Changing values of yes/no binary variables from 1/2 to 1/0 using loop

[Tom Scott]

Hello,

I believe this is a common problem but I have not found an answer online. I have a large dataset with multiple numeric binary variables with "no" response options coded as 2 instead of 0. Additionally, there are no value labels in the dataset. Fortunately, when a variable has values equal to either 1 or 2, it is always a yes/no question. Therefore, I would like to create a loop that changes the value 2 to 0 only for binary variables with response options equal to 1 or 2 (or have a max value of 2). I would also like to create value labels. I provided the code I have so far and an example dataset below. In the example dataset, all the variables are binary except for 'numkids', so I would not want to change the values of that variable. Could you help me fix the blue part of my code?

Thank you,
Tom

local imputedvars *_i
foreach var of `imputedvars’ {
recode `var' (2=0) if `var' has values equal to 1, 2, or .
label define yesno 0 “No” 1 “Yes”
label values `var’ yesno if `var' has values equal to 0, 1, or .
}

* Example generated by -dataex-. To install: ssc install dataex
clear
input byte(id male white older21 numkids married)
1 1 2 1 0 1
2 2 1 1 3 2
3 1 2 2 2 1
4 2 2 2 0 1
5 1 1 2 1 2
6 2 2 1 1 1
7 1 1 2 2 2
8 2 1 1 0 1
end

[Nick Cox]

findname from the Stata Journal can help here. I note that you want to exclude the identifier too and that you should define the value labels just once, outside the loop (indeed, you'll get an error trying to redefine it otherwise).

* Example generated by -dataex-. To install: ssc install dataex
clear
input byte(id male white older21 numkids married)
1 1 2 1 0 1
2 2 1 1 3 2
3 1 2 2 2 1
4 2 2 2 0 1
5 1 1 2 1 2
6 2 2 1 1 1
7 1 1 2 2 2
8 2 1 1 0 1
end

findname, all(inlist(@, 1, 2)) local(wanted) 
label define yesno 0 "No" 1 "Yes" 

foreach v of local wanted { 
    replace `v' = `v' == 1 
    label val `v' yesno 
}

[Tom Scott]

Hi Nick,

Thank you very much. I will have to remember that great command. I tried your code but the number of cases in the yes and no conditions were wrong. It was my fault for not including missing values in my example dataset. I used the following code and it seems to have worked. Thanks again!

findname, all(inlist(@, ., 1, 2)) local(wanted)
label define yesno 0 "No" 1 "Yes"

foreach v of local wanted {
replace `v' = 0 if `v' == 2
label val `v' yesno
}

[Nick Cox]

Indeed: missing values would oblige a change in the recipe.

[smith Jason]

Indeed: missing values would oblige a change in the recipe.

Could you please show us how to solve this problem when missing values take place?
* Example generated by -dataex-.
To install: ssc install dataex
clear
input byte(id male white older21 numkids married)
1 1 2 1 0 1
2 2 1 . 3 2
3 1 2 2 . 1
4 . . . 0 1
5 1 1 2 1 2
6 2 2 1 . 1
7 . 1 2 . 2
8 2 . . 0 1
end

Thank you!

[Joseph Coveney]

Could you please show us how to solve this problem when missing values take place?

I believe that Nick was alluding to the approach shown immediately before, in #3. Have you tried it?

[Nick Cox]

The mapping for whatever 2 to 0, 1 to 1 is accomplished by 2 - whatever. Notice that this works for missings too as 2 - . is returned as . (missing).

Here is some other technique too. See also https://journals.sagepub.com/doi/pdf...36867X19830921

Clearly one technique that works and is congenial is enough.

clear
input byte(id male white older21 numkids married)
1 1 2 1 0 1
2 2 1 . 3 2
3 1 2 2 . 1
4 . . . 0 1
5 1 1 2 1 2
6 2 2 1 . 1
7 . 1 2 . 2
8 2 . . 0 1
end

findname , all(inlist(@, 1, 2, .)) local(tofix) 

foreach v of local tofix { 
    gen `v'_1 = 2 - `v' 
    gen `v'_2 = cond(missing(`v'), ., `v' == 1)
    assert `v'_1 == `v'_2 
}

label def yesno 1 "yes" 0 "no"
label val *_? yesno 

list *_1 , sep(0)

     +----------------------------------------+
     | male_1   white_1   older~_1   marrie~1 |
     |----------------------------------------|
  1. |    yes        no        yes        yes |
  2. |     no       yes          .         no |
  3. |    yes        no         no        yes |
  4. |      .         .          .        yes |
  5. |    yes       yes         no         no |
  6. |     no        no        yes        yes |
  7. |      .       yes         no         no |
  8. |     no         .          .        yes |
     +----------------------------------------+

[smith Jason]

Thank you very much!