从具有相同国家列表的多个变量创建新变量

Question

在 Stata 中，我拆分了一个变量，其中最多 20 个国家用逗号分隔，现在我有 20 个不同的变量（country1 到 country20），但同一个国家列在多个变量 country1 到 country20.

例如，乌干达可能位于 country1、country2 和 country5。现在，我想为每个国家/地区创建一个变量（如果为真，则为 1，如果为假，则为 0）。所以，基本上我想为 20 个国家/地区中的每一个国家分配一个变量。我试过了，但没有用。

local N = _N
forvalues i = 1/`N' {
    local s1 = Countryies1 [`i']
    local s2 = Countryies2   [`i']   
    local s3 = Countryies3 [`i']
    local s4 = Countryies4   [`i'] 
    local s5 = Countryies5 [`i']
    local s6 = Countryies6   [`i'] 
    local s7 = Countryies7 [`i']
    local s8 = Countryies8   [`i'] 
    local s9 = Countryies9 [`i']
    local s10 = Countryies10   [`i'] 
    local s11 = Countryies11 [`i']
    local s12 = Countryies12   [`i']   
    local s13 = Countryies13 [`i']
    local s14 = Countryies14   [`i'] 
    local s15 = Countryies15 [`i']
    local s16 = Countryies16   [`i'] 
    local s17 = Countryies17 [`i']
    local s18 = Countryies18   [`i'] 
    local s19 = Countryies19 [`i']
    local s20 = Countryies20   [`i'] 
    
    local intersection: list s1 & s2 & s3 & s4 & s5 & s6 & s7 & s8 & s9 & s10 & s11 & s12 & s13 & s14 & s15 & s16 & s17 & s18 & s19 & s20
    replace country ="`intersection'" in `i'
}

Answer 1

这似乎可行 -- 并且在任何意义上都不排除其他解决方案。

clear 
input str42 countries 
"Uganda"
"Uganda, Kenya"
"Uganda, Kenya, Tanzania"
"South Africa"
end 

gen id = _n 
save datasofar, replace 

keep id countries 
split countries, parse(,)
drop countries 
reshape long countries, i(id) j(which)
drop if missing(countries)
replace countries = trim(countries)
gen name = strtoname(countries)
levelsof name, local(names) 

gen new_id = _n 

foreach n of local names { 
    gen is_`n' = name == "`n'" 
    su new_id if is_`n', meanonly 
    label var is_`n' "`=countries[r(min)]'"
    local vars `vars' is_`n' 
}

collapse  (max) `vars', by(id)

merge 1:1 id using datasofar 

     +----------------------------------------------------------------------------------------+
     | id   is_Kenya   is_Sou~a   is_Tan~a   is_Uga~a                 countries        _merge |
     |----------------------------------------------------------------------------------------|
  1. |  1          0          0          0          1                    Uganda   Matched (3) |
  2. |  2          1          0          0          1             Uganda, Kenya   Matched (3) |
  3. |  3          1          0          1          1   Uganda, Kenya, Tanzania   Matched (3) |
  4. |  4          0          1          0          0              South Africa   Matched (3) |
     +----------------------------------------------------------------------------------------+

另一种解决方案是只遍历名称，所以

foreach c in Uganda Kenya Tanzania "South Africa" { 
     local C = strtoname("`c'") 
     gen is_`C' = strpos(countries, "`c'") > 0 
}

但要小心——拼写的变化会伤害你。他们也会咬住早期的代码。

从具有相同国家列表的多个变量创建新变量

Creating new variable from several variables with same list of countries

variables

split

stata

intersection