python - 您如何遍历 Pandas 数据框中的组，如何对每个组进行操作，然后将值分配给原始数据框？

    yearCount = df[['antibiotic', 'order_date', 'antiYearCount']]

    yearGroups = yearCount.groupby('order_date')

    for year in yearGroups:
        yearCount['antiYearCount'] =year.groupby('antibiotic'['antibiotic'].transform(pd.Series.value_counts)

在本例中，yearCount是一个包含“order\u date”、“anticle”、“antiYearCount”的数据帧。我已清除“订单日期”以仅包含订单年份。我要按“顺序日期”中的年份对yearCount进行分组，计算每个“年份组”中每个“抗生素”出现的次数，然后将该值赋给yearCount的“antiYearCount”变量。谢谢你的帮助！

最佳答案

我认为您需要将新列order_date添加到groupby中，然后也可以使用size代替pd.Series.value_counts来获得相同的输出：

df = pd.DataFrame({'antibiotic':list('accbbb'),
                   'antiYearCount':[4,5,4,5,5,4],
                   'C':[7,8,9,4,2,3],
                   'D':[1,3,5,7,1,0],
                   'E':[5,3,6,9,2,4],
                   'order_date': pd.to_datetime(['2012-01-01']*3+['2012-01-02']*3)})

print (df)
   C  D  E  antiYearCount antibiotic order_date
0  7  1  5              4          a 2012-01-01
1  8  3  3              5          c 2012-01-01
2  9  5  6              4          c 2012-01-01
3  4  7  9              5          b 2012-01-02
4  2  1  2              5          b 2012-01-02
5  3  0  4              4          b 2012-01-02

#copy for remove warning
#https://stackoverflow.com/a/45035966/2901002
yearCount = df[['antibiotic', 'order_date', 'antiYearCount']].copy()
yearCount['antiYearCount'] = yearCount.groupby(['order_date','antibiotic'])['antibiotic'] \
                                      .transform('size')
print (yearCount)
  antibiotic order_date  antiYearCount
0          a 2012-01-01              1
1          c 2012-01-01              2
2          c 2012-01-01              2
3          b 2012-01-02              3
4          b 2012-01-02              3
5          b 2012-01-02              3

yearCount['antiYearCount'] = yearCount.groupby(['order_date','antibiotic'])['antibiotic'] \
                                      .transform(pd.Series.value_counts)
print (yearCount)
  antibiotic order_date  antiYearCount
0          a 2012-01-01              1
1          c 2012-01-01              2
2          c 2012-01-01              2
3          b 2012-01-02              3
4          b 2012-01-02              3
5          b 2012-01-02              3