Python / Pandas：计算每行中缺失/ NaN的数量

I've got a dataset with a big number of rows. Some of the values are NaN, like this:

我有一个包含大量行的数据集。一些值是NaN,如下所示:

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:

我想计算每个字符串中的NaN值的数量,它将是这样的:

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?

最好和最快的方法是什么?

1 个解决方案

#1

You could first find if element is NaN or not by isnull() and then take row-wise sum(axis=1)

您可以首先通过isnull()找到元素是否为NaN,然后采用行方式求和(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

而且,如果您希望输出为列表,则可以

In [196]: df.isnull().sum(axis=1).tolist()
Out[196]: [0, 0, 0, 3, 0, 0]

Or use count like

或者使用计数

In [130]: df.shape[1] - df.count(axis=1)
Out[130]:
0    0
1    0
2    0
3    3
4    0
5    0
dtype: int64

#1