python - 用 Pandas 替换数据框中不同列的值

Question

我有一个数据帧df_in定义为:

import pandas as pd

dic_in = {'A': ['ff','rr' ,'nn' ,'qq','tt' ,'pp','uu'],
      'B1': ['33',r'\N','39' ,'22',r'\N','56','90'],
      'C1': ['44',r'\N','74' ,'34',r'\N','89','99'],
      'B2': ['33','63' ,r'\N','22','71' ,'56','90'],
      'C2': ['44','85' ,r'\N','34','52' ,'89','99']}
df_in = pd.DataFrame(dic_in,columns=['A','B1','C1','B2','C2'])

如果我在控制台上打印它，它看起来像这样:

In [28]:df_in
Out[28]: 
    A  B1  C1  B2  C2
0  ff  33  44  33  44
1  rr  \N  \N  63  85
2  nn  39  74  \N  \N
3  qq  22  34  22  34
4  tt  \N  \N  71  52
5  pp  56  89  56  89
6  uu  90  99  90  99

我想做的是调查列 B1 和 C1 的每一行:如果通用行包含 \N在两列中，它需要分别用 B2 和 C2 的内容替换其值。这样，输出 (df_out) 应如下所示:

In [28]:df_in                In[30]:df_out
Out[28]:                     Out[30]:
    A  B1  C1  B2  C2            A   B   C
0  ff  33  44  33  44        0  ff  33  44
1  rr  \N  \N  63  85 -----> 1  rr  63  85
2  nn  39  74  \N  \N -----> 2  nn  39  74
3  qq  22  34  22  34        3  qq  22  34
4  tt  \N  \N  71  52 -----> 4  tt  71  52
5  pp  56  89  56  89        5  pp  56  89
6  uu  90  99  90  99        6  uu  90  99

我能够使用这些代码行实现我的目标:

df_out = pd.DataFrame()
for index, row in df_in.iterrows():
    if row['B1']!=r'\N' and row['C1']!=r'\N':
        dic = {'A': [row['A']], 'B': [row['B1']], 'C': [row['C1']]}
        df_out = pd.concat([df_out,pd.DataFrame(dic)], ignore_index=True)
    else:
        dic = {'A': [row['A']], 'B': [row['B2']], 'C': [row['C2']]}
        df_out = pd.concat([df_out,pd.DataFrame(dic)], ignore_index=True)

你能建议我一个聪明的方法来实现这样的结果吗？

Answer 1

你可以先replace \N 到 NaN 然后 combine_first或fillna :

df_out = df_in.replace({'\\N': np.nan})
df_out['B']= df_out.B1.combine_first(df_out.B2)
df_out['C'] = df_out.C1.combine_first(df_out.C2)
df_out = df_out[['A','B','C']]
print (df_out)
    A   B   C
0  ff  33  44
1  rr  63  85
2  nn  39  74
3  qq  22  34
4  tt  71  52
5  pp  56  89
6  uu  90  99

如果需要按子集 B1 和 C1 将值添加到 B2 和 C2:

df_out = df_in.replace({'\\N': np.nan})
df_out[['B', 'C']] = df_out[['B1', 'C1']].fillna(df_out[['B2', 'C2']]
                                         .rename(columns={'B2':'B1','C2':'C1'}))
df_out = df_out[['A','B','C']]
print (df_out)
    A   B   C
0  ff  33  44
1  rr  63  85
2  nn  39  74
3  qq  22  34
4  tt  71  52
5  pp  56  89
6  uu  90  99