选择数据集中的多项选择

Multiple choices in a choice data set

原始数据包含有关 consumeridcars 他们 purchased 的信息。

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

由于这是购买数据,因此需要扩展数据集以描述消费者每次购买时的完整汽车选择集。最终的数据集应该是这样的(截取自 Stata 手册 www.stata.com/manuals/cm.pdf on p. 97 in "Example 4: Multiple choices per case"):

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 consumerid-carnumber 组合生成单个 purchase=1 值(即由于扩展,购买值是重复)。

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


    

您可以使用 reshape 获得每辆购买的汽车的所有组合 consumerid/car。此示例假设原始数据集中的排序顺序定义了哪辆车是 carnumber 1,carnumber 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 |
     +----------------------------------------------+