[FIXED] groupby streak of numbers and one row after it then check the first value of a column for each group and create a new column

Issue

This is my dataframe:

df = pd.DataFrame(
    {
        'a': [0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0],
        'b': [1, -1, -1, -1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1]    
    }
)

And this is the output that I want. I want to create column c:

   a  b  c
0   0  1  0
1   0 -1  0
2   1 -1  0
3   1 -1  0
4   1  1  1
5   1 -1  1
6   0  1  0
7   0  1  0
8   0 -1  0
9   1  1  1
10  1  1  1
11  1 -1  1
12  0  1  0
13  0  1  0
14  1 -1  0
15  1 -1  1
16  0  1  0

This is basically an extension to this post. The highlighted rows below summarize the way that this needs to be done.

First of all in column a, groups are created by streak of 1s and one row after the streak ends. The highlighted rows in column a are these groups. The solution for this step is here.

Now what I need is to check column b for each group in a. Find the first value that is 1 in b for each group. And then any value that comes before this becomes 0. This is how column c is created.

For example for first group in a, the first value that column b is 1 is row number 4. The previous values in that group becomes 0. And the result is the first highlighted group in column c.

Note that if for a group, all the values in b are NOT 1, the corresponding group in c will become all 0s.

This is what I have tried but I can't find the complete solution:

g = df.loc[::-1, 'a'].eq(0).cumsum()
x = df.groupby(g).filter(lambda x: x.b.iloc[0] == 1)

Solution

A variation of the linked answer using groupby.cummax on df['b'].eq(1) and an intermediate mask derived from the grouper:

m = df.loc[::-1, 'a'].eq(0)
g = m.cumsum()

df['c'] = np.where(df['b'].eq(1).groupby(g).cummax() & ~m, 1, 0)

Output and intermediates:

    a  b  c      m     ~m  cummax  cummax&~m
0   0  1  0   True  False    True      False
1   0 -1  0   True  False   False      False
2   1 -1  0  False   True   False      False
3   1 -1  0  False   True   False      False
4   1  1  1  False   True    True       True
5   1 -1  1  False   True    True       True
6   0  1  0   True  False    True      False
7   0  1  0   True  False    True      False
8   0 -1  0   True  False   False      False
9   1  1  1  False   True    True       True
10  1  1  1  False   True    True       True
11  1 -1  1  False   True    True       True
12  0  1  0   True  False    True      False
13  0  1  0   True  False    True      False
14  1 -1  0  False   True   False      False
15  1  1  1  False   True    True       True
16  0  1  0   True  False    True      False

Answered By - mozway