('B.floordiv(2).le(3)')
# for pandas < 1.4, need `.values`
df.query('B.floordiv(2).values <= 3')
1 Benchmark code using a frame with 80k rows
import numpy as np
df = pd.DataFrame({'A': 'foo bar foo baz foo bar foo foo'.split()*10000,
'B': np.random.rand(80000)})
%timeit df[df.A == 'foo']
# 8.5 ms ± 104.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit df.query("A == 'foo'")
# 6.36 ms ± 95.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit df[((df.A == 'foo') & (df.A != 'bar')) | ((df.A != 'baz') & (df.A != 'buz'))]
# 29 ms ± 554 µs per loop (mean ± std. dev. of 10 runs, 100 loops each)
%timeit df.query("A == 'foo' & A != 'bar' | A != 'baz' & A != 'buz'")
# 16 ms ± 339 µs per loop (mean ± std. dev. of 10 runs, 100 loops each)
%timeit df[(df.B % 5) **2 < 0.1]
# 5.35 ms ± 37.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit df.query("(B % 5) **2 < 0.1")
# 4.37 ms ± 46.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
2 Benchmark code using a frame with 800k rows
df = pd.DataFrame({'A': 'foo bar foo baz foo bar foo foo'.split()*100000,
'B': np.random.rand(800000)})
%timeit df[df.A == 'foo']
# 87.9 ms ± 873 µs per loop (mean ± std. dev. of 10 runs, 100 loops each)
%timeit df.query("A == 'foo'")
# 54.4 ms ± 726 µs per loop (mean ± std. dev. of 10 runs, 100 loops each)
%timeit df[((df.A == 'foo') & (df.A != 'bar')) | ((df.A != 'baz') & (df.A != 'buz'))]
# 310 ms ± 3.4 ms per loop (mean ± std. dev. of 10 runs, 100 loops each)
%timeit df.query("A == 'foo' & A != 'bar' | A != 'baz' & A != 'buz'")
# 132 ms ± 2.43 ms per loop (mean ± std. dev. of 10 runs, 100 loops each)
%timeit df[(df.B % 5) **2 < 0.1]
# 54 ms ± 488 µs per loop (mean ± std. dev. of 10 runs, 100 loops each)
%timeit df.query("(B % 5) **2 < 0.1")
# 26.3 ms ± 320 µs per loop (mean ± std. dev. of 10 runs, 100 loops each)
3: Code used to produce the performance graphs of the two methods for strings and numbers.
from perfplot import plot
constructor = lambda n: pd.DataFrame({'A': 'foo bar foo baz foo bar foo foo'.split()*n, 'B': np.random.rand(8*n)})
plot(
setup=constructor,
kernels=[lambda df: df[(df.B%5)**2<0.1], lambda df: df.query("(B%5)**2<0.1")],
labels= ['df[(df.B % 5) **2 < 0.1]', 'df.query("(B % 5) **2 < 0.1")'],
n_range=[2**k for k in range(4, 24)],
xlabel='Rows in DataFrame',
title='Multiple mathematical operations on numbers',
equality_check=pd.DataFrame.equals);
plot(
setup=constructor,
kernels=[lambda df: df[df.A == 'foo'], lambda df: df.query("A == 'foo'")],
labels= ["df[df.A == 'foo']", """df.query("A == 'foo'")"""],
n_range=[2**k for k in range(4, 24)],
xlabel='Rows in DataFrame',
title='Comparison operation on strings',
equality_check=pd.DataFrame.equals);
If you want to make query to your dataframe repeatedly and speed is important to you, the best thing is to convert your dataframe to dictionary and then by doing this you can make query thousands of times faster.
my_df = df.set_index(column_name)
my_dict = my_df.to_dict('index')
After make my_dict dictionary you can go through:
if some_value in my_dict.keys():
my_result = my_dict[some_value]
If you have duplicated values in column_name you can't make a dictionary. but you can use:
my_result = my_df.loc[some_value]
SQL statements on DataFrames to select rows using DuckDB
With DuckDB we can query pandas DataFrames with SQL statements, in a highly performant way.
Since the question is How do I select rows from a DataFrame based on column values?, and the example in the question is a SQL query, this answer looks logical in this topic.
