熊猫非常简单来自分组依据的总大小的百分比

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['col1'].size() . My DF after grouping looks like this and I just want a percent of total. I remember using a variation of this statement in the past but cannot get this to work now: percent = totals.div(totals.sum(1), axis=0)

Original 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  44  95  27

Result:

df1.groupby('A').size() / df1.groupby('A').size().sum()

    A
    34    0.1
    44    0.1
    72    0.1
    77    0.7

Here is what I came up with so far which seems pretty reasonable way to do this:

df.groupby('col1').size().apply(lambda x: float(x) / df.groupby('col1').size().sum()*100)

Answer 1

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 2

通过使用以下df.groupby('col1').size().apply(lambda x: float(x) / df.groupby('col1').size().sum()*100)在形状为（3e6,59）的DF上获得良好的性能（3.73s）： df.groupby('col1').size().apply(lambda x: float(x) / df.groupby('col1').size().sum()*100)

Answer 3

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