从每个客户 ID 中识别零值，然后从前一行下一行 Pandas 中获取值

Question

Input:

df = pd.DataFrame([[101, 1, 'reg'],
               [101, 1, '1098'],
               [101, 0, 'Reg'],
               [102, 1, 'Paymode'],
               [102, 0, 'Reg'],
               [103, 1, 'reg'],
               [103, 0.0, 'reg'],
               [103, 0.0, 'reg']
              ]
              , columns=['cus_ID', 'Paperlessmode', 'types of paper'])

output:

df=pd.DataFrame([[101, 1, 'reg','1098'],
               [101, 1, '1098','1098'],
               [101, 0, 'Reg','1098'],
               [102, 1, 'Paymode','Paymode'],
               [102, 0, 'Reg','Paymode'],
               [103, 1, 'reg','reg'],
               [103, 0.0, 'reg','reg'],
               [103, 0.0, 'reg','reg']
              ]
              , columns=['cus_ID', 'Paperlessmode', 'types of paper','last occurance_paper'])

I want to identify the types of paper which is presence before zero in Paperlessmode for each customer id in Python 3.6

Answer 1

You can use Series.map by shifted values with cumulative sum and compared by 1 :

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)
   cus_ID  Paperlessmode types of paper last occurance_paper
0     101            1.0            reg                 1098
1     101            1.0           1098                 1098
2     101            0.0            Reg                 1098
3     102            1.0        Paymode              Paymode
4     102            0.0            Reg              Paymode
5     103            1.0            reg                  reg
6     103            0.0            reg                  reg
7     103            0.0            reg                  reg

Alternative:

d = df[df['Paperlessmode'].eq(0).groupby(df['cus_ID']).shift(-1, fill_value=False)].set_index('cus_ID')['types of paper'].to_dict()
df['last occurance_paper'] = df['cus_ID'].map(d)