wl2776 wl2776 - 5 months ago 184
Python Question

Reverse column values in pandas.DataFrame

I've got a pandas DataFrame with boolean column, sorted by another column and need to calculate reverse cumulative sum, 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 anount of True values on this row and rows below.

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

Think, I'm missing some obvious thing. I've starting using Pandas only yesterday.

Answer

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

df['C'] = df.ix[::-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.ix[::-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.ix[::-1, 'A'].cumsum()[::-1]
1000 loops, best of 3: 701 ┬Ás per loop