使用熊猫的条件数据框操作

Question

I am doing some calculations on a dataFrame A : I want to add a new column RESULT , and do the following calculation:

There usually are multiple rows with the same key1 values and their key2 can be either X or Y . For each group having the same key1 : If key2 = X , then RESULT = 0 , else, RESULT = (C1 | key2= Y)+ (C2| key2= Y)+ (C2| key2= X) . See A_MODIFIED .

    A =
        key1   key2  C1    C2    
    0   A      X     5     2     
    1   A      Y     3     2     
    2   B      X     6     1     
    3   B      Y     1     3     
    4   C      Y     1     4     
    5   D      X     2     3     
    6   D      Y     1     3     

   A_MODIFIED =
       key1   key2  C1    C2    RESULT
   0   A      X     5     2     0
   1   A      Y     3     2     7
   2   B      X     6     1     0
   3   B      Y     1     3     5
   4   C      Y     1     4     5
   5   D      X     2     3     0
   6   D      Y     1     3     7

This is what I did:

import pandas as pd
import numpy as np

df1 = pd.DataFrame(A.groupby('key1', sort = False).sum().ix[:, ['C2']].sum(axis=1), columns=['C2_T']).reset_index(level=1)
df2 = A[A['key2'] == 'Y']
df3 = pd.merge(df1, df2, how = 'left').set_index(df1.index)
df3.RESULT = df3.C1+ df3.C2_T

But now I don't know how to merge it with the original A .

Answer 1

You can apply function f for each group.

Function f sum all values of column C2 , because there not depends on value of key2 . Values of C1 depends on key2 , so there are selected only value with df['key2'] == 'Y' .

Last if df['key2'] == 'X' output is set to 0 .

print A
#  key1 key2  C1  C2
#0    A    X   5   2
#1    A    Y   3   2
#2    B    X   6   1
#3    B    Y   1   3
#4    C    Y   1   4
#5    D    X   2   3
#6    D    Y   1   3

def f(df):
    df['RESULT'] = df['C2'].sum() + df['C1'].loc[df['key2'] == 'Y'].sum()
    df['RESULT'].loc[df['key2'] == 'X'] = 0
    return df

df = A.groupby('key1', sort = False).apply(f)
print df
#  key1 key2  C1  C2  RESULT
#0    A    X   5   2       0
#1    A    Y   3   2       7
#2    B    X   6   1       0
#3    B    Y   1   3       5
#4    C    Y   1   4       5
#5    D    X   2   3       0
#6    D    Y   1   3       7