I have two pandas dataframes and i want to find all entries of the second dataframe where a specific value occurs.
As an example:
df1:
NID
0 1
1 2
2 3
3 4
4 5
df2:
EID N1 N2 N3 N4
0 1 1 2 13 12
1 2 2 3 14 13
2 3 3 4 15 14
3 4 4 5 16 15
4 5 5 6 17 16
5 6 6 7 18 17
6 7 7 8 19 18
7 8 8 9 20 19
8 9 9 10 21 20
9 10 10 11 22 21
Now, what i basically want, is a list of lists with the values EID (from df2) where the values NID (from df1) occur in any of the columns N1,N2,N3,N4:
Solution would be:
sol = [[1], [1, 2], [2, 3], [3, 4], [4, 5]]
The desired solution explained:
The solution has 5 entries (len(sol = 5)) since I have 5 entries in df1.
The first entry in sol is 1 because the value NID = 1 only appears in the columns N1,N2,N3,N4 for EID=1 in df2.
The second entry in sol refers to the value NID=2 (of df1) and has the length 2 because NID=2 can be found in column N1 (for EID=2) and in column N2 (for EID=1). Therefore, the second entry in the solution is [1,2] and so on.
What I tried so far is looping for each element in df1 and then looping for each element in df2 to see if NID is in any of the columns N1,N2,N3,N4. This solution works but for huge dataframes (each df can have up to some thousand entries) this solution becomes extremely time-consuming.
Therefore I was looking for a much more efficient solution.
My code as implemented:
Input data:
import pandas as pd
df1 = pd.DataFrame({'NID':[1,2,3,4,5]})
df2 = pd.DataFrame({'EID':[1,2,3,4,5,6,7,8,9,10],
'N1':[1,2,3,4,5,6,7,8,9,10],
'N2':[2,3,4,5,6,7,8,9,10,11],
'N3':[13,14,15,16,17,18,19,20,21,22],
'N4':[12,13,14,15,16,17,18,19,20,21]})
solution acquired using looping:
sol= []
for idx,node in df1.iterrows():
x = []
for idx2,elem in df2.iterrows():
if node['NID'] == elem['N1']:
x.append(elem['EID'])
if node['NID'] == elem['N2']:
x.append(elem['EID'])
if node['NID'] == elem['N3']:
x.append(elem['EID'])
if node['NID'] == elem['N4']:
x.append(elem['EID'])
sol.append(x)
print(sol)
If anyone has a solution where I do not have to loop, I would be very happy. Maybe using a numpy function or something like cKDTrees but unfortunately I have no idea on how to get this problem solved in a faster way.
Thank you in advance!
You can reshape with melt, filter with loc, and groupby.agg as list. Then reindex and convert tolist:
out = (df2
.melt('EID') # reshape to long form
# filter the values that are in df1['NID']
.loc[lambda d: d['value'].isin(df1['NID'])]
# aggregate as list
.groupby('value')['EID'].agg(list)
# ensure all original NID are present in order
# and convert to list
.reindex(df1['NID']).tolist()
)
Alternative with stack:
df3 = df2.set_index('EID')
out = (df3
.where(df3.isin(df1['NID'].tolist())).stack()
.reset_index(name='group')
.groupby('group')['EID'].agg(list)
.reindex(df1['NID']).tolist()
)
Output:
[[1], [2, 1], [3, 2], [4, 3], [5, 4]]
Related
Lets say you have a dataframe like this:
df = pd.DataFrame({'A': [3, 1, 2, 3],
'B': [5, 6, 7, 8]})
df
A B
0 3 5
1 1 6
2 2 7
3 3 8
Now I want to skew and calculate on each column. I put the values as I want them skewed in the index:
range_span = range(4)
result = pd.DataFrame(index=range_span)
Then I try to pupulate result with the following:
for c in df.columns:
for i in range_span:
result.iloc[i][c] = df[c].shift(i).max()
result
This only returns the index. I expected something like this:
You've got 3 critical issues:
issue #1
At this line
result.iloc[i][c] = df[c].shift(i).max()
Raises warning that help understand why result is empty.
...\pandas\core\indexing.py:670: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame
See the caveats in the documentation: https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#returning-a-view-versus-a-copy
According to their document:
dfmi['one']['second'] = value
# becomes
dfmi.__getitem__('one').__setitem__('second', value)
As iloc[i] will return slice - aka copy - of that rows, you couldn't set original dataframe result. Further, this is why iloc didn't raised issue when it got str index. Explained in #2.
Instead you use iloc - potentially loc with str - like this:
>>> df
A B C
0 1 10 100
1 2 20 200
2 3 30 300
>>> df.iloc[1, 2]
200
>>>df.iloc[[1, 2], [1, 2]]
B C
1 20 200
2 30 300
>>> df.iloc[1:3, 1:3]
B C
1 20 200
2 30 300
>>> df.iloc[:, 1:3]
B C
0 10 100
1 20 200
2 30 300
# ..and so on
issue #2
If you fix issue #1 then you'll see following error:
result.iloc[[i][c]] = df[c].shift(i).max()
TypeError: list indices must be integers or slices, not str
Also from their document:
property DataFrame.iloc: Purely integer-location based indexing for selection by position.
At for c in df.columns: You're passing column name A, B which is str, not int. Use loc instead for str column indices.
This didn't raise TypeError due to issue #1 - as c was passed as argument of __setitem__().
Issue #3
Normally dataframe cannot be enlarged without special functions like combine.
# using same df from #1
>>> df.iloc[1, 3] = 300
Traceback (most recent call last):
File "~\pandas\core\indexing.py", line 1394, in _has_valid_setitem_indexer
raise IndexError("iloc cannot enlarge its target object")
IndexError: iloc cannot enlarge its target object
Easier fix would be using dict and convert to DataFrame when manipulation is complete. Or just creating DataFrame to match or have a larger size at firsthand:
>>> df2 = pd.DataFrame(index=range(4), columns=range(3))
>>> df2
0 1 2
0 NaN NaN NaN
1 NaN NaN NaN
2 NaN NaN NaN
3 NaN NaN NaN
Combining all, correct fix would be:
import pandas as pd
df = pd.DataFrame({'A': [3, 1, 2, 3],
'B': [5, 6, 7, 8]})
result = pd.DataFrame(index=df.index, columns=df.columns)
for col in df.columns:
for index in df.index:
result.loc[index, col] = df[col].shift(index).max()
print(result)
Output:
A B
0 3 8
1 3 7
2 3 6
3 3 5
Good afternoon,
I'm iterating through a huge Dataframe (104062 x 20) with the following code:
import pandas as pd
df_tot = pd.read_csv("C:\\Users\\XXXXX\\Desktop\\XXXXXXX\\LOGS\\DF_TOT.txt", header=None)
df_tot = df_tot.replace("\[", "", regex=True)
df_tot = df_tot.replace("\]", "", regex=True)
df_tot = df_tot.replace("\'", "", regex=True)
i = 0
while i < len(df_tot):
to_compare = df_tot.iloc[i].tolist()
for j in range(len(df_tot)):
if to_compare == df_tot.iloc[j].tolist():
if i == j:
print('Matched itself.')
else:
print('MATCH FOUND - row: {} --- match row: {}'.format(i,j))
i += 1
I am looking to optimize time spent for each iteration as much as possible, since this code iterates 104062(^2) times. (More or less ten billions iterations).
With my computing power, time spent comparing to_compare in the whole DF is around 26 seconds.
I want to clarify that in case it would be needed, the whole code could be changed with faster constructs.
As usual, Thanks in advance.
as far as i understand, You just want to find duplicated rows.
Sample data(2 last rows are duplicated):
In [1]: df = pd.DataFrame([[1,2], [3,4], [5,6], [7,8], [1,2], [5,6]], columns=['a', 'b'])
df
Out[1]:
a b
0 1 2
1 3 4
2 5 6
3 7 8
4 1 2
5 5 6
This will return all duplicated rows:
In [2]: df[df.duplicated(keep=False)]
Out[2]:
a b
0 1 2
2 5 6
4 1 2
5 5 6
And indexes, grouped by duplicated row:
In [3]: df[df.duplicated(keep=False)].reset_index().groupby(list(df.columns), as_index=False)['index'].apply(list)
Out[3]: a b
1 2 [0, 4]
5 6 [2, 5]
You can also just remove duplicates from dataframe:
In [4]: df.drop_duplicates()
Out[4]:
a b
0 1 2
1 3 4
2 5 6
3 7 8
I have a very large dataframe with 1,000 columns. The first few columns occur only once, denoting a customer. The next few columns are representative of multiple encounters with the customer, with an underscore and the number encounter. Every additional encounter adds a new column, so there is NOT a fixed number of columns -- it'll grow with time.
Sample dataframe header structure excerpt:
id dob gender pro_1 pro_10 pro_11 pro_2 ... pro_9 pre_1 pre_10 ...
I'm trying to re-order the columns based on the number after the column name, so all _1 should be together, all _2 should be together, etc, like so:
id dob gender pro_1 pre_1 que_1 fre_1 gen_1 pro2 pre_2 que_2 fre_2 ...
(Note that the re-order should order the numbers correctly; the current order treats them like strings, which orders 1, 10, 11, etc. rather than 1, 2, 3)
Is this possible to do in pandas, or should I be looking at something else? Any help would be greatly appreciated! Thank you!
EDIT:
Alternatively, is it also possible to re-arrange column names based on the string part AND number part of the column names? So the output would then look similar to the original, except the numbers would be considered so that the order is more intuitive:
id dob gender pro_1 pro_2 pro_3 ... pre_1 pre_2 pre_3 ...
EDIT 2.0:
Just wanted to thank everyone for helping! While only one of the responses worked, I really appreciate the effort and learned a lot about other approaches / ways to think about this.
Here is one way you can try:
# column names copied from your example
example_cols = 'id dob gender pro_1 pro_10 pro_11 pro_2 pro_9 pre_1 pre_10'.split()
# sample DF
df = pd.DataFrame([range(len(example_cols))], columns=example_cols)
df
# id dob gender pro_1 pro_10 pro_11 pro_2 pro_9 pre_1 pre_10
#0 0 1 2 3 4 5 6 7 8 9
# number of columns excluded from sorting
N = 3
# get a list of columns from the dataframe
cols = df.columns.tolist()
# split, create an tuple of (column_name, prefix, number) and sorted based on the 2nd and 3rd item of the tuple, then retrieved the first item.
# adjust "key = lambda x: x[2]" to group cols by numbers only
cols_new = cols[:N] + [ a[0] for a in sorted([ (c, p, int(n)) for c in cols[N:] for p,n in [c.split('_')]], key = lambda x: (x[1], x[2])) ]
# get the new dataframe based on the cols_new
df_new = df[cols_new]
# id dob gender pre_1 pre_10 pro_1 pro_2 pro_9 pro_10 pro_11
#0 0 1 2 8 9 3 6 7 4 5
Luckily there is a one liner in python that can fix this:
df = df.reindex(sorted(df.columns), axis=1)
For Example lets say you had this dataframe:
import pandas as pd
import numpy as np
df = pd.DataFrame({'Name': [2, 4, 8, 0],
'ID': [2, 0, 0, 0],
'Prod3': [10, 2, 1, 8],
'Prod1': [2, 4, 8, 0],
'Prod_1': [2, 4, 8, 0],
'Pre7': [2, 0, 0, 0],
'Pre2': [10, 2, 1, 8],
'Pre_2': [10, 2, 1, 8],
'Pre_9': [10, 2, 1, 8]}
)
print(df)
Output:
Name ID Prod3 Prod1 Prod_1 Pre7 Pre2 Pre_2 Pre_9
0 2 2 10 2 2 2 10 10 10
1 4 0 2 4 4 0 2 2 2
2 8 0 1 8 8 0 1 1 1
3 0 0 8 0 0 0 8 8 8
Then used
df = df.reindex(sorted(df.columns), axis=1)
Then the dataframe will then look like:
ID Name Pre2 Pre7 Pre_2 Pre_9 Prod1 Prod3 Prod_1
0 2 2 10 2 10 10 2 10 2
1 0 4 2 0 2 2 4 2 4
2 0 8 1 0 1 1 8 1 8
3 0 0 8 0 8 8 0 8 0
As you can see, the columns without underscore will come first, followed by an ordering based on the number after the underscore. However this also sorts of the column names, so the column names that come first in the alphabet will be first.
