Question

我的数据格式如下：

Month       Date        Time1 Time2 Time3
  Jan      01/01/2018   1234  2345   3456
  Jan      02/01/2018   1234  2346   3457
  Jan      03/02/2018   1234  2346   3145
  Feb      03/02/2018   1234  2347   3458
  Feb      04/02/2018   1235  2398   3456
March      04/03/2018   1235  2347   3456
March      01/03/2018   1236  2345   3456
March      02/03/2018   1234  2567   3897
  Feb      02/02/2018   1236  2346   3478

我想随机选择观察值，以便从每个月和每个时隙中获得两个样本日期和数字。例如，对于Jan，Time 1，我想获取：01/01/2018，1234和02/01/2018，1234。

如何在Stata中做到这一点？

Answer 1

您可以如下随机选择观测值：

clear

input str5 Month str10 Date Time1 Time2 Time3
Jan      "01/01/2018"   1234  2345   3456
Jan      "02/01/2018"   1234  2346   3457
Jan      "03/02/2018"   1234  2346   3145
Feb      "03/02/2018"   1234  2347   3458
Feb      "04/02/2018"   1235  2398   3456
March    "04/03/2018"   1235  2347   3456
March    "01/03/2018"   1236  2345   3456
March    "02/03/2018"   1234  2567   3897
Feb      "02/02/2018"   1236  2346   3478
end

bysort Month (Time1): generate random = runiform()
bysort Month (Time1 random): keep if _n < 3

list, sepby(Month)

     +-------------------------------------------------------+
     | Month         Date   Time1   Time2   Time3     random |
     |-------------------------------------------------------|
  1. |   Feb   03/02/2018    1234    2347    3458   .3576297 |
  2. |   Feb   04/02/2018    1235    2398    3456   .4004426 |
     |-------------------------------------------------------|
  3. |   Jan   02/01/2018    1234    2346    3457   .2076905 |
  4. |   Jan   01/01/2018    1234    2345    3456   .5597356 |
     |-------------------------------------------------------|
  5. | March   02/03/2018    1234    2567    3897   .0286627 |
  6. | March   04/03/2018    1235    2347    3456   .6889245 |
     +-------------------------------------------------------+

随机选择观察结果

1 个答案: