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    <p>I rearranged my approach to be more clear. You mentioned that you can't recode your variables, but I'm not sure if there's a way around that (I think any solution here recodes either explicitly or implicitly). Of course, you'll need to replace "4" with "20" throughout.</p> <pre><code>* generate some projects and members clear set obs 5 generate int project = _n generate person_1 = "Tom" generate person_2 = "Dick" if (_n &gt;= 3) generate person_3 = "Harry" if (_n &gt;=5) replace person_1 = "Jane" if inlist(_n, 2, 4) tempfile orig save `orig' * reshape to long reshape long person_, i(project) string drop _j drop if missing(person) sort project person egen id = group(person) drop if missing(id) reshape wide person, i(project) j(id) * recode to allow easier group identification forvalues i = 1/4 { levelsof person_`i', local(name) clean generate byte d_person_`i' = cond(missing(person_`i'), 0, 1) label define d_person_`i'_lbl 1 "`name'" 0 "" label values d_person_`i' d_person_`i'_lbl } * determine number of workers on project egen gp_size = rowtotal(d_person_*) * unique id for each group composition generate int id = 0 forvalues i = 1/4 { local two_i = 2^(`i' - 1) replace id = id + d_person_`i' * `two_i' } * group members generate str mbrs = "" forvalues i = 1/4 { local name: label d_person_`i'_lbl 1 replace mbrs = mbrs + "/" + "`name'" if (d_person_`i' == 1) } * there's always a leading "/" to remove with this approach replace m = substr(m, 2, .) * merge back your orig data merge 1:1 project using `orig', nogenerate replace update </code></pre> <p>This yields:</p> <pre><code>. list +---------------------------------------------------------------------------------------------------------------------------------+ | project person_1 person_2 person_3 person_4 d_pers~1 d_pers~2 d_pers~3 d_pers~4 gp_size id mbrs | |---------------------------------------------------------------------------------------------------------------------------------| 1. | 1 Tom Tom Tom 1 8 Tom | 2. | 2 Jane Jane Jane 1 4 Jane | 3. | 3 Tom Dick Tom Dick Tom 2 9 Dick/Tom | 4. | 4 Jane Dick Jane Dick Jane 2 5 Dick/Jane | 5. | 5 Tom Dick Harry Tom Dick Harry Tom 3 11 Dick/Harry/Tom | +---------------------------------------------------------------------------------------------------------------------------------+ </code></pre>
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    1. USRichard Herron
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    1. COI don't see how this code addresses the permutations and alphabetic ordering. Besides, my understanding is that the OP's data structure is different: `person_1` will always be non-empty, `person_2` will be non-empty for projects with 2+ people who worked on it, etc.
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    2. CO@richardh Thanks for the code, however, StasK is correct, this does not address my main questions in (2) and (3), and my data is structured in a way that person_1 is always non-empty and if there is a name in person_4 that means it is a 4 person project. Any ideas on addