在python中对分组行执行操作

Question

我有一个数据帧,其中pic_code值可能重复.如果重复,我想为最接近其mpe_wgt的pic_code设置变量"keep"为"t".

例如,第二pic_code将"keep"设置为t,因为它具有最接近其对应的"mpe_weight"的"权重".我的代码导致"保持"为所有人保持'f'并且"差异"为所有人保持"100".

df['keep']='f'
df['diff']=100

def cln_df(data):
    if pd.unique(data['mpe_wgt']).shape==(1,):
        data['keep'][0:1]='t'
    elif pd.unique(data['mpe_wgt']).shape!=(1,): 
        data['diff']=abs(data['weight']-(data['mpe_wgt']/100))
        data['keep'][data['diff']==min(data['diff'])]='t'
    return data

df=df.groupby('pic_code').apply(cln_df)

df之前

  pic_code      weight      mpe_wgt    keep    diff
  1234          45          34         f       100
  1234          32          23         f       100
  45344         54          35         f       100
  234           76          98         f       100
  234           65          12         f       100

df输出应该是

  pic_code      weight      mpe_wgt    keep    diff
  1234          45          34         f       11
  1234          32          23         t       9
  45344         54          35         t       100
  234           76          98         t       22
  234           65          12         f       53

我是python的新手,所以请保持解决方案尽可能简单.我真的想让我的方法工作所以请不要太花哨.在此先感谢您的帮助.

Answer 1

这是一种方式.注意我使用布尔值True/ False代替字符串"t"和"f".这只是一种很好的做法.

请注意,以下所有操作都是矢量化的,而groupby.apply使用自定义功能肯定不是.

建立

print(df)

   pic_code  weight  mpe_wgt
0      1234      45       34
1      1234      32       23
2     45344      54       35
3       234      76       98
4       234      65       12

解

# calculate difference
df['diff'] = (df['weight'] - df['mpe_wgt']).abs()

# sort by pic_code, then by diff
df = df.sort_values(['pic_code', 'diff'])

# define keep column as True only for non-duplicates by pic_code
df['keep'] = ~df.duplicated('pic_code')

结果

print(df)

   pic_code  weight  mpe_wgt  diff   keep
3       234      76       98    22   True
4       234      65       12    53  False
1      1234      32       23     9   True
0      1234      45       34    11  False
2     45344      54       35    19   True