group by rank continuous date by pandas

Question

I refer this post . But My goal is something different.

Example

ID    TIME
01    2018-07-11
01    2018-07-12
01    2018-07-13
01    2018-0         


        

Answer 1

Use DataFrameGroupBy.diff for difference per groups of TIME column, compare days for not equal 1 and create groups by cumulative sums, last pass to GroupBy.cumcount:

df['TIME'] = pd.to_datetime(df['TIME'])

new = df.groupby('ID', group_keys=False)['TIME'].diff().dt.days.ne(1).cumsum()
df['rank'] = df.groupby(['ID',new]).cumcount().add(1)
print (df)
   ID       TIME  rank
0   1 2018-07-11     1
1   1 2018-07-12     2
2   1 2018-07-13     3
3   1 2018-07-15     1
4   1 2018-07-16     2
5   1 2018-07-17     3
6   2 2019-09-11     1
7   2 2019-09-12     2
8   2 2019-09-15     1
9   2 2019-09-16     2

Answer 2

First we check the difference between the dates, which are > 1 day. Then we groupby on ID and the cumsum of these differences and cumulative count each group`

# df['TIME'] = pd.to_datetime(df['TIME'])
s = df['TIME'].diff().fillna(pd.Timedelta(days=1)).ne(pd.Timedelta(days=1))
df['RANK'] = s.groupby([df['ID'], s.cumsum()]).cumcount().add(1)


   ID       TIME  RANK
0   1 2018-07-11     1
1   1 2018-07-12     2
2   1 2018-07-13     3
3   1 2018-07-15     1
4   1 2018-07-16     2
5   1 2018-07-17     3
6   2 2019-09-11     1
7   2 2019-09-12     2
8   2 2019-09-15     1
9   2 2019-09-16     2