Example:
In [1]: import duckdb
In [2]: import pandas as pd
In [3]: con = duckdb.connect()
In [4]: df = pd.DataFrame({"A": range(11), "B": range(11, 22)})
In [5]: df
Out[5]:
A B
0 0 11
1 1 12
2 2 13
3 3 14
4 4 15
5 5 16
6 6 17
7 7 18
8 8 19
9 9 20
10 10 21
In [6]: results = con.execute("SELECT * FROM df where A > 2").df()
In [7]: results
Out[7]:
A B
0 3 14
1 4 15
2 5 16
3 6 17
4 7 18
5 8 19
6 9 20
7 10 21
You can use loc (square brackets) with a function:
# Series
s = pd.Series([1, 2, 3, 4])
s.loc[lambda x: x > 1]
# s[lambda x: x > 1]
Output:
1 2
2 3
3 4
dtype: int64
or
# DataFrame
df = pd.DataFrame({'A': [1, 2, 3], 'B': [10, 20, 30]})
df.loc[lambda x: x['A'] > 1]
# df[lambda x: x['A'] > 1]
Output:
A B
1 2 20
2 3 30
Great answers. Only, when the size of the dataframe approaches million rows, many of the methods tend to take ages when using df[df['col']==val]. I wanted to have all possible values of "another_column" that correspond to specific values in "some_column" (in this case in a dictionary). This worked and fast.
s=datetime.datetime.now()
my_dict={}
for i, my_key in enumerate(df['some_column'].values):
if i%100==0:
print(i) # to see the progress
if my_key not in my_dict.keys():
my_dict[my_key]={}
my_dict[my_key]['values']=[df.iloc[i]['another_column']]
else:
my_dict[my_key]['values'].append(df.iloc[i]['another_column'])
e=datetime.datetime.now()
print('operation took '+str(e-s)+' seconds')```
I have a very large dataframe (around 1 million rows) with data from an experiment (60 respondents).
I would like to split the dataframe into 60 dataframes (a dataframe for each participant).
In the dataframe, data, there is a variable called 'name', which is the unique code for each participant.
I have tried the following, but nothing happens (or execution does not stop within an hour). What I intend to do is to split the data into smaller dataframes, and append these to a list (datalist):
import pandas as pd
def splitframe(data, name='name'):
n = data[name][0]
df = pd.DataFrame(columns=data.columns)
datalist = []
for i in range(len(data)):
if data[name][i] == n:
df = df.append(data.iloc[i])
else:
datalist.append(df)
df = pd.DataFrame(columns=data.columns)
n = data[name][i]
df = df.append(data.iloc[i])
return datalist
I do not get an error message, the script just seems to run forever!
Is there a smart way to do it?
Can I ask why not just do it by slicing the data frame. Something like
#create some data with Names column
data = pd.DataFrame({'Names': ['Joe', 'John', 'Jasper', 'Jez'] *4, 'Ob1' : np.random.rand(16), 'Ob2' : np.random.rand(16)})
#create unique list of names
UniqueNames = data.Names.unique()
#create a data frame dictionary to store your data frames
DataFrameDict = {elem : pd.DataFrame() for elem in UniqueNames}
for key in DataFrameDict.keys():
DataFrameDict[key] = data[:][data.Names == key]
Hey presto you have a dictionary of data frames just as (I think) you want them. Need to access one? Just enter
DataFrameDict['Joe']
Firstly your approach is inefficient because the appending to the list on a row by basis will be slow as it has to periodically grow the list when there is insufficient space for the new entry, list comprehensions are better in this respect as the size is determined up front and allocated once.
However, I think fundamentally your approach is a little wasteful as you have a dataframe already so why create a new one for each of these users?
I would sort the dataframe by column 'name', set the index to be this and if required not drop the column.
Then generate a list of all the unique entries and then you can perform a lookup using these entries and crucially if you only querying the data, use the selection criteria to return a view on the dataframe without incurring a costly data copy.
