stata - 选择数据集中的多项选择

Question

原始数据包含有关 consumerid 的信息和 cars他们purchased .

clear
    input consumerid car    purchase
    6   American    1
    6   Japanese    0
    6   European    0
    7   American    0
    7   Japanese    0
    7   European    1
    7   Korean      1
end

由于这是一个购买数据，因此需要以某种方式扩展数据集，以在消费者每次购买时描述完整的汽车选择集。最终的数据集应如下所示(取自 Stata 手册 www.stata.com/manuals/cm.pdf 第 97 页的“示例 4:每个案例的多项选择”中的屏幕截图):

我已经生成了几个代码(如下所示)几乎可以让我到达我需要的地方，但是我无法为每个消费者 ID-车号组合生成一个购买值=1(即由于扩展，购买值重复)。

egen sumpurchase=total(purchase), by(id)
expand sumpurchase
bysort id car (purchase): gen carnumber=_n


    

Answer 1

您可以使用 reshape获取每辆车购买的消费者 ID/汽车的所有组合。此示例假设原始数据集中的排序顺序定义了汽车编号 1、汽车编号 2 等。

* Example generated by -dataex-. To install: ssc install dataex
clear
input byte consumerid str8 car byte purchase
6 "American" 1
6 "Japanese" 0
6 "European" 0
7 "American" 0
7 "Japanese" 0
7 "European" 1
7 "Korean"   1
end

// Generate carnumber
bys consumerid: gen carnumber = cond(purchase != 0, sum(purchase), 0)

// To wide
reshape wide purchase, i(consumerid car) j(carnumber)

// Keep purchased cars only
drop purchase0

// Back to long
reshape long

// Drop if no cars purchased for consumerid/carnumber
bysort consumerid carnumber (purchase) : drop if missing(purchase[1])

// Replace missing with 0 for non-purchased cars
mvencode purchase, mv(0)

// Sort and see results
sort consumerid carnumber car
list, sepby(consumerid carnumber) abbr(14)

结果:

. list, sepby(consumerid carnumber) abbr(14)

     +----------------------------------------------+
     | consumerid        car   carnumber   purchase |
     |----------------------------------------------|
  1. |          6   American           1          1 |
  2. |          6   European           1          0 |
  3. |          6   Japanese           1          0 |
     |----------------------------------------------|
  4. |          7   American           1          0 |
  5. |          7   European           1          1 |
  6. |          7   Japanese           1          0 |
  7. |          7     Korean           1          0 |
     |----------------------------------------------|
  8. |          7   American           2          0 |
  9. |          7   European           2          0 |
 10. |          7   Japanese           2          0 |
 11. |          7     Korean           2          1 |
     +----------------------------------------------+