You need to split you column on '_' then convert to int:
c = ['A_1','A_10','A_2','A_3','B_1','B_10','B_2','B_3']
df = pd.DataFrame(np.random.randint(0,100,(2,8)), columns = c)
df.reindex(sorted(df.columns, key = lambda x: int(x.split('_')[1])), axis=1)
Output:
A_1 B_1 A_2 B_2 A_3 B_3 A_10 B_10
0 68 11 59 69 37 68 76 17
1 19 37 52 54 23 93 85 3
Next case, you need human sorting:
import re
def atoi(text):
return int(text) if text.isdigit() else text
def natural_keys(text):
'''
alist.sort(key=natural_keys) sorts in human order
http://nedbatchelder.com/blog/200712/human_sorting.html
(See Toothy's implementation in the comments)
'''
return [ atoi(c) for c in re.split(r'(\d+)', text) ]
df.reindex(sorted(df.columns, key = lambda x:natural_keys(x)), axis=1)
Output:
A_1 A_2 A_3 A_10 B_1 B_2 B_3 B_10
0 68 59 37 76 11 69 68 17
1 19 52 23 85 37 54 93 3
Try this.
To re-order the columns based on the number after the column name
cols_fixed = df.columns[:3] # change index no based on your df
cols_variable = df.columns[3:] # change index no based on your df
cols_variable = sorted(cols_variable, key=lambda x : int(x.split('_')[1])) # split based on the number after '_'
cols_new = cols_fixed + cols_variable
new_df = pd.DataFrame(df[cols_new])
To re-arrange column names based on the string part AND number part of the column names
cols_fixed = df.columns[:3] # change index no based on your df
cols_variable = df.columns[3:] # change index no based on your df
cols_variable = sorted(cols_variable)
cols_new = cols_fixed + cols_variable
new_df = pd.DataFrame(df[cols_new])
I have a huge dataframe with product_id and their property_id's. Note that for each property starts with new index. I need to filter simultaneously by different property_id values for each product_id. Is there any way to do it fast?
out_df
product_id property_id
0 3588 1
1 3588 2
2 3588 5
3 3589 1
4 3589 3
5 3589 5
6 3590 1
7 3590 2
8 3590 5
For example want kinda that to filter for each product_id by two properties that are assigned at different rows like out_df.loc[(out_df['property_id'] == 1) & (out_df['property_id'] == 2)] but instead of it).
I need something like that but working at the same time for all rows of each product_id column.
I know that it can be done via groupby into lists
3587 [2, 1, 5]
3588 [1, 3, 5]
3590 [1, 2, 5]
and finding intersections inside lists.
gp_df.apply(lambda r: {1, 2} < (set(r['property_id'])), axis=1)
But it takes time and at the same time Pandas common filtering is greatly optimized for speed (believe in using some tricky right and inverse indexes inside what do search engines like ElasticSearch, Sphinx etc) .
Expected output: where both {1 and 2} are having.