Use pandas.DataFrame.sort_values and pandas.DataFrame.set_index:
# sort the dataframe
df.sort_values(by='name', axis=1, inplace=True)
# set the index to be this and don't drop
df.set_index(keys=['name'], drop=False,inplace=True)
# get a list of names
names=df['name'].unique().tolist()
# now we can perform a lookup on a 'view' of the dataframe
joe = df.loc[df.name=='joe']
# now you can query all 'joes'
You can convert groupby object to tuples and then to dict:
df = pd.DataFrame({'Name':list('aabbef'),
'A':[4,5,4,5,5,4],
'B':[7,8,9,4,2,3],
'C':[1,3,5,7,1,0]}, columns = ['Name','A','B','C'])
print (df)
Name A B C
0 a 4 7 1
1 a 5 8 3
2 b 4 9 5
3 b 5 4 7
4 e 5 2 1
5 f 4 3 0
d = dict(tuple(df.groupby('Name')))
print (d)
{'b': Name A B C
2 b 4 9 5
3 b 5 4 7, 'e': Name A B C
4 e 5 2 1, 'a': Name A B C
0 a 4 7 1
1 a 5 8 3, 'f': Name A B C
5 f 4 3 0}
print (d['a'])
Name A B C
0 a 4 7 1
1 a 5 8 3
It is not recommended, but possible create DataFrames by groups:
for i, g in df.groupby('Name'):
globals()['df_' + str(i)] = g
print (df_a)
Name A B C
0 a 4 7 1
1 a 5 8 3
Easy:
[v for k, v in df.groupby('name')]
Groupby can helps you:
grouped = data.groupby(['name'])
Then you can work with each group like with a dataframe for each participant. And DataFrameGroupBy object methods such as (apply, transform, aggregate, head, first, last) return a DataFrame object.
Or you can make list from grouped and get all DataFrame's by index:
l_grouped = list(grouped)
l_grouped[0][1] - DataFrame for first group with first name.
In addition to Gusev Slava's answer, you might want to use groupby's groups:
{key: df.loc[value] for key, value in df.groupby("name").groups.items()}
This will yield a dictionary with the keys you have grouped by, pointing to the corresponding partitions. The advantage is that the keys are maintained and don't vanish in the list index.
The method in the OP works, but isn't efficient. It may have seemed to run forever, because the dataset was long.
Use .groupby on the 'method' column, and create a dict of DataFrames with unique 'method' values as the keys, with a dict-comprehension.
.groupby returns a groupby object, that contains information about the groups, where g is the unique value in 'method' for each group, and d is the DataFrame for that group.
The value of each key in df_dict, will be a DataFrame, which can be accessed in the standard way, df_dict['key'].
The original question wanted a list of DataFrames, which can be done with a list-comprehension
df_list = [d for _, d in df.groupby('method')]
import pandas as pd
import seaborn as sns # for test dataset
# load data for example
df = sns.load_dataset('planets')
# display(df.head())
method number orbital_period mass distance year
0 Radial Velocity 1 269.300 7.10 77.40 2006
1 Radial Velocity 1 874.774 2.21 56.95 2008
2 Radial Velocity 1 763.000 2.60 19.84 2011
3 Radial Velocity 1 326.030 19.40 110.62 2007
4 Radial Velocity 1 516.220 10.50 119.47 2009
# Using a dict-comprehension, the unique 'method' value will be the key
df_dict = {g: d for g, d in df.groupby('method')}
print(df_dict.keys())
[out]:
dict_keys(['Astrometry', 'Eclipse Timing Variations', 'Imaging', 'Microlensing', 'Orbital Brightness Modulation', 'Pulsar Timing', 'Pulsation Timing Variations', 'Radial Velocity', 'Transit', 'Transit Timing Variations'])
# or a specific name for the key, using enumerate (e.g. df1, df2, etc.)
df_dict = {f'df{i}': d for i, (g, d) in enumerate(df.groupby('method'))}
print(df_dict.keys())
[out]:
dict_keys(['df0', 'df1', 'df2', 'df3', 'df4', 'df5', 'df6', 'df7', 'df8', 'df9'])
df_dict['df1].head(3) or df_dict['Astrometry'].head(3)
There are only 2 in this group
method number orbital_period mass distance year
113 Astrometry 1 246.36 NaN 20.77 2013
537 Astrometry 1 1016.00 NaN 14.98 2010
df_dict['df2].head(3) or df_dict['Eclipse Timing Variations'].head(3)
method number orbital_period mass distance year
32 Eclipse Timing Variations 1 10220.0 6.05 NaN 2009
37 Eclipse Timing Variations 2 5767.0 NaN 130.72 2008
38 Eclipse Timing Variations 2 3321.0 NaN 130.72 2008
df_dict['df3].head(3) or df_dict['Imaging'].head(3)
method number orbital_period mass distance year
29 Imaging 1 NaN NaN 45.52 2005
30 Imaging 1 NaN NaN 165.00 2007
31 Imaging 1 NaN NaN 140.00 2004
For more information about the seaborn datasets
NASA Exoplanets
Alternatively
This is a manual method to create separate DataFrames using pandas: Boolean Indexing
This is similar to the accepted answer, but .loc is not required.
