Function changes all column data of DF to the same value and ignores that conditions are different for different rows. And .Apply() gives Error

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Please, give me an advice about how to .apply( ) in correct way to get correct result.

I have 2 Pandas dataframes with 'a','b','c' columns. And i want to change some 'c'-column data of the second dataframe df_2. It is needed to change df_2 'c'-data for those rows only where 'a' is equal to 1. Zero 0 in those 'c'-rows has to be changed to median value counted on those 'c'-rows of the first dataframe df_1 where 'a' is 1. I wrote a function to do it. It is applied to df_2 and uses df_1 data.

The problem is:

  1. If function is applied like this: 'df_2['c'] = set_c(df_1, df_2)', all 'c'-column of df_2 gets new price, no matter is 'a' == 1 or not. It is incorrect.

  2. If function is applied like this: 'df_2['c'] = df_2.apply(set_c(df_2, df_1))', error occures: 'AssertionError:' and no additional comments.

Code is:

import pandas as pd

df_1 = pd.DataFrame({'a': [1,2,1], 'b': [4,5,6], 'c': [7,100,9]}) # From C
df_2 = pd.DataFrame({'a': [1,2,3], 'b': [4,50,6], 'c': [0,0,0]}) # To C

display('df_1', df_1)
display('df_2', df_2)

def set_c(df1, df2):
    
    mask = ( df1.loc[:, 'a'] == df2.loc[0, 'a'] )
    final_c = df1[mask]['c'].median()
    
    display("df2.loc[0, 'a']", df2.loc[0, 'a'])
    display('df1[mask]', df1[mask])
    print('final_c median', final_c)
    
    return final_c

df_2['c'] = set_c(df_1, df_2)

display(df_2)

df_2 and df_1 are global dataframes outside the function, df2 and df1 are dataframes inside the function used as function parameters.

My function shows all the process of its working. For variant 1 it shows such steps:

'df_1'
    a   b   c
0   1   4   7
1   2   5   100
2   1   6   9

'df_2'
    a   b   c
0   1   4   0
1   2   50  0
2   3   6   0

"df2.loc[0, 'a']" # 'a'=1 is a mask base for counting 'c'-median on df_1 data
1
 
'df1[mask]' # Rows of df_1 with 'a'=1 were found!
    a   b   c
0   1   4   7
2   1   6   9

final_c median 8.0 # It is median between 7 and 9 of df1

'df_2 result'
    a   b   c
0   1   4   8.0
1   2   50  8.0
2   3   6   8.0

Could you please show me how is it correct to apply this function for it to give such a result of df_2 with new 'c'=8.0 **in row [0], but in row[1] - 100 ('a'=2), in row[2] - 0 (no 'a'=3 in df_1):

'df_2 result'
    a   b   c
0   1   4   8.0
1   2   50  100.0
2   3   6   0.0

Thank you very much!

pandas apply dataframe function python
1 Answers

If I got your meaning correctly, I suggest using the apply method with axis=1 and transferring to the function the index row, then using it as an index for the loc.

import pandas as pd

df_1 = pd.DataFrame({'a': [1, 2, 1], 'b': [4, 5, 6], 'c': [7, 100, 9]})  # From C
df_2 = pd.DataFrame({'a': [1, 2, 3], 'b': [4, 50, 6], 'c': [0, 0, 0]})  # To C


def set_c(df1, df2, i):
    mask = (df1.loc[:, 'a'] == df2.loc[i, 'a'])
    final_c = df1[mask]['c'].median()

    return final_c


df_2['c'] = df_1.apply(lambda x: set_c(df_1, df_2, x.name), axis=1).fillna(0)
print(df_2)

OUTPUT:

   a   b      c
0  1   4    8.0
1  2  50  100.0
2  3   6    0.0
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