Reversed cumulative sum of a column in pandas.DataFrame

Question

I’ve got a pandas DataFrame with a boolean column sorted by another column and
need to calculate reverse cumulative sum of the boolean column, that is,
amount of true values from current row to bottom.

Example

In [13]: df = pd.DataFrame({'A': [True] * 3 + [False] * 5, 'B': np.random.rand(8) })

In [15]: df = df.sort_values('B')

In [16]: df
Out[16]:
       A         B
6  False  0.037710
2   True  0.315414
4  False  0.332480
7  False  0.445505
3  False  0.580156
1   True  0.741551
5  False  0.796944
0   True  0.817563

I need something that will give me a new column with values

3
3
2
2
2
2
1
1

That is, for each row it should contain amount of True values on this row and
rows below.

I’ve tried various methods using .iloc[::-1] but result is not that is
desired.

It looks like I’m missing some obvious bit of information. I’ve starting using
Pandas only yesterday.

Answer 1

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