获取按值分组的每行的百分比答案

【问题标题】：Get percentage of each row grouped by a value获取按值分组的每行的百分比
【发布时间】：2021-06-17 00:07:34
【问题描述】：

我有以下df：

df3 = pd.DataFrame(np.array([['Iza', 'Tuesday'],['Martin', 'Friday'],['John', 'Monday'],['Iza', 'Tuesday'],['Iza', 'Tuesday'],['Iza', 'Wednesday'],['Sara', 'Friday'], ['Sara', 'Friday'], ['Sara', 'Sunday'],['Silvia', 'Monday'],['Silvia', 'Wednesday'],['Paul', 'Monday'],['Paul', 'Tuesday'],['Paul', 'Wednesday']]),
                   columns=['Name', 'Day'])

df3:

    Name    Day
0   Iza     Tuesday
1   Martin  Friday
2   John    Monday
3   Iza     Tuesday
4   Iza     Tuesday
5   Iza     Wednesday
6   Sara    Friday
7   Sara    Friday
8   Sara    Sunday
9   Silvia  Monday
10  Silvia  Wednesday
11  Paul    Monday
12  Paul    Tuesday
13  Paul    Wednesday

我得到了每个用户的天数：

oo = df3.groupby(['Name','Day'])['Day'].size().reset_index(name='counts')

结果：

    Name    Day      counts
0   Iza    Tuesday      3
1   Iza    Wednesday    1
2   John    Monday      1
3   Martin  Friday      1
4   Paul    Monday      1
5   Paul    Tuesday     1
6   Paul    Wednesday   1
7   Sara    Friday      2
8   Sara    Sunday      1
9   Silvia  Monday      1
10  Silvia  Wednesday   1

删除了只有一天记录的不受欢迎的用户；

uniq_us = oo[oo.duplicated(['Name'], keep=False)]

结果：

    Name    Day       counts
0   Iza     Tuesday     3
1   Iza     Wednesday   1
4   Paul    Monday      1
5   Paul    Tuesday     1
6   Paul    Wednesday   1
7   Sara    Friday      2
8   Sara    Sunday      1
9   Silvia  Monday      1
10  Silvia  Wednesday   1

现在我想按名称获取每个分组天数的百分比：

uniq_us.groupby(['Name','Day'])['counts'].apply(lambda x: x.value_counts(normalize=True)) * 100

我明白了：

Name    Day         
Iza     Tuesday    3    100.0
        Wednesday  1    100.0
Paul    Monday     1    100.0
        Tuesday    1    100.0
        Wednesday  1    100.0
Sara    Friday     2    100.0
        Sunday     1    100.0
Silvia  Monday     1    100.0
        Wednesday  1    100.0
Name: counts, dtype: float64

我不知道如何按分组名称计算它

期望的输出：

     Name    Day         

    Iza     Tuesday    3    75.0
            Wednesday  1    25.0
    Paul    Monday     1    33.33
            Tuesday    1    33.33
            Wednesday  1    33.33
    Sara    Friday     2    66.66
            Sunday     1    33.34
    Silvia  Monday     1    50.0
            Wednesday  1    50.0
    Name: counts, dtype: float64

【问题讨论】：

标签： python pandas dataframe numpy

【解决方案1】：

您可以通过transform 将计数归一化：

uniq_us["pcnt"] = uniq_us.groupby("Name").counts.transform(lambda x: x / x.sum())

得到

>>> uniq_us

      Name        Day  counts      pcnt
0      Iza    Tuesday       3  0.750000
1      Iza  Wednesday       1  0.250000
4     Paul     Monday       1  0.333333
5     Paul    Tuesday       1  0.333333
6     Paul  Wednesday       1  0.333333
7     Sara     Friday       2  0.666667
8     Sara     Sunday       1  0.333333
9   Silvia     Monday       1  0.500000
10  Silvia  Wednesday       1  0.500000

您可以将100 * 和round(2) 放入lambda 并设置Name 和Day 作为索引以匹配输出：

...transform(lambda x: (100 * x / x.sum()).round(2))

uniq_us = uniq_us.set_index(["Name", "Day"])

得到

                  counts   pcnt
Name   Day
Iza    Tuesday         3  75.00
       Wednesday       1  25.00
Paul   Monday          1  33.33
       Tuesday         1  33.33
       Wednesday       1  33.33
Sara   Friday          2  66.67
       Sunday          1  33.33
Silvia Monday          1  50.00
       Wednesday       1  50.00

【讨论】：

【解决方案2】：

你快到了。试试：

>>> uniq_us.groupby(["Name", "Day"]).sum()/uniq_us.groupby("Name").sum()
                    counts
Name   Day                
Iza    Tuesday    0.750000
       Wednesday  0.250000
Paul   Monday     0.333333
       Tuesday    0.333333
       Wednesday  0.333333
Sara   Friday     0.666667
       Sunday     0.333333
Silvia Monday     0.500000
       Wednesday  0.500000

【讨论】：

【解决方案3】：

另一种选择是在早期标准化计数：

(df3.groupby('Name')
    .Day
    .value_counts(normalize=True)
    .mul(100)
    .rename('Counts')
    .reset_index()
    .pipe(lambda x: x[x.duplicated(['Name'], keep=False)]))

#      Name        Day     Counts
#0      Iza    Tuesday  75.000000
#1      Iza  Wednesday  25.000000
#4     Paul     Monday  33.333333
#5     Paul    Tuesday  33.333333
#6     Paul  Wednesday  33.333333
#7     Sara     Friday  66.666667
#8     Sara     Sunday  33.333333
#9   Silvia     Monday  50.000000
#10  Silvia  Wednesday  50.000000

【讨论】：

这是一个非常令人印象深刻的单行代码，我甚至无法理解。谢谢 Psidom