Pandas Very Simple Percent of total size from Group by

Question

I\'m having trouble for a seemingly incredibly easy operation. What is the most succint way to just get a percent of total from a group by operation such as df.groupby

Answer 1

How about:

df = pd.DataFrame({'A': {0: 77, 1: 77, 2: 77, 3: 77, 4: 77, 5: 77, 6: 77, 7: 72, 8: 34, 9: None},
                   'B': {0: 3, 1: 52, 2: 58, 3: 3, 4: 31, 5: 53, 6: 2, 7: 25, 8: 41, 9: 95},
                   'C': {0: 98, 1: 99, 2: 61, 3: 93, 4: 99, 5: 51, 6: 9, 7: 78, 8: 34, 9: 27}})

>>> df.groupby('A').size().divide(sum(df['A'].notnull()))
A
34    0.111111
72    0.111111
77    0.777778
dtype: float64

>>> df
    A   B   C
0  77   3  98
1  77  52  99
2  77  58  61
3  77   3  93
4  77  31  99
5  77  53  51
6  77   2   9
7  72  25  78
8  34  41  34
9 NaN  95  27

Answer 2

I don't know if I'm missing something, but looks like you could do something like this:

df.groupby('A').size() * 100 / len(df)

or

df.groupby('A').size() * 100 / df.shape[0]

Answer 3

Getting good performance (3.73s) on DF with shape (3e6,59) by using: df.groupby('col1').size().apply(lambda x: float(x) / df.groupby('col1').size().sum()*100)