从下面的单元格中减去两个字符串，并分别为熊猫索引或用户ID。

Question

输入：

wdf = pd.DataFrame([[101, 1, 'reg'],
                    [101,0,'VA,reg,pca'],
                    [102,1, 'Paymode,reg,va,fa'],
                    [102, 0, 'roy,pca'],
                    [103, 1, 'reg,pca,va,GA'],
                    [103, 0, 'PCA,reg,va,RF']
                  ]
                  , columns=['cus_ID','Paperlessmode', 'types of paper']) 

产出：

out_df = pd.DataFrame([[101, 1, 'reg','VA,pca'],
                [101,0,'VA,reg,pca','VA,pca'],
                [102,1, 'Paymode,reg,va,fa','roy,pca'],
                [102, 0, 'roy,pca','roy,pca'],
                [103, 1, 'reg,pca,va,GA','RF'],
                [103, 0, 'PCA,reg,va,RF','RF']
              ]
              , columns=['cus_ID','Paperlessmode', 'types of paper','result'])

我期待着这样的事情：

wdf的结果‘=wdf’无纸模式‘== 1 (-) wdf’‘==0，映射为(cus_ID)

我想识别额外类型的纸张，当与Python3.6中每个客户id在无纸化模式下的1相比时，这种纸张在无纸化模式中为零。

如果有任何关心，请告诉我

Answer 1

数据：

wdf = pd.DataFrame([[101, 1, 'reg'],
                    [101,0,'VA,reg,pca'],
                    [102,1, 'Paymode,reg,va,fa'],
                    [102, 0, 'roy,pca'],
                    [103, 1, 'reg,pca,va,GA'],
                    [103, 0, 'PCA,reg,va,RF']
                  ]
                  , columns=['cus_ID','Paperlessmode', 'types of paper']) 

wdf：

   cus_ID  Paperlessmode     types of paper
0     101              1                reg
1     101              0         VA,reg,pca
2     102              1  Paymode,reg,va,fa
3     102              0            roy,pca
4     103              1      reg,pca,va,GA
5     103              0      PCA,reg,va,RF

期望产出：

out_df = pd.DataFrame([[101, 1, 'reg','VA,pca'],
                [101,0,'VA,reg,pca','VA,pca'],
                [102,1, 'Paymode,reg,va,fa','roy,pca'],
                [102, 0, 'roy,pca','roy,pca'],
                [103, 1, 'reg,pca,va,GA','RF'],
                [103, 0, 'PCA,reg,va,RF','RF']
              ]
              , columns=['cus_ID','Paperlessmode', 'types of paper','result'])

out_df：

   cus_ID  Paperlessmode     types of paper   result
0     101              1                reg   VA,pca
1     101              0         VA,reg,pca   VA,pca
2     102              1  Paymode,reg,va,fa  roy,pca
3     102              0            roy,pca  roy,pca
4     103              1      reg,pca,va,GA       RF
5     103              0      PCA,reg,va,RF       RF

dic = {}
for i in [0,1]:
    f = wdf.groupby(["cus_ID", "Paperlessmode"]).apply(lambda x:x.to_dict()).unstack()[i].apply(lambda x:x['types of paper']).to_list()
    dic[i] = [list(i.values())[0] for i in f]

L = []
for f_0, f_1 in zip(dic.get(0), dic.get(1)):
    i_0 = f_0.split(",")
    i_1 = f_1.split(",")
    L += [','.join([i for i in i_0 if not i.lower() in [ii.lower() for ii in i_1]])]*2
    
wdf['result'] = L

wdf：

   cus_ID  Paperlessmode     types of paper   result
0     101              1                reg   VA,pca
1     101              0         VA,reg,pca   VA,pca
2     102              1  Paymode,reg,va,fa  roy,pca
3     102              0            roy,pca  roy,pca
4     103              1      reg,pca,va,GA       RF
5     103              0      PCA,reg,va,RF       RF