假设我有以下数据:
表1(我的主数据集):
clear
input str1 Name str9 (Day Time)
A 24-Mar-08 "10:30:01"
A 24-Mar-08 "10:30:01"
B 24-Mar-08 "10:30:01"
B 24-Mar-08 "11:03:12"
B 24-Mar-08 "11:03:12"
C 25-Mar-08 "10:30:01"
D 25-Mar-08 "10:30:01"
D 25-Mar-08 "11:13:59"
E 25-Mar-08 "11:13:59"
F 25-Mar-08 "11:13:59"
F 25-Mar-08 "11:13:59"
F 25-Mar-08 "11:59:01"
end
表2:
clear
input str1 Insrument str9 (Day Time) float price
A 30-Mar-10 "09:29:34" 10
E 03-Mar-09 "08:23:19" 14
E 20-Mar-12 "12:15:11" 29
F 01-Mar-11 "10:30:01" 12
end
我想删除基于
name
表1中的变量。我在Stata中打开表1,需要向Stata介绍表2,然后从表1中删除两个表中名称相同的相同观测值。
最后,结果是:
+------+------------+----------+
| name | date | time |
+------+------------+----------+
| B | 24-Mar-08 | 10:30:01 |
| B | 24-Mar-08 | 11:03:12 |
| B | 24-Mar-08 | 11:03:12 |
| C | 25-Mar-08 | 10:30:01 |
| D | 25-Mar-08 | 10:30:01 |
| D | 25-Mar-08 | 11:13:59 |
+------+------------+----------+
SAS中的SQL代码为:
proc SQL;
create table table3 as
select * from table1
where name not in (select Instrument from table2);
quit;
如何使用Stata使用此程序?