This is an acceptable method for creating a couple extra DataFrames.
The pythonic way to create multiple objects, is by placing them in a container (e.g. dict, list, generator, etc.), as shown above.
df1 = df[df.method == 'Astrometry']
df2 = df[df.method == 'Eclipse Timing Variations']
In [28]: df = DataFrame(np.random.randn(1000000,10))
In [29]: df
Out[29]:
<class 'pandas.core.frame.DataFrame'>
Int64Index: 1000000 entries, 0 to 999999
Data columns (total 10 columns):
0 1000000 non-null values
1 1000000 non-null values
2 1000000 non-null values
3 1000000 non-null values
4 1000000 non-null values
5 1000000 non-null values
6 1000000 non-null values
7 1000000 non-null values
8 1000000 non-null values
9 1000000 non-null values
dtypes: float64(10)
In [30]: frames = [ df.iloc[i*60:min((i+1)*60,len(df))] for i in xrange(int(len(df)/60.) + 1) ]
In [31]: %timeit [ df.iloc[i*60:min((i+1)*60,len(df))] for i in xrange(int(len(df)/60.) + 1) ]
1 loops, best of 3: 849 ms per loop
In [32]: len(frames)
Out[32]: 16667
Here's a groupby way (and you could do an arbitrary apply rather than sum)
In [9]: g = df.groupby(lambda x: x/60)
In [8]: g.sum()
Out[8]:
<class 'pandas.core.frame.DataFrame'>
Int64Index: 16667 entries, 0 to 16666
Data columns (total 10 columns):
0 16667 non-null values
1 16667 non-null values
2 16667 non-null values
3 16667 non-null values
4 16667 non-null values
5 16667 non-null values
6 16667 non-null values
7 16667 non-null values
8 16667 non-null values
9 16667 non-null values
dtypes: float64(10)
Sum is cythonized that's why this is so fast
In [10]: %timeit g.sum()
10 loops, best of 3: 27.5 ms per loop
In [11]: %timeit df.groupby(lambda x: x/60)
1 loops, best of 3: 231 ms per loop
The method based on list comprehension and groupby- Which stores all the split dataframe in list variable and can be accessed using the index.
Example
ans = [pd.DataFrame(y) for x, y in DF.groupby('column_name', as_index=False)]
ans[0]
ans[0].column_name
You can use the groupby command, if you already have some labels for your data.
out_list = [group[1] for group in in_series.groupby(label_series.values)]
Here's a detailed example:
Let's say we want to partition a pd series using some labels into a list of chunks
For example, in_series is:
2019-07-01 08:00:00 -0.10
2019-07-01 08:02:00 1.16
2019-07-01 08:04:00 0.69
2019-07-01 08:06:00 -0.81
2019-07-01 08:08:00 -0.64
Length: 5, dtype: float64
And its corresponding label_series is:
2019-07-01 08:00:00 1
2019-07-01 08:02:00 1
2019-07-01 08:04:00 2
2019-07-01 08:06:00 2
2019-07-01 08:08:00 2
Length: 5, dtype: float64
Run
out_list = [group[1] for group in in_series.groupby(label_series.values)]
which returns out_list a list of two pd.Series:
[2019-07-01 08:00:00 -0.10
2019-07-01 08:02:00 1.16
Length: 2, dtype: float64,
2019-07-01 08:04:00 0.69
2019-07-01 08:06:00 -0.81
2019-07-01 08:08:00 -0.64
Length: 3, dtype: float64]
Note that you can use some parameters from in_series itself to group the series, e.g., in_series.index.day
here's a small function which might help some (efficiency not perfect probably, but compact + more or less easy to understand):
def get_splited_df_dict(df: 'pd.DataFrame', split_column: 'str'):
"""
splits a pandas.DataFrame on split_column and returns it as a dict
"""
df_dict = {value: df[df[split_column] == value].drop(split_column, axis=1) for value in df[split_column].unique()}
return df_dict
it converts a DataFrame to multiple DataFrames, by selecting each unique value in the given column and putting all those entries into a separate DataFrame.
the .drop(split_column, axis=1) is just for removing the column which was used to split the DataFrame. the removal is not necessary, but can help a little to cut down on memory usage after the operation.
