我有两个数据集，一个具有每周日期时间，另一个具有每小时日期时间。 我的数据集看起来像这样：-
df1
Week_date w_values
21-04-2019 20:00:00 10
28-04-2019 20:00:00 20
05-05-2019 20:00:00 30
df2
hour_date h_values
19-04-2019 08:00:00 a
21-04-2019 07:00:00 b
21-04-2019 20:00:00 c
22-04-2019 06:00:00 d
23-04-2019 05:00:00 e
28-04-2019 19:00:00 f
28-04-2019 20:00:00 g
28-04-2019 21:00:00 h
29-04-2019 20:00:00 i
05-05-2019 20:00:00 j
06-05-2019 23:00:00 k
尝试合并，但未能获得所需的输出
输出数据集应如下所示
week_date w_values hour_date h_values
21-04-2019 20:00:00 10 21-04-2019 20:00:00 c
21-04-2019 20:00:00 10 22-04-2019 06:00:00 d
21-04-2019 20:00:00 10 23-04-2019 05:00:00 e
21-04-2019 20:00:00 10 28-04-2019 19:00:00 f
28-04-2019 20:00:00 20 28-04-2019 20:00:00 g
28-04-2019 20:00:00 20 28-04-2019 21:00:00 h
28-04-2019 20:00:00 20 29-04-2019 20:00:00 i
05-05-2019 20:00:00 30 05-05-2019 20:00:00 j
05-05-2019 20:00:00 30 06-05-2019 23:00:00 k
仅当星期日期等于小时日期时，每周日期才会更改。...否则它将采用上一个星期日期。...
答案 0 :(得分：3)
使用'merge_asof'函数。来自pandas文档，“此合并类似于左联接，但我们匹配的是最接近的键而不是相等的键。”
df_week['Week_date']=pd.to_datetime(df_week['Week_date'])
df_hour['hour_date']=pd.to_datetime(df_hour['hour_date'])
df_week_sort=df_week.sort_values(by='Week_date')
df_hour_sort=df_hour.sort_values(by='hour_date')
df_week_sort.rename(columns={'Week_date':'Merge_date'},inplace=True)
df_hour_sort.rename(columns={'hour_date':'Merge_date'},inplace=True)
df_merged=pd.merge_asof(df_hour_sort,df_week_sort,on='Merge_date')
确保按日期戳对两个框架进行排序
答案 1 :(得分：1)
应该执行以下操作（假设Week_date和hour_date是日期时间）
(df2.merge(df1, how='left', right_on='Week_date', left_on='hour_date')
.ffill()
.dropna())
确保两个df都已排序
>>> df1 = df1.sort_values('Week_date')
>>> df2 = df2.sort_values('hour_date')
进行合并
>>> df3 = df2.merge(df1, how='left', right_on='Week_date', left_on='hour_date')
>>> df3
hour_date h_values Week_date w_values
0 2019-04-19 08:00:00 a NaT NaN
1 2019-04-21 07:00:00 b NaT NaN
2 2019-04-21 20:00:00 c 2019-04-21 20:00:00 10.0
3 2019-04-22 06:00:00 d NaT NaN
4 2019-04-23 05:00:00 e NaT NaN
5 2019-04-28 19:00:00 f NaT NaN
6 2019-04-28 20:00:00 g 2019-04-28 20:00:00 20.0
7 2019-04-28 21:00:00 h NaT NaN
8 2019-04-29 20:00:00 i NaT NaN
9 2019-05-05 20:00:00 j 2019-05-05 20:00:00 30.0
10 2019-06-05 23:00:00 k NaT NaN
向前填补空白
>>> df3 = df3.ffill()
>>> df3
hour_date h_values Week_date w_values
0 2019-04-19 08:00:00 a NaT NaN
1 2019-04-21 07:00:00 b NaT NaN
2 2019-04-21 20:00:00 c 2019-04-21 20:00:00 10.0
3 2019-04-22 06:00:00 d 2019-04-21 20:00:00 10.0
4 2019-04-23 05:00:00 e 2019-04-21 20:00:00 10.0
5 2019-04-28 19:00:00 f 2019-04-21 20:00:00 10.0
6 2019-04-28 20:00:00 g 2019-04-28 20:00:00 20.0
7 2019-04-28 21:00:00 h 2019-04-28 20:00:00 20.0
8 2019-04-29 20:00:00 i 2019-04-28 20:00:00 20.0
9 2019-05-05 20:00:00 j 2019-05-05 20:00:00 30.0
10 2019-06-05 23:00:00 k 2019-05-05 20:00:00 30.0
删除其余的NaN
>>> df3 = df3.dropna()
>>> df3
hour_date h_values Week_date w_values
2 2019-04-21 20:00:00 c 2019-04-21 20:00:00 10.0
3 2019-04-22 06:00:00 d 2019-04-21 20:00:00 10.0
4 2019-04-23 05:00:00 e 2019-04-21 20:00:00 10.0
5 2019-04-28 19:00:00 f 2019-04-21 20:00:00 10.0
6 2019-04-28 20:00:00 g 2019-04-28 20:00:00 20.0
7 2019-04-28 21:00:00 h 2019-04-28 20:00:00 20.0
8 2019-04-29 20:00:00 i 2019-04-28 20:00:00 20.0
9 2019-05-05 20:00:00 j 2019-05-05 20:00:00 30.0
10 2019-06-05 23:00:00 k 2019-05-05 20:00:00 30.0