按2列分组，并打印每个组合的数据

Question

我想用{}和{}将下面的{}分组：

            Id          Timestamp               Data    ItemId      Date
2012-04-21  19389576    2012-04-21 00:04:03.533 39.0    1              2012-04-21
2012-04-21  19389577    2012-04-21 00:04:04.870 38.5    1              2012-04-21
2012-04-21  19389608    2012-04-21 00:07:03.450 38.0    1              2012-04-21
                        ...

2012-04-22  19389609    2012-04-21 00:03:04.817 37.5    2              2012-04-21
2012-04-22  19389620    2012-04-21 00:10:04.400 37.0    2              2012-04-21
                        ...       

要获取Date和ItemId的所有组合，请使用Date和ItemId的每个组合从原始数据帧df中进行选择，例如Date== 2012-04-21 and ItemId==1、Date== 2012-04-21 and ItemId==2

如何在for循环中同时使用两列选择数据

Answer 1

IIUC，如果您只想打印每个组的数据，请使用：

for key, group in df.groupby(['ItemId', 'Date']): 
    print(key)
    print(group)

这张照片是：

(1, '2012-04-21')
                  Id                Timestamp  Data  ItemId        Date
2012-04-21  19389576  2012-04-21 00:04:03.533  39.0       1  2012-04-21
2012-04-21  19389577  2012-04-21 00:04:04.870  38.5       1  2012-04-21
2012-04-21  19389608  2012-04-21 00:07:03.450  38.0       1  2012-04-21

(2, '2012-04-21')
                  Id                Timestamp  Data  ItemId        Date
2012-04-22  19389609  2012-04-21 00:03:04.817  37.5       2  2012-04-21
2012-04-22  19389620  2012-04-21 00:10:04.400  37.0       2  2012-04-21

Answer 2

因为使用group by时，每行索引将是一个元组(2012-04-21,1)，(2012-04-21,2)，(2012-04-22,1)：

from datetime import datetime
import pandas as pd 
import io
s_e="""    Id    Timestamp    Data    ProductId    Date
    2012-04-21  19389576    2012-04-21 00:04:03.533    39.0    1    2012-04-21
    2012-04-21  19389577    2012-04-21 00:04:04.870    38.5    1    2012-04-21
    2012-04-21  19389608    2012-04-21 00:07:03.450    38.0    1    2012-04-22
    2012-04-22  19389609    2012-04-21 00:03:04.817    37.5    2    2012-04-21
    2012-04-22  19389620    2012-04-21 00:10:04.400    37.0    2    2012-04-22

    """
pd.set_option('display.max_columns', None )
df = pd.read_csv(io.StringIO(s_e), sep='    ', parse_dates=[1,4], engine='python')
df=df.groupby(['Date','ProductId']).agg(list)
print('df:\n',df)
print('df.index.values:\n',df.index.values)

输出：

>>>df:
                                                               Timestamp          Data  
Date       ProductId                                                                          
2012-04-21 1          [2012-04-21 00:04:03.533000, 2012-04-21 00:04:04.870000]  [39.0, 38.5]  
           2                                      [2012-04-21 00:03:04.817000]        [37.5]  
2012-04-22 1                                      [2012-04-21 00:07:03.450000]        [38.0]  
           2                                      [2012-04-21 00:10:04.400000]        [37.0] 


>>>df.index.values:
 [(Timestamp('2012-04-21 00:00:00'), 1)
 (Timestamp('2012-04-21 00:00:00'), 2)
 (Timestamp('2012-04-22 00:00:00'), 1)
 (Timestamp('2012-04-22 00:00:00'), 2)]

您可以尝试以下方法来选择特定的组合，例如Date== 2012-04-21 and ItemId==1组合：

datetoselect=(datetime.strptime('2012-04-21','%Y-%m-%d'),2)   #Date== 2012-04-21 and ItemId==1
print(df[[i==datetoselect for i in df.index.values]])

输出：

                                          Id                     Timestamp    Data
Date       ProductId                                                              
2012-04-21 2          [2012-04-22  19389609]  [2012-04-21 00:03:04.817000]  [37.5]

Answer 3

尝试使用双选择器，将每个选择器添加到一组括号中，并在括号之间添加一个与&；：

df[(df[“Date”] == “2020-04-21”)& (df[“ItemId”] == 2)]