Pandas groupby 将非空值计数为百分比
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Pandas groupby count non-null values as percentage
提问by J. Paul
Given this dataset, I would like to count missing, NaN, values:
鉴于此数据集，我想计算缺失的 NaN 值：
df = pd.DataFrame({'A' : [1, np.nan, 2 , 55, 6, np.nan, -17, np.nan],
'Team' : ['one', 'one', 'two', 'three','two', 'two', 'one', 'three'],
'C' : [4, 14, 3 , 8, 8, 7, np.nan, 11],
'D' : [np.nan, np.nan, -12 , 12, 12, -12, np.nan, np.nan]})
Specifically I want to count (as a percentage) per group in the 'Team' column. I can get the raw count by this:
具体来说，我想在“团队”列中对每个组进行计数（以百分比形式）。我可以通过以下方式获得原始计数：
df.groupby('Team').count()
This will get the number of nonmissing numbers. What I would like to do is create a percentage, so instead of getting the raw number I would get it as a percentage of the total entries in each group (I don't know the size of the groups which are all uneven). I've tried using .agg(), but I can't seem to get what I want. How can I do this?
这将获得非缺失数字的数量。我想要做的是创建一个百分比，因此不是获取原始数字，而是将其作为每个组中总条目的百分比来获取（我不知道所有不均匀的组的大小）。我试过使用 .agg()，但我似乎无法得到我想要的。我怎样才能做到这一点？
回答by Andy Hayden
You can take the meanof the notnullBoolean DataFrame:
In [11]: df.notnull()
Out[11]:
A C D Team
0 True True False True
1 False True False True
2 True True True True
3 True True True True
4 True True True True
5 False True True True
6 True False False True
7 False True False True
In [12]: df.notnull().mean()
Out[12]:
A 0.625
C 0.875
D 0.500
Team 1.000
dtype: float64
and with the groupby:
并与 groupby：
In [13]: df.groupby("Team").apply(lambda x: x.notnull().mean())
Out[13]:
A C D Team
Team
one 0.666667 0.666667 0.0 1.0
three 0.500000 1.000000 0.5 1.0
two 0.666667 1.000000 1.0 1.0
It may be faster to do this without an apply using set_indexfirst:
在没有申请的情况下，set_index先使用以下方法可能会更快：
In [14]: df.set_index("Team").notnull().groupby(level=0).mean()
Out[14]:
A C D
Team
one 0.666667 0.666667 0.0
three 0.500000 1.000000 0.5
two 0.666667 1.000000 1.0
回答by YOBEN_S
Base on your own code add div(df.groupby('Team').size(),0)
根据您自己的代码添加 div(df.groupby('Team').size(),0)
df.groupby('Team').count().div(df.groupby('Team').size(),0)
Out[190]:
A C D
Team
one 0.666667 0.666667 0.0
three 0.500000 1.000000 0.5
two 0.666667 1.000000 1.0