the result of get_splited_df_dict is a dict, meaning one can access each DataFrame like this:
splitted = get_splited_df_dict(some_df, some_column)
# accessing the DataFrame with 'some_column_value'
splitted[some_column_value]
The existing answers cover all good cases and explains fairly well how the groupby object is like a dictionary with keys and values that can be accessed via .groups. Yet more methods to do the same job as the existing answers are:
Create a list by unpacking the groupby object and casting it to a dictionary:
dict([*df.groupby('Name')]) # same as dict(list(df.groupby('Name')))
Create a tuple + dict (this is the same as #jezrael's answer):
dict((*df.groupby('Name'),))
If we only want the DataFrames, we could get the values of the dictionary (created above):
[*dict([*df.groupby('Name')]).values()]
I had similar problem. I had a time series of daily sales for 10 different stores and 50 different items. I needed to split the original dataframe in 500 dataframes (10stores*50stores) to apply Machine Learning models to each of them and I couldn't do it manually.
This is the head of the dataframe:
I have created two lists;
one for the names of dataframes
and one for the couple of array [item_number, store_number].
list=[]
for i in range(1,len(items)*len(stores)+1):
global list
list.append('df'+str(i))
list_couple_s_i =[]
for item in items:
for store in stores:
global list_couple_s_i
list_couple_s_i.append([item,store])
And once the two lists are ready you can loop on them to create the dataframes you want:
for name, it_st in zip(list,list_couple_s_i):
globals()[name] = df.where((df['item']==it_st[0]) &
(df['store']==(it_st[1])))
globals()[name].dropna(inplace=True)
In this way I have created 500 dataframes.
Hope this will be helpful!
import pandas as pd
data = ["X", "Y", "Z", "A", "B"]
label = ['a','b','c','d','e']
df = pd.DataFrame(data, columns=label)
print(df)
I want to get the dataframe to be:
a b c d e
X Y Z A B
I am getting
ValueError: Shape of passed values is (1, 5), indices imply (5, 5)
How to fix this to get the desired dataframe ?
Pass it as a list of list.
In [439]: pd.DataFrame([data], columns=label)
Out[439]:
a b c d e
0 X Y Z A B
You can use a bit complicated, but very fast solution if large data - convert list to numpy array and then reshape:
df = pd.DataFrame(np.array(data).reshape(-1, len(data)), columns=label)
print(df)
a b c d e
0 X Y Z A B
Timings:
N = 100
data = ["X", "Y", "Z", "A", "B"] * N
label = ['a','b','c','d','e'] * N
In [30]: %timeit pd.DataFrame([data], columns=label)
10 loops, best of 3: 178 ms per loop
In [31]: %timeit pd.DataFrame(np.array(data).reshape(-1, len(data)), columns=label)
1000 loops, best of 3: 1.06 ms per loop
N = 1000
In [35]: %timeit pd.DataFrame([data], columns=label)
1 loop, best of 3: 1.7 s per loop
In [36]: %timeit pd.DataFrame(np.array(data).reshape(-1, len(data)), columns=label)
100 loops, best of 3: 3.83 ms per loop
I have a dataframe in which one columns called 'label' holds values like 'b', 'm', 'n' etc.
I want 'label' to instead hold the ascii equivalent of the letter.
How do I do it?
IIUC:
In [81]:
df = pd.DataFrame({'label':list('bmn')})
df
Out[81]:
label
0 b
1 m
2 n
In [82]:
df['ascii'] = df['label'].apply(ord)
df
Out[82]:
label ascii
0 b 98
1 m 109
2 n 110
It maybe quicker to do a list comprehension:
In [83]:
df['ascii'] = [ord(x) for x in df['label']]
df
Out[83]:
label ascii
0 b 98
1 m 109
2 n 110
You can also use map:
In [85]:
df['ascii'] = df['label'].map(ord)
df
Out[85]:
label ascii
0 b 98
1 m 109
2 n 110
Timings
for a small df:
In [87]:
%timeit [ord(x) for x in df['label']]
%timeit df['label'].map(ord)
%timeit df['label'].apply(ord)
100000 loops, best of 3: 14 µs per loop
10000 loops, best of 3: 123 µs per loop
10000 loops, best of 3: 146 µs per loop
For a 3K df:
In [89]:
%timeit [ord(x) for x in df['label']]
%timeit df['label'].map(ord)
%timeit df['label'].apply(ord)
1000 loops, best of 3: 246 µs per loop
1000 loops, best of 3: 1 ms per loop
1000 loops, best of 3: 1.02 ms per loop
So here the list comprehension scales better than the other methods
e.g. "a"=97 in ascii}
write print(ord("a"))
print(ord("a"))
answer would be 97