如何选择熊猫NaN前后的行

Question

我有一个数据帧，看起来像这样：

Name      Age       Job         
0   Alex      20        Student
1   Sara      21        Doctor
2   john      23        NaN
3   kevin     22        Teacher
4   Rosa      20        senior manager
5   johanes   25        Dentist
6   lina      23        Student
7   yaser     25        Pilot
8   jason     20        Manager
9   Ali       23        NaN
10  Ahmad     21        Professor
11  Joe       24        NaN
12  Donald    29        Waiter
.
.
.
.

我想选择行本身在Job列中具有NaN值的行之前和之后的行。为此，我有以下代码：

Rows = df[df. Shift(1, fill_value="dummy").Job. isna() | df.Job. isna()| df. Shift(-1, fill_value="dummy"). df. isna()]
print(Rows)

结果是：

1   Sara      21        Doctor
2   john      23        NaN
3   kevin     22        Teacher
8   jason     20        Manager
9   Ali       23        NaN
10  Ahmad     21        Professor
11  Joe       24        NaN
12  Donald    29        Waiter

这里唯一的问题是行号10，它在结果中应该是双倍的，因为这一行是NaN之后的一行，即编号9，同时是NaN值之前的行，即行号11(该行位于具有NaN值的两行之间(。所以最后我想要这个：

1   Sara      21        Doctor
2   john      23        NaN
3   kevin     22        Teacher
8   jason     20        Manager
9   Ali       23        NaN
10  Ahmad     21        Professor
10  Ahmad     21        Professor
11  Joe       24        NaN
12  Donald    29        Waiter

因此，在具有NaN值的两行之间的每一行在结果中也应该是两次(或者应该是重复的(。有办法做到这一点吗？任何帮助都将不胜感激。

Answer 1

将concat与前、后和匹配条件一起使用：

m = df.Job.isna()
df = pd.concat([df[m.shift(fill_value=False)],
df[m.shift(-1, fill_value=False)],
df[m]]).sort_index()
print (df)
Name  Age        Job
1     Sara   21     Doctor
2     john   23        NaN
3    kevin   22    Teacher
8    jason   20    Manager
9      Ali   23        NaN
10   Ahmad   21  Professor
10   Ahmad   21  Professor
11     Joe   24        NaN
12  Donald   29     Waiter