python - 乘以包含 NaN 的 Pandas 系列行

Question

给出这个数据框:

import pandas as pd
import numpy as np

data = {'column1': [True,False, False, True, True],
        'column2' : [np.nan,0.21, np.nan, 0.2222, np.nan],
        'column3': [1000, 0, 0, 0, 0 ]}


df = pd.DataFrame.from_dict(data)

print(df)

---

   column1  column2  column3
0     True      NaN     1000
1    False   0.2100        0
2    False      NaN        0
3     True   0.2222        0
4     True      NaN        0

当 column2 行不是 NaN 时，如何将 column2 的结果与 column3 的先前值相乘，否则只返回column3 的先前值 ?

结果应该是这样的:

   column1  column2  column3
0     True      NaN     1000
1    False   0.2100        210
2    False      NaN        210
3     True   0.2222        46.662
4     True      NaN        46.662

我一直在浏览类似的问题，但我就是无法理解它..

非常感谢您的意见:)

Answer 1

你可以试一试:

#replace 0 with nan and create a copy of the df
m=df.assign(column3=df.column3.replace(0,np.nan))
#ffill on axis 1 where column2 is not null , and filter the last col then cumprod
final=(df.assign(column3=m.mask(m.column2.notna(),m.ffill(1)).iloc[:,-1].cumprod().ffill()))

---

   column1  column2   column3
0     True      NaN  1000.000
1    False   0.2100   210.000
2    False      NaN   210.000
3     True   0.2222    46.662
4     True      NaN    46.662