错误: pandas.core.indexing.IndexingError:太多索引器

Question

我试图在不同的集合中匹配多个列，并用所有不匹配的列名更新另一列，

用不匹配的列名更新结果列

输入：

  A  B  C  D  E
0  f  e  b  a  d
1  c  b  a  c  b
2  f  f  a  b  c
3  d  c  c  d  c
4  f  b  b  b  e
5  b  a  f  c  d

预期输出

   A  B  C  D  E           MATCHES
0  f  e  b  a  d  AD, BC Unmatched
1  c  b  a  c  b      BC Unmatched
2  f  f  a  b  c  AD, BC Unmatched
3  d  c  c  d  c       ALL MATCHED
4  f  b  b  b  e      AD Unmatched
5  b  a  f  c  d  AD, BC Unmatched

下面的代码在函数内部使用时会出现错误，否则，如果我单独使用而不使用任何函数，它的工作情况很好。

def test(x):
    try:
       for idx in df.index:
           unmatch_list = []
           if not df.loc[idx, 'A'] == df.loc[idx, 'D']:
              unmatch_list.append('AD')
           if not df.loc[idx, 'B'] == df.loc[idx, 'C']:
              unmatch_list.append('BC')
           # etcetera...
           if len(unmatch_list):
              unmatch_string = ', '.join(unmatch_list) + ' Unmatched'
           else:
              unmatch_string = 'ALL MATCHED'
           df.loc[idx, 'MATCHES'] = unmatch_string
 
     except ValueError:

它在尝试处理时会产生错误：

if not df.loc[idx, 'A'] == df.loc[idx, 'D']:
Error: pandas.core.indexing.IndexingError: Too many indexers

需要建议：

Answer 1

怎么叫函数？

对于我来说，如果添加return df并将DataFrame传递给函数，那么我就可以工作了：

def test(x):
    try:
       for idx in df.index:
           unmatch_list = []
           if not df.loc[idx, 'A'] == df.loc[idx, 'D']:
              unmatch_list.append('AD')
           if not df.loc[idx, 'B'] == df.loc[idx, 'C']:
              unmatch_list.append('BC')
           # etcetera...
           if len(unmatch_list):
              unmatch_string = ', '.join(unmatch_list) + ' Unmatched'
           else:
              unmatch_string = 'ALL MATCHED'
           df.loc[idx, 'MATCHES'] = unmatch_string
          
 
    except ValueError:
         print ('error')
    return df

df = test(df)
print (df)
   A  B  C  D  E           MATCHES
0  f  e  b  a  d  AD, BC Unmatched
1  c  b  a  c  b      BC Unmatched
2  f  f  a  b  c  AD, BC Unmatched
3  d  c  c  d  c       ALL MATCHED
4  f  b  b  b  e      AD Unmatched
5  b  a  f  c  d  AD, BC Unmatched

使用apply解决方案是可能的，但必要的更改功能如下：

def test(x):
    try:
        unmatch_list = []
        if not x['A'] == x['D']:
           unmatch_list.append('AD')
        if not x['B'] == x['C']:
           unmatch_list.append('BC')
        # etcetera...
        if len(unmatch_list):
           unmatch_string = ', '.join(unmatch_list) + ' Unmatched'
        else:
           unmatch_string = 'ALL MATCHED'
 
    except ValueError:
         print ('error')
    return unmatch_string

df['MATCHES'] = df.apply(test, axis=1)
print (df)
   A  B  C  D  E           MATCHES
0  f  e  b  a  d  AD, BC Unmatched
1  c  b  a  c  b      BC Unmatched
2  f  f  a  b  c  AD, BC Unmatched
3  d  c  c  d  c       ALL MATCHED
4  f  b  b  b  e      AD Unmatched
5  b  a  f  c  d  AD, BC Unmatched