pandas groupby and find max no. of consecutive occurences of 1s in dataframe

Question

I have a dataframe with Ids and a target value per each Id. my goal is to find the max no. of consecutive occurrences of 1's in each group of the data frame.

original dataframe is as below

df = pd.DataFrame({
'Id': {0: 1,1: 1,2: 1,3: 2,4: 2,5: 2,6: 2,7: 2, 8: 2,9: 2, 10: 2, 11: 2, 12: 2},
't_year': {0: 2021, 1: 2022, 2: 2022, 3: 2021,4: 2021,5: 2021,6: 2021,7: 2021,8: 2021,9: 2021,10: 2021,11: 2022,12: 2022},
 't_month_prx': {0: 10.0,1: 1.0,2: 2.0,3: 1.0,4: 2.0,5: 3.0,6: 6.0,7: 7.0,
 8: 8.0,9: 9.0,10: 10.0,11: 1.0,12: 2.0},
'values': {0: 1.0, 1: 1.0, 2: 1.0,3: 1.0,4: 1.0,5: 1.0,6: 3.0,7: 1.0,8: 1.0,9: 1.0,10: 1.0,11: 1.0, 12: 1.0}
})

and my desired output is

pd.DataFrame({
'Id': {0: 1,1: 2},
'counts': {0: 3.0,1: 6.0}
})

Answer 1

You could create a mask with a unique value for each consecutive group of numbers (cumsum + ne/!==), and then groupby that and the ID, sum the numbers, and get the the max:

df.groupby([df['Id'], df['values'].ne(df.groupby('Id')['values'].shift(1)).cumsum()])['values'].sum().groupby(level=0).max().reset_index()

Output:

>>> df
   Id  values
0   1     3.0
1   2     6.0

Answer 2

You can first group consecutive values in the dataframe, then filter the dataframe where values contain 1 and then groupby to proceed further.

 df['tmp'] = (df['values'] != df['values'].shift()).cumsum() #Group Consecutive values in Dataframe
 df = df[df['values'].eq(1)].groupby(['Id', 'tmp'], as_index=False).count().groupby('Id', as_index=False)['values'].max()

OUTPUT After First Line

    Id  values  tmp
0    1     1.0    1
1    1     1.0    1
2    1     1.0    1
3    2     1.0    1
4    2     1.0    1
5    2     1.0    1
6    2     3.0    2
7    2     1.0    3
8    2     1.0    3
9    2     1.0    3
10   2     1.0    3
11   2     1.0    3
12   2     1.0    3

Final OUTPUT

   Id  values
0   1       3
1   2       6