Python/Pandas：计算每行中缺失/NaN 的数量
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Python/Pandas: counting the number of missing/NaN in each row
提问by Chernyavski.aa
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。其中一些值为 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 NaN 值的数量，就像这样：
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？
回答by Zero
You could first find if element is NaNor not by isnull()and then take row-wise sum(axis=1)
您可以先查找元素是否为NaNby isnull()，然后按行进行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
而且，如果您希望将输出作为列表，您可以
In [196]: df.isnull().sum(axis=1).tolist()
Out[196]: [0, 0, 0, 3, 0, 0]
Or use countlike
或者使用count像
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
