根据pandas中另一列的值对列执行操作

让我们尝试布尔掩蔽：

# select the columns
c = pd.Index(['col1','col2','col3'])

# create boolean mask
m = df[c].eq('').to_numpy()

# mask the values in `_num` and `_deno` like columns
df[c + '_num'] = df[c + '_num'].mask(m,'')
df[c + '_deno'] = df[c + '_deno'].mask(m,'')

>>> df

  id  date col1 col2 col3 col1_num col1_deno col3_num col3_deno col2_num col2_deno
0  A     1   98   88           567       453                         323       756
1  B     1   99                231       232                                      
2  C     1   97   67   23      543       458      456       876      935       876
3  B     1        79   84                         678       452      545       453
4  A     1   45        58      334       778      234       983                   
5  C     1   23   55           183       565                         234       234

@shubham 的回答简单明了，我相信也更快；这只是一个选项，您可能无法（或想要）列出所有列

获取需要更改的列列表：

cols = [col for col in df if col.startswith('col')]

['col1','col3','col1_num','col1_deno','col3_num','col3_deno','col2_num','col2_deno']

创建一个字典，将 col1 与要更改的列配对，对于 col2 等也是如此：

from collections import defaultdict
d = defaultdict(list)

for col in cols:
    if "_" in col:
        d[col.split("_")[0]].append(col)

d

defaultdict(list,{'col1': ['col1_num','col1_deno'],'col3': ['col3_num','col3_deno'],'col2': ['col2_num','col2_deno']})

遍历字典以分配新值：

for key,val in d.items():
    df.loc[df[key].eq(""),val] = ""




 id  date col1 col2 col3 col1_num col1_deno col3_num col3_deno col2_num col2_deno
0  A     1   98   88           567       453                         323       756
1  B     1   99                231       232                                      
2  C     1   97   67   23      543       458      456       876      935       876
3  B     1        79   84                         678       452      545       453
4  A     1   45        58      334       778      234       983                   
5  C     1   23   55           183       565                         234       234

MultiIndex 的解决方案：

#first convert not processing and testing columns to index
df1 = df.set_index(['id','date'])
cols = df1.columns
#split columns by _ for MultiIndex
df1.columns = df1.columns.str.split('_',expand=True)

#compare columns without _ (with NaN in second level) by empty string
m = df1.xs(np.nan,axis=1,level=1).eq('')
#create mask by all columns
mask = m.reindex(df1.columns,level=0)
#set new values by mask,overwrite columns names
df1 = df1.mask(mask,'').set_axis(cols,axis=1).reset_index()
print (df1)
  id  date col1 col2 col3 col1_num col1_deno col3_num col3_deno col2_num  \
0  A     1   98   88           567       453                         323   
1  B     1   99                231       232                               
2  C     1   97   67   23      543       458      456       876      935   
3  B     1        79   84                         678       452      545   
4  A     1   45        58      334       778      234       983            
5  C     1   23   55           183       565                         234   

  col2_deno  
0       756  
1            
2       876  
3       453  
4            
5       234