pandas groupby：如何计算总数的百分比？
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pandas groupby: how to calculate percentage of total?
提问by Pythonista anonymous
How can I calculate a column showing the % of total in a groupby?
如何计算显示 groupby 中总数百分比的列？
One way to do it is to calculate it manually after the gorupby, as in the last line of this example:
一种方法是在 gorupby 之后手动计算它，如本示例的最后一行：
import numpy as np
import pandas as pd
df= pd.DataFrame(np.random.randint(5,8,(10,4)), columns=['a','b','c','d'])
g = df.groupby('a').agg({'b':['sum','mean'], 'c':['sum'], 'd':['sum']})
g.columns = g.columns.map('_'.join)
g['b %']=g['b_sum']/g['b_sum'].sum()
However, in my real data I have many more columns, and I'd need the % right after the sum, so with this approach I'd have to manually change the order of the columns.
但是，在我的真实数据中，我有更多的列，并且在求和之后我需要 %，因此使用这种方法我必须手动更改列的顺序。
Is there a more direct way of doing it so that the % is the column right after the sum? Note that I need the agg(), or something equivalent, because in all my groupbys I apply different aggregate functions to different columns (e.g. sum and avg of x, but only the min of y, etc.).
有没有更直接的方法可以使 % 成为总和之后的列？请注意，我需要 agg() 或等效的东西，因为在我所有的 groupbys 中，我将不同的聚合函数应用于不同的列（例如 x 的 sum 和 avg，但只有 y 的 min 等）。
Thanks!
谢谢！
采纳答案by jezrael
I think you need lambda functionin aggand then replacecolumn names to %:
我认为您需要lambda function输入agg然后replace列名%：
np.random.seed(78)
df= pd.DataFrame(np.random.randint(5,8,(10,4)), columns=['a','b','c','d'])
g =(df.groupby('a')
.agg({'b':['sum',lambda x: x.sum()/ df['b'].sum(),'mean'],
'c':['sum'],
'd':['sum']}))
g.columns = g.columns.map('_'.join).str.replace('<lambda>','%')
print (g)
d_sum c_sum b_sum b_% b_mean
a
5 25 24 24 0.387097 6
6 11 11 14 0.225806 7
7 22 23 24 0.387097 6
