Reversed cumulative sum of a column in pandas.DataFrame

Reverse column A, take the cumsum, then reverse again:

df['C'] = df.loc[::-1, 'A'].cumsum()[::-1]

import pandas as pd
df = pd.DataFrame(
    {'A': [False, True, False, False, False, True, False, True],
     'B': [0.03771, 0.315414, 0.33248, 0.445505, 0.580156, 0.741551, 0.796944, 0.817563],},
     index=[6, 2, 4, 7, 3, 1, 5, 0])
df['C'] = df.loc[::-1, 'A'].cumsum()[::-1]
print(df)

yields

       A         B  C
6  False  0.037710  3
2   True  0.315414  3
4  False  0.332480  2
7  False  0.445505  2
3  False  0.580156  2
1   True  0.741551  2
5  False  0.796944  1
0   True  0.817563  1

Alternatively, you could count the number of Trues in column A and subtract the (shifted) cumsum:

In [113]: df['A'].sum()-df['A'].shift(1).fillna(0).cumsum()
Out[113]: 
6    3
2    3
4    2
7    2
3    2
1    2
5    1
0    1
Name: A, dtype: object

But this is significantly slower. Using IPython to perform the benchmark:

In [116]: df = pd.DataFrame({'A':np.random.randint(2, size=10**5).astype(bool)})

In [117]: %timeit df['A'].sum()-df['A'].shift(1).fillna(0).cumsum()
10 loops, best of 3: 19.8 ms per loop

In [118]: %timeit df.loc[::-1, 'A'].cumsum()[::-1]
1000 loops, best of 3: 701 µs per loop

Similar to unutbus first suggestion, but without the deprecated ix:

df['C']=df.A[::-1].cumsum()

If wanting to reverse cumulative sum column-wise:

(-df).cumsum(axis=1).add(1).shift(1,axis=1,fill_value=1.0)

This works but is slow... like @unutbu answer. True resolves to 1. Fails on False, or any other value though.

df[2] = df.groupby('A').cumcount(ascending=False)+1
df[1] = np.where(df['A']==True,df[2],None)
df[1] = df[1].fillna(method='bfill').fillna(0)
del df[2]

      A         B    1
# 3  False  0.277557  3.0
# 7  False  0.400751  3.0
# 6  False  0.431587  3.0
# 5  False  0.481006  3.0
# 1   True  0.534364  3.0
# 2   True  0.556378  2.0
# 0   True  0.863192  1.0
# 4  False  0.916247  0.0