Python Pandas 根据多列条件替换值

Question

我有一个数据框

data_in = {'A':['A1', '', '', 'A4',''],
        'B':['', 'B2', 'B3', '',''],
        'C':['C1','C2','','','C5']}
df_in = pd.DataFrame(data)

print(df_in)

    A   B   C
0  A1      C1
1      B2  C2
2      B3    
3  A4        
4          C5

如果 C 列不为空且 A 或 B 不为空，我将尝试替换 A 或 B 列。 替换后，我需要清除C列中的值。

我期望这个输出

    A   B   C
0   C1      
1       C2  
2       B3  
3   A4      
4           C5

我尝试了几种方法，最接近的是

df_in['A'] = np.where(
   (df_in['A'] !='') & (df_in['C'] != '') , df_in['A'], df_in['C']
   )

df_in['B'] = np.where(
   (df_in['B'] !='') & (df_in['C'] != '') , df_in['B'], df_in['C']
   )

但这也清楚了其他值，我失去了 A4 和 B3，并且我不清除 C1 和 C2

我得到了什么

    A   B   C
0   C1      C1
1       C2  C2
2           
3           
4           C5

谢谢

Answer 1

你已经非常接近了，但是你在np.where中切换了参数，语法是np.where(cond, if_cond_True, if_cond_False)。如果满足条件 (if_cond_True)，列 A 和 B 应该具有列的值，否则它们将保留其原始值 (if_cond_False)。

import pandas as pd
import numpy as np 

data_in = {'A':['A1', '', '', 'A4',''],
        'B':['', 'B2', 'B3', '',''],
        'C':['C1','C2','','','C5']}

df_in = pd.DataFrame(data_in)

maskA = df_in['A'] != ''   # A not empty
maskB = df_in['B'] != ''   # B not empty
maskC = df_in['C'] != ''   # C not empty

# If the column havs NaNs instead of '' then use : 
#
# maskA = df_in['A'].notnull()   # A not empty
# maskB = df_in['B'].notnull()   # B not empty
# maskC = df_in['C'].notnull()   # C not empty

# If A and C are not empty, A = C, else A keep its value 
df_in['A'] = np.where(maskA & maskC, df_in['C'], df_in['A'])

# If B and C are not empty, B = C, else B keep its value
df_in['B'] = np.where(maskB & maskC, df_in['C'], df_in['B'])

# If (A and C are not empty) or (B and C are not empty),
# C should be empty, else C keep its value
df_in['C'] = np.where((maskA & maskC) | (maskB & maskC), "", df_in['C'])

输出

>>> df_in 

    A   B   C
0  C1        
1      C2    
2      B3    
3  A4        
4          C5