stata - 逗号在ado程序中通过语法varlist传递

Question

我有一个简短的 .ado 文件来标准化来自多个来源的字符串变量。该程序采用一个字符串变量，用后缀重命名它，然后用标准化名称替换原始变量。

但是 syntax 没有正确解析字符串变量后面的逗号。也就是说，不是传递 country，而是传递 country, 并抛出错误。

varname 和, 之间是否需要空格？还是我对应该如何使用 syntax 和 varlist 有误解？

clear
set obs 10
generate country = "United States of America"

* runs fine without `suffix()` options
country_names country
list

/*
. list

     +------------------------------------------+
     |                country_0         country |
     |------------------------------------------|
  1. | United States of America   United States |
  2. | United States of America   United States |
  3. | United States of America   United States |
  4. | United States of America   United States |
  5. | United States of America   United States |
     |------------------------------------------|
  6. | United States of America   United States |
  7. | United States of America   United States |
  8. | United States of America   United States |
  9. | United States of America   United States |
 10. | United States of America   United States |
     +------------------------------------------+
*/

* but gives error with `suffix()` 
country_names country, suffix(_orig)

/*
. country_names country, suffix(_orig)
, invalid name
r(198);
*/

* `set trace on` reveals comma passed as part of `varlist`
set trace on
country_names country, suffix(_orig)

/*
. country_names country, suffix(_orig)
  ---------------------------------------------------------------------------------------------- begin country_names ---
  - version 11.2
  - syntax varname(string) [, Suffix(string) ]
  - quietly {
  - if "`suffix'" == "" local suffix "_0"
  = if "_orig" == "" local suffix "_0"
  - rename `1' `1'`suffix'
  = rename country, country,_orig
, invalid name
    generate `1' = proper(`1'`suffix')
    replace `1' = "United States" if inlist(`1', "United States Of America")
    local name: variable label `1'`suffix'
    label variable `1' "`name'"
    label variable `1'`suffix' "`name' (orig)"
    }
  ------------------------------------------------------------------------------------------------ end country_names ---
r(198);
*/

* if I leave space before comma, then program works
country_names country , suffix(_orig)
list

/*
. list

     +----------------------------------------------------------+
     |                country_0    country_orig         country |
     |----------------------------------------------------------|
  1. | United States of America   United States   United States |
  2. | United States of America   United States   United States |
  3. | United States of America   United States   United States |
  4. | United States of America   United States   United States |
  5. | United States of America   United States   United States |
     |----------------------------------------------------------|
  6. | United States of America   United States   United States |
  7. | United States of America   United States   United States |
  8. | United States of America   United States   United States |
  9. | United States of America   United States   United States |
 10. | United States of America   United States   United States |
     +----------------------------------------------------------+
*/

这是 .ado 文件。

*! 0.1 Richard Herron 2/11/2014

/* use to standardize country names across several data sources */

program country_names
    version 11.2
    syntax varname(string) [, Suffix(string) ]

    quietly {

        /* default suffix */
        if "`suffix'" == "" local suffix "_0"

        /* save original as new variable w/ suffix */
        rename `1' `1'`suffix'

        /* first standardize capitalization */
        generate `1' = proper(`1'`suffix')

        /* -if- picks bad names from several sources */
        replace `1' = "United States" ///
            if inlist(`1', "United States Of America")

        /* fix labels */ 
        local name: variable label `1'`suffix'
        label variable `1' "`name'"
        label variable `1'`suffix' "`name' (orig)"

    }

    end

Answer 1

当且仅当您提供有效的变量名时，您的syntax 语句才会生成局部宏varlist。

你的程序的问题是你没有使用那个本地的。

相反，您使用名称为 1 的本地宏。无论 语法如何，默认情况下本地宏 0 是在命令名称和本地宏 1、 之后键入的整个命令行2 等等是命令行中的第一个、第二个等等“标记”。

在决定什么是标记时，在这种情况下，这里的重要细节是 Stata 在空间上进行解析。因此，在您的示例中，标记 1 是 country,(包括逗号)等(假设 suffix 是 _orig)

rename `1' `1'`suffix'

被解释为

rename country, country,_orig 

如果您set trace on 我预计您会看到这一行，这就是让您失望的原因。就 rename 而言，country(一个有效的变量名)后面的是逗号，它不是一个有效的变量名。

简短的总结很贴心:您在应该使用 varlist 的地方使用了对 1 的引用。

注意:虽然你的syntax语句确实指定了varname，一个单一的变量名，你键入的变量名仍然放在局部宏varlist。

注意:您可以通过始终在变量名后加一个空格来解决这个问题，但我强烈建议不要这样做。

注意:syntax 在这里是无可指责的。