如何获得groupby大小的百分比 [英] how to get percentage for groupby size
本文介绍了如何获得groupby大小的百分比的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我正在寻找一种获取百分比的方法
I am looking for a way to get percentages
df.groupby(['state', 'approved_or_not']).size()
Output:
school_state project_is_approved
AK 0 55
1 290
AL 0 256
1 1506
AR 0 177
1 872
AZ 0 347
1 1800
这很好,但我想要的是百分比而不是计数.
which is good but what I want is percentages instead of counts.
school_state project_is_approved
AK 0 0.16
1 0.84
AL 0 0.14
1 0.86
我尝试过,但找不到办法.感谢有人可以提供帮助吗?
I tried and couldn't figure out a way. Appreciate if someone can help?
推荐答案
使用 SeriesGroupBy.value_counts
带参数 normalize=True
:
df.groupby('state')['approved_or_not'].value_counts(normalize=True)
示例:
np.random.seed(2019)
L = list('ABC')
df = pd.DataFrame({'state':np.random.choice(L, size=10),
'approved_or_not':np.random.choice([0,1], size=10)})
print (df)
state approved_or_not
0 A 0
1 C 0
2 B 1
3 A 0
4 C 1
5 C 1
6 A 0
7 B 0
8 A 0
9 C 1
<小时>
a = df.groupby(['state', 'approved_or_not']).size()
print (a)
A 0 4
B 0 1
1 1
C 0 1
1 3
dtype: int64
a = df.groupby('state')['approved_or_not'].value_counts(normalize=True)
print (a)
state approved_or_not
A 0 1.00
B 0 0.50
1 0.50
C 1 0.75
0 0.25
Name: approved_or_not, dtype: float64
您可以除以 Series.div
with sum
每个第一级state
:
You can divide by Series.div
with sum
per first level state
:
a = df.groupby(['state', 'approved_or_not']).size()
a = a.div(a.sum(level=0), level=0)
print (a)
state approved_or_not
A 0 1.00
B 0 0.50
1 0.50
C 0 0.25
1 0.75
dtype: float64
这篇关于如何获得groupby大小的百分比的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文