您可以先找到 NaN > isnull（），然后按行 sum（axis = 1）

 在[195]：df.isnull（）。sum（axis = 1）
 Out [195]：
 0 0 
 1 0 
 2 0 
 3 3 
 4 0 
 5 0 
 dtype：int64 
 

如果你想将输出作为列表，你可以

  .isnull（）。sum（axis = 1）.tolist（）
 Out [196]：[0，0，0，3，0，0] 
 pre> 
I've got a dataset with a big number of rows. Some of the values are NaN, like this:
In [91]: df
Out[91]:
 1    3      1      1      1
 1    3      1      1      1
 2    3      1      1      1
 1    1    NaN    NaN    NaN
 1    3      1      1      1
 1    1      1      1      1
And I want to count the number of NaN values in each string, it would be like this:
In [91]: list = <somecode with df>
In [92]: list
    Out[91]:
     [0,
      0,
      0,
      3,
      0,
      0]
What is the best and fastest way to do it?
You could first find if element is NaN or not by isnull() and then take row-wise sum(axis=1)
In [195]: df.isnull().sum(axis=1)
Out[195]:
0    0
1    0
2    0
3    3
4    0
5    0
dtype: int64
And, if you want the output as list, you can
In [196]: df.isnull().sum(axis=1).tolist()
Out[196]: [0, 0, 0, 3, 0, 0]