Python / Pandas:计算每行中丢失/ NaN的数量 [英] Python/Pandas: counting the number of missing/NaN in each row
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问题描述
我有一个包含大量行的数据集。一些值是NaN,像这样:
在[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
我想计算每个字符串中NaN值的数量, this:
在[91]:list =< somecode with df>
在[92]:list
Out [91]:
[0,
0,
0,
3,
0,
0]
最好最快的方法是什么?
解决方案
您可以先找到
,然后按行 NaN
> isnull() 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()
pre>
Out [196]:[0,0,0,3,0,0]
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?
解决方案You could first find if element is
NaN
or not byisnull()
and then take row-wisesum(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]
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