Generate new variable by counting

[Maike Schlssr]

ID	Best Friend 1	Best Friend 2	Best friend 3
200102	200105	200110	200104
200103	200102	200104	.
200104	.	.	.
200105	200104	.	.
200106	200110	200105	200104
200107	200110	200105	200104
200108	200102	200109	.
200109	200110	200102	200109
200110	.	.

My data looks similar to the table above. Individuals are identified by a unique ID. Eveyone was asked to nominate other people from the same dataset as their friends.
I would like to generate a (or several) new variable(s) that show how many times each individual was nominated as best friend by the other individuals. For example ID 200102 was named two times under Best Friend 1.
So the new variable "Number of nominations as BF1" would simply show the value 2 in the row of ID 200102.
Does anyone have an idea how to solve this problem. Every help is highly appreciated. Thanks in advance.

[Nick Cox]

A loop over observations is usually despised, and I will join in if there is a neater way to do it, which I did try to think up.

* Example generated by -dataex-. For more info, type help dataex
clear
input long(id bestfriend1 bestfriend2 bestfriend3)
200102 200105 200110 200104
200103 200102 200104      .
200104      .      .      .
200105 200104      .      .
200106 200110 200105 200104
200107 200110 200105 200104
200108 200102 200109      .
200109 200110 200102 200109
200110      .      .      .
end

gen count = 0

quietly forval i = 1/`=_N' {
    forval j = 1/3 {
        count if bestfriend`j' == id[`i']
        replace count = count + r(N) in `i'
    }
}

list 

     +-------------------------------------------------+
     |     id   bestfr~1   bestfr~2   bestfr~3   count |
     |-------------------------------------------------|
  1. | 200102     200105     200110     200104       3 |
  2. | 200103     200102     200104          .       0 |
  3. | 200104          .          .          .       5 |
  4. | 200105     200104          .          .       3 |
  5. | 200106     200110     200105     200104       0 |
     |-------------------------------------------------|
  6. | 200107     200110     200105     200104       0 |
  7. | 200108     200102     200109          .       0 |
  8. | 200109     200110     200102     200109       2 |
  9. | 200110          .          .          .       4 |
     +-------------------------------------------------+

.

[William Lisowski]

I think post #2 isn't quite what post #1 asks for, which seems to me to be three separate counters. I think the following will do the job, as long as you are running Stata 16 or later.

* Example generated by -dataex-. For more info, type help dataex
clear
input long(id bestfriend1 bestfriend2 bestfriend3)
200102 200105 200110 200104
200103 200102 200104      .
200104      .      .      .
200105 200104      .      .
200106 200110 200105 200104
200107 200110 200105 200104
200108 200102 200109      .
200109 200110 200102 200109
200110      .      .      .
end
frame copy default bff, replace

frame bff {
    reshape long bestfriend, i(id) j(rank)
    drop if missing(bestfriend)
    sort bestfriend
    collapse (count) nbf=id, by(bestfriend rank)
    reshape wide nbf, i(bestfriend) j(rank)
}

frlink 1:1 id, frame(bff bestfriend) generate(bflink)
frget nbf*, from(bflink)
drop bflink
mvencode nbf*, mv(0)
list, clean noobs abbreviate(12)

. list, clean noobs abbreviate(12)

        id   bestfriend1   bestfriend2   bestfriend3   nbf1   nbf2   nbf3  
    200102        200105        200110        200104      2      1      0  
    200103        200102        200104             .      0      0      0  
    200104             .             .             .      1      1      3  
    200105        200104             .             .      1      2      0  
    200106        200110        200105        200104      0      0      0  
    200107        200110        200105        200104      0      0      0  
    200108        200102        200109             .      0      0      0  
    200109        200110        200102        200109      0      1      1  
    200110             .             .             .      3      1      0

[Nick Cox]

The question in #1 was

a (or several) new variable(s) that show how many times each individual was nominated as best friend by the other individuals

and I went for the first request, a variable with a total. It is easy enough to break down the total into separate variables.

* Example generated by -dataex-. For more info, type help dataex
clear
input long(id bestfriend1 bestfriend2 bestfriend3)
200102 200105 200110 200104
200103 200102 200104      .
200104      .      .      .
200105 200104      .      .
200106 200110 200105 200104
200107 200110 200105 200104
200108 200102 200109      .
200109 200110 200102 200109
200110      .      .      .
end

quietly forval j = 1/3 { 
    gen count`j' = 0 
    forval i = 1/`=_N' { 
    
        count if bestfriend`j' == id[`i']
        replace count`j' = count`j' + r(N) in `i'
    }
}

gen count = count1 + count2 + count3 

list 

     +----------------------------------------------------------------------------+
     |     id   bestfr~1   bestfr~2   bestfr~3   count1   count2   count3   count |
     |----------------------------------------------------------------------------|
  1. | 200102     200105     200110     200104        2        1        0       3 |
  2. | 200103     200102     200104          .        0        0        0       0 |
  3. | 200104          .          .          .        1        1        3       5 |
  4. | 200105     200104          .          .        1        2        0       3 |
  5. | 200106     200110     200105     200104        0        0        0       0 |
     |----------------------------------------------------------------------------|
  6. | 200107     200110     200105     200104        0        0        0       0 |
  7. | 200108     200102     200109          .        0        0        0       0 |
  8. | 200109     200110     200102     200109        0        1        1       2 |
  9. | 200110          .          .          .        3        1        0       4 |
     +----------------------------------------------------------------------------+

[Maike Schlssr]

Thanks so much! it turned out to be exactly what I was looking for.