Python 计算 Dataframe 每一列中非 NaN 条目的数量

Count number of non-NaN entries in every column of Dataframe

提问by cryp

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.

我有一个非常大的 DataFrame，我想知道是否有短（一个或两个班轮）方法来获取 DataFrame 中非 NaN 条目的数量。我不想一次做一列，因为我有近 1000 列。

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:

输出：

a: 3
b: 2
d: 1

回答by Alex Riley

The count()method returns the number of non-NaNvalues in each column:

该count()方法返回NaN每列中非值的数量：

>>> df1.count()
a    3
b    2
d    1
dtype: int64

Similarly, count(axis=1)returns the number of non-NaNvalues in each row.

同样，count(axis=1)返回NaN每行中非值的数量。

回答by hemanta

If you want to sum the total count values which are not NAN, one can do;

如果要对非 NAN 的总计数值求和，可以这样做；

np.sum(df.count())