python - 如果 pandas 中两个日期之间的日期，则查找每个 ID 的值总和

Question

data={'id':[1,1,1,1,2,2,2,2],
      'date1':[datetime.date(2016,1,1),datetime.date(2016,1,2),datetime.date(2016,1,2),datetime.date(2016,1,4),
               datetime.date(2016,1,2),datetime.date(2016,1,4),datetime.date(2016,1,3),datetime.date(2016,1,1)],
      'date2':[datetime.date(2016,1,5),datetime.date(2016,1,3),datetime.date(2016,1,5),datetime.date(2016,1,5),
               datetime.date(2016,1,4),datetime.date(2016,1,5),datetime.date(2016,1,4),datetime.date(2016,1,1)],
      'score1':[5,7,3,2,9,3,8,3],
      'score2':[1,3,0,5,2,20,7,7]}
df=pd.DataFrame.from_dict(data)

数据框 df 看起来像这样:

   id       date1       date2  score1  score2
0   1  2016-01-01  2016-01-05       5       1
1   1  2016-01-02  2016-01-03       7       3
2   1  2016-01-02  2016-01-05       3       0
3   1  2016-01-04  2016-01-05       2       5
4   2  2016-01-02  2016-01-04       9       2
5   2  2016-01-04  2016-01-05       3      20
6   2  2016-01-03  2016-01-04       8       7
7   2  2016-01-01  2016-01-01       3       7

另一个包含 ID、usetdate 的数据帧 UF

   id      usetdate  
0   1  2016-01-01    
1   1  2016-01-03    
2   2  2016-01-04    
3   2  2016-01-02    

如果我传递的日期(usetdate)在 date1 和 date2 之间，我想根据 UF 的每个 ID 找到 Score1 和 Score2 的总和

   id      usetdate  score1  score2
0   1  2016-01-01      5       1
1   1  2016-01-03     17       9
2   2  2016-01-04     20       29
3   2  2016-01-02     9        2

Answer 1

您可以先通过 date_range 创建包含所有日期时间的 Series ，将索引与 Series 和 DataFrame.join 中的值交换原始的、最后的聚合总和:

s = pd.concat([pd.Series(r.Index,pd.date_range(r.date1, r.date2)) for r in df.itertuples()])
s = pd.Series(s.index, index=s, name='usetdate')

df = df.drop(['date1','date2'],axis=1).join(s).groupby(['id','usetdate'], as_index=False).sum()
print (df)
   id   usetdate  score1  score2
0   1 2016-01-01       5       1
1   1 2016-01-02      15       4
2   1 2016-01-03      15       4
3   1 2016-01-04      10       6
4   1 2016-01-05      10       6
5   2 2016-01-01       3       7
6   2 2016-01-02       9       2
7   2 2016-01-03      17       9
8   2 2016-01-04      20      29
9   2 2016-01-05       3      20

编辑:

L = [(i, d, s1, s2) for i, d1, d2, s1, s2 in df.values for d in pd.date_range(d1, d2)]
df = (pd.DataFrame(L, columns=['id','usetdate','score1','score2'])
        .groupby(['id','usetdate'], as_index=False).sum())
print (df)
   id   usetdate  score1  score2
0   1 2016-01-01       5       1
1   1 2016-01-02      15       4
2   1 2016-01-03      15       4
3   1 2016-01-04      10       6
4   1 2016-01-05      10       6
5   2 2016-01-01       3       7
6   2 2016-01-02       9       2
7   2 2016-01-03      17       9
8   2 2016-01-04      20      29
9   2 2016-01-05       3      20

编辑:

您可以在聚合之前使用左连接合并值:

df1['userdate'] = pd.to_datetime(df1['userdate'])
print (df1)
   id   userdate
0   1 2016-01-01
1   1 2016-01-03
2   2 2016-01-04
3   2 2016-01-02

L = [(i, d, s1, s2) for i, d1, d2, s1, s2 in df.values for d in pd.date_range(d1, d2)]
df = (pd.DataFrame(L, columns=['id','userdate','score1','score2'])
        .merge(df1)
        .groupby(['id','userdate'], as_index=False)
        .sum())
print (df)
   id   userdate  score1  score2
0   1 2016-01-01       5       1
1   1 2016-01-03      15       4
2   2 2016-01-02       9       2
3   2 2016-01-04      20      29

编辑1:

您可以过滤列表理解中转换为元组的值:

df1['userdate'] = pd.to_datetime(df1['userdate'])
print (df1)
   id   userdate
0   1 2016-01-01
1   1 2016-01-03
2   2 2016-01-04
3   2 2016-01-02

a = [tuple(x) for x in df1.values]
print (a)
[(1, Timestamp('2016-01-01 00:00:00')), (1, Timestamp('2016-01-03 00:00:00')), 
 (2, Timestamp('2016-01-04 00:00:00')), (2, Timestamp('2016-01-02 00:00:00'))]
L = [(i, d, s1, s2) for i, d1, d2, s1, s2 in df.values 
                    for d in pd.date_range(d1, d2) 
                    if (i, d) in a]
df = (pd.DataFrame(L, columns=['id','userdate','score1','score2'])
        .groupby(['id','userdate'], as_index=False)
        .sum())
print (df)
   id   userdate  score1  score2
0   1 2016-01-01       5       1
1   1 2016-01-03      15       4
2   2 2016-01-02       9       2
3   2 2016-01-04      20      29