CODEWITHSUNDEEP
pythonpandas
J. Paul
J. Paul

Reputation: 391

Pandas groupby count non-null values as percentage

Given this dataset, I would like to count missing, NaN, values:

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:

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?

Upvotes: 15

Views: 23834

Answers (2)

BENY
BENY

Reputation: 323276

Base on your own code add 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

Upvotes: 3

Andy Hayden
Andy Hayden

Reputation: 375535

You can take the mean of the notnull Boolean 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:

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_index first:

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

Upvotes: 31

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