在Dataframe的每一列中计算非NaN条目的数量 [英] Count number of non-NaN entries in every column of Dataframe
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问题描述
df1 = pd.DataFrame([(1,2,无),(无,4,无)没有,7),(5,None,None)],
columns = ['a','b','d'],index = ['A','B','C' D'])
abd
A 1 2 NaN
B NaN 4 NaN
C 5 NaN 7
D 5 NaN NaN
输出:
a :3
b:2
d:1
解决方案
count()
方法返回每列中非 NaN
值的数量:
>>> df1.count()
a 3
b 2
d 1
dtype:int64
同样, count(axis = 1)
返回每个 NaN
中的值行。
I have a really big DataFrame and I was wondering if there was short (one or two liner) way to get the a count of non-NaN entries in a DataFrame. I don't want to do this one column at a time as I have close to 1000 columns.
df1 = pd.DataFrame([(1,2,None),(None,4,None),(5,None,7),(5,None,None)],
columns=['a','b','d'], index = ['A', 'B','C','D'])
a b d
A 1 2 NaN
B NaN 4 NaN
C 5 NaN 7
D 5 NaN NaN
Output:
a: 3
b: 2
d: 1
解决方案
The count()
method returns the number of non-NaN
values in each column:
>>> df1.count()
a 3
b 2
d 1
dtype: int64
Similarly, count(axis=1)
returns the number of non-NaN
values in each row.
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