从每个客户 ID 的下一列中识别最后一个是值和 0，然后从前一行下一列熊猫中获取值

Question

Input:

df = pd.DataFrame([[101,'y', 1, 'reg'],
                [101,'y', 1, '1098'],
                [101, 'y',0, 'Reg'],
                [101, 'y',0, 'sed'],
                [101,'n',0,'VA'],
                [102, 'y',1, 'Paymode'],
                [102, 'y',0, 'roy'],
                [102, 'n',0, 'Reg'],
                [103, 'y',1, 'reg'],
                [103, 'n',0, 'PCA'],
                [103, 'n',0, 'FXD']
              ]
              , columns=['cus_ID', 'emailflg','Paperlessmode', 'types of paper'])

output:

df= pd.DataFrame([[101,'y', 1, 'reg','1098'],
                [101,'y', 1, '1098','1098'],
                [101, 'y',0, 'Reg','1098'],
                [101, 'y',0, 'sed','1098'],
                [101,'n',0,'VA','1098'],
                [102, 'y',1, 'Paymode','roy'],
                [102, 'y',0, 'roy','roy'],
                [102, 'n',0, 'Reg','roy'],
                [103, 'y',1, 'reg','reg'],
                [103, 'n',0, 'PCA','reg'],
                [103, 'n',0, 'FXD','reg']
              ]
              , columns=['cus_ID','emailflg', 'Paperlessmode', 'types of paper','last occurance_paper'])

condtion 1= df['Emailflg' =='y'] & df['Paperlessmode'].eq(0) & df['types of paper!='Reg','PCA']

if Condition 1 exits then take same row value from type of paper
OR
identify the types of paper which is presence before zero in Paperlessmode and 'n' in Email flag for each customer id in Python 3.6
Kindly let me know if any concern and plz don't devote

Answer 1

I don't really understand how your conditions are done, but for the solution that you should follow, it doesn't rally matter. I have listed several pseudo-conditions which you can adapt to your own real conditions. The idea is to list the conditions and choices for what the values of the new column should be. In this example, 4 conditions, 4 potential values. Then use np.select in the following way:

conditions = [((df['emailflg'] =='y') & (df['Paperlessmode']==0)), 
               ((df['types of paper'] =='reg') & (df['Paperlessmode']==0)),
             ((df['types of paper'] =='Reg') & (df['Paperlessmode']==0)),
              ((df['types of paper'] =='VA') & (df['Paperlessmode']==0))]
choices = ['REG','reg','Paymode', 'VA']


import numpy as np

df['last occurance_paper'] = np.select(conditions, choices, default='XXX')

which gives

   cus_ID emailflg  Paperlessmode types of paper last occurance_paper
0     101        y              1            reg                  XXX
1     101        y              1           1098                  XXX
2     101        y              0            Reg                  REG
3     101        n              0             VA                   VA
4     102        y              1        Paymode                  XXX
5     102        n              0            Reg              Paymode
6     103        y              1            reg                  XXX
7     103        n              0            PCA                  XXX
8     103        n              0            FXD                  XXX

Answer 2

I think you need:

m1 = df['Paperlessmode'].eq(0)
m2 = df['emailflg'].eq('y')
m3 = df['emailflg'].eq('n') 
 
d = df[m1 & m2].set_index('cus_ID')['types of paper'].to_dict()
s = df[(m1 & m3).groupby(df['cus_ID']).transform(lambda x: x.shift(-1).cumsum().eq(1))].set_index('cus_ID')['types of paper']

df['last occurance_paper'] = df['cus_ID'].map(d).fillna(df['cus_ID'].map(s))
print (df)
       cus_ID emailflg  Paperlessmode types of paper last occurance_paper
0     101        y              1            reg                  Reg
1     101        y              1           1098                  Reg
2     101        y              0            Reg                  Reg
3     101        n              0             VA                  Reg
4     102        y              1        Paymode              Paymode
5     102        n              0            Reg              Paymode
6     103        y              1            reg                  reg
7     103        n              0            PCA                  reg
8     103        n              0            FXD                  reg

Answer 3

s = df[df['Paperlessmode'].eq(0).groupby(df['cus_ID']).transform(lambda x: x.shift(-1).cumsum().eq(1))].set_index('cus_ID')['types of paper']
df['last occurance_paper'] = df['cus_ID'].map(s)
print (df)