how can I keep the first entry by ID? pandas python

user_13491046 0
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pk_cid means client id Each pk_partition represents the current state of the month of em_acount(which is a product sold). So if a client has 1 in january and also has 1 in February that means that he had only buy once in January. I would like to change the rest of the months to have a 0 if the previous month has a 1 to see exacly the time where people buy only keeping the first one.

if its possible on the other hand if a person left to have a -1

This is what i have

you can create a pandas.Series:

a = pd.Series({(15891, '2018-07-28'): 1,
 (15891, '2018-08-28'): 0,
 (16063, '2018-11-28'): 0,
 (16063, '2018-12-28'): 0,
 (16063, '2019-01-28'): 0,
 (16063, '2019-02-28'): 0,
 (16063, '2019-03-28'): 0,
 (16063, '2019-04-28'): 0,
 (16063, '2019-05-28'): 0,
 (16203, '2018-12-28'): 0,
 (16203, '2019-01-28'): 1,
 (16203, '2019-02-28'): 1,
 (16203, '2019-03-28'): 1,
 (16203, '2019-04-28'): 1,
 (16203, '2019-05-28'): 1,
 (16502, '2018-09-28'): 0,
 (16502, '2018-10-28'): 1,
 (16502, '2018-11-28'): 1,
 (16502, '2018-12-28'): 1,
 (16502, '2019-01-28'): 1})

Desired output

enter image description here

Thanks

pandas dataframe diff group-by python
2 Answers

I think this logic is correct, but it does assume that your dates are ordered in your index.

s = pd.Series( {(15891, '2018-07-28'): 1, (15891, '2018-08-28'): 0,
                (16063, '2018-11-28'): 0, (16063, '2018-12-28'): 0,
                (16063, '2019-01-28'): 0, (16063, '2019-02-28'): 0, 
                (16063, '2019-03-28'): 0, (16063, '2019-04-28'): 0, 
                (16063, '2019-05-28'): 0, (16203, '2018-12-28'): 0, 
                (16203, '2019-01-28'): 1, (16203, '2019-02-28'): 1, 
                (16203, '2019-03-28'): 1, (16203, '2019-04-28'): 1, 
                (16203, '2019-05-28'): 1, (16502, '2018-09-28'): 0, 
                (16502, '2018-10-28'): 1, (16502, '2018-11-28'): 1, 
                (16502, '2018-12-28'): 1, (16502, '2019-01-28'): 1})
# groupby level 0
g = s.groupby(level=0)
# create a mask based on your logic
mask = ((g.shift(0)==1) & (g.shift(-1)==1) & (g.shift()==1)) | \
       ((g.shift(0)==1) & (g.shift(-1).isna()) & (g.shift()==1)) | \
       ((g.shift(0)==1) & (g.shift(-1)==0) & (g.shift()==1))
# assign all true values to 0
s[mask] = 0
print(s)

15891  2018-07-28    1
       2018-08-28    0
16063  2018-11-28    0
       2018-12-28    0
       2019-01-28    0
       2019-02-28    0
       2019-03-28    0
       2019-04-28    0
       2019-05-28    0
16203  2018-12-28    0
       2019-01-28    1
       2019-02-28    0
       2019-03-28    0
       2019-04-28    0
       2019-05-28    0
16502  2018-09-28    0
       2018-10-28    1
       2018-11-28    0
       2018-12-28    0
       2019-01-28    0
user_91778770

Sort the index, then groupby user_id get pct_change and set the values to 0 if the change is <= 0:

serie = pd.Series( {(15891, '2018-07-28'): 1, (15891, '2018-08-28'): 0,
                    (16063, '2018-11-28'): 0, (16063, '2018-12-28'): 0,
                    (16063, '2019-01-28'): 0, (16063, '2019-02-28'): 0, 
                    (16063, '2019-03-28'): 0, (16063, '2019-04-28'): 0, 
                    (16063, '2019-05-28'): 0, (16203, '2018-12-28'): 0, 
                    (16203, '2019-01-28'): 1, (16203, '2019-02-28'): 1, 
                    (16203, '2019-03-28'): 1, (16203, '2019-04-28'): 1, 
                    (16203, '2019-05-28'): 1, (16502, '2018-09-28'): 0, 
                    (16502, '2018-10-28'): 1, (16502, '2018-11-28'): 1, 
                    (16502, '2018-12-28'): 1, (16502, '2019-01-28'): 1})

serie.sort_index(inplace=True)
serie [serie.groupby(level=0).pct_change() <= 0 ] = 0
serie

# 15891 2018-07-28 1
#       2018-08-28 0
# 16063 2018-11-28 0
#       2018-12-28 0
#       2019-01-28 0
#       2019-02-28 0
#       2019-03-28 0
#       2019-04-28 0
#       2019-05-28 0
# 16203 2018-12-28 0
#       2019-01-28 1
#       2019-02-28 0
#       2019-03-28 0
#       2019-04-28 0
#       2019-05-28 0
# 16502 2018-09-28 0
#       2018-10-28 1
#       2018-11-28 0
#       2018-12-28 0
#       2019-01-28 0
user_66957930
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