反转A列,取总和,然后再次反转:
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
另外,您可以计算Trues
列中的s
数量,A
然后减去(移位的)总和:
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
但这要慢得多。使用IPython的,以执行基准:
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