pandas 非常简单来自分组依据的总大小的百分比 [英] Pandas Very Simple Percent of total size from Group by

问题描述

我在看似非常简单的操作上遇到了麻烦.最简洁的方法是通过诸如df.groupby['col1'].size()之类的操作从一个组中获得总数的百分比.分组后我的DF看起来像这样，我只想占总数的百分之一.我记得以前使用过此语句的变体，但现在无法使它起作用:percent = totals.div(totals.sum(1), axis=0)

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)

原始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

结果:

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

    A
    34    0.1
    44    0.1
    72    0.1
    77    0.7

到目前为止，这是我想出的方法，这似乎很合理:

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)

推荐答案

通过使用以下方法在形状为(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)

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)

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