我创建了两个数据框，让我们调用它们，df1 df2其中
print(df1)
a b
1 0.375241 1.0
2 NaN 2.0
3 0.448792 3.0
4 NaN 4.0
df1 = df1[np.isfinite(df1['a'])]
print(df1)
a b
1 0.375241 1.0
3 0.448792 3.0
print(df2)
aa bb
1 0.047606 1.0
2 0.202927 1.0
3 0.205663 1.0
4 NaN 1.0
5 1.388080 1.0
6 0.097084 1.0
7 0.136873 1.0
8 NaN 1.0
9 NaN 1.0
10 NaN 1.0
12 0.084676 2.0
13 0.236850 2.0
14 0.532835 2.0
15 NaN 2.0
16 NaN 2.0
17 0.035106 2.0
18 NaN 2.0
19 NaN 2.0
20 NaN 2.0
22 0.419956 3.0
23 0.267132 3.0
24 0.944217 3.0
25 0.403024 3.0
26 NaN 3.0
27 0.184425 3.0
28 0.473998 3.0
29 NaN 3.0
30 NaN 3.0
31 NaN 3.0
33 0.465454 4.0
34 0.240867 4.0
35 NaN 4.0
36 NaN 4.0
37 0.323195 4.0
38 0.193764 4.0
39 NaN 4.0
40 NaN 4.0
41 NaN 4.0
42 NaN 4.0
基于df1['b']的结果，其中我只有1＆amp; 3现在作为有效数字，我将如何保持df2['bb'] 1＆amp; 3，并将df2中的所有其他值设置为np.nan，以便这将是最终产品
df2 = df2[np.isfinite(df2['aa'])]
print(df2)
aa bb
1 0.047606 1.0
2 0.202927 1.0
3 0.205663 1.0
5 1.388080 1.0
6 0.097084 1.0
7 0.136873 1.0
22 0.419956 3.0
23 0.267132 3.0
24 0.944217 3.0
25 0.403024 3.0
27 0.184425 3.0
28 0.473998 3.0