3587 [2, 1, 5]
3590 [1, 2, 5]
Since this is just as much a performance as a functional question, I would go with an intersection approach like this:
df = pd.DataFrame({'product_id': [3588, 3588, 3588, 3589, 3589, 3589, 3590, 3590,3590],
'property_id': [1, 2, 5, 1, 3, 5, 1, 2, 5]})
df = df.set_index(['property_id'])
print("The full DataFrame:")
print(df)
start = time()
for i in range(1000):
s1 = df.loc[(1), 'product_id']
s2 = df.loc[(2), 'product_id']
s_done = pd.Series(list(set(s1).intersection(set(s2))))
print("Overlapping product_id's")
print(time()-start)
Iterating the lookup 1000 times takes 0.93 seconds on my ThinkPad T450s. I took the liberty to test #jezrael's two suggestions and they come in at 2.11 and 2.00 seconds, the groupby approach is, software engineering wise, more elegant though.
Depending on the size of your data set and the importance of performance, you can also switch to more simple datatypes, like classic dictionaries and gain further speed.
Jupyter Notebook can be found here: pandas_fast_lookup_using_intersection.ipynb
do you mean something like this?
result = out_df.loc[out_df['property_id'].isin([1,2]), :]
If you want you can then drop duplicates based on product_id...
The simpliest is use GroupBy.transform with compare sets:
s = {1, 2}
a = df[df.groupby('product_id')['property_id'].transform(lambda r: s < set(r))]
print (a)
product_id property_id
0 3588 1
1 3588 2
2 3588 5
6 3590 1
7 3590 2
8 3590 5
Another solution is filter only values of sets, removing duplicates first:
df1 = df[df['property_id'].isin(s) & ~df.duplicated(['product_id', 'property_id'])]
Then is necessary check if lengths of each group is same as length of set with this solution:
f, u = df1['product_id'].factorize()
ids = df1.loc[np.bincount(f)[f] == len(s), 'product_id'].unique()
Last filter all rows with product_id by condition:
a = df[df['product_id'].isin(ids)]
print (a)
product_id property_id
0 3588 1
1 3588 2
2 3588 5
6 3590 1
7 3590 2
8 3590 5
I am analyzing a dataset that has an Origin ID (Column A), a Destination ID (Column B), and how many trips have happened between them (Column Count). Now I want to sum the A-B trips with the B-A trips. This sum is the total number of trips between A and B.
Here is how my data looks like (it is not necessarily ordered in the same way):
In [1]: group_station = pd.DataFrame([[1, 2, 100], [2, 1, 200], [4, 6, 5] , [6, 4, 10], [1, 4, 70]], columns=['A', 'B', 'Count'])
Out[2]:
A B Count
0 1 2 100
1 2 1 200
2 4 6 5
3 6 4 10
4 1 4 70
And I want the following output:
A B C
0 1 2 300
1 4 6 15
4 1 4 70
I have tried groupby and setting the index to both variables with no success. Right now I am doing a very inefficient double loop, that is too slow for the size of my dataset.
If it helps this is the code for the double loop (I removed some efficiency modifications to make it more clear):
# group_station is the dataframe
collapsed_group_station = np.zeros(len(group_station), 3))
for i, row in enumerate(group_station.iterrows()):
start_id = row[0][0]
end_id = row[0][1]
count = row[1][0]
for check_row in group_station.iterrows():
check_start_id = check_row[0][0]
check_end_id = check_row[0][1]
check_time = check_row[1][0]
if start_id == check_end_id and end_id == check_start_id:
new_group_station[i][0] = start_id
new_group_station[i][1] = end_id
new_group_station[i][2] = time + check_time
break
I have ideas of how to make this code more efficient, but I wanted to know if there is a way of doing it without looping.
You can using np.sort with groupby.sum()
import numpy as np; import pandas as pd
group_station[['A','B']]=np.sort(group_station[['A','B']],axis=1)
group_station.groupby(['A','B'],as_index=False).Count.sum()
Out[175]:
A B Count
0 1 2 300
1 1 4 70
2 4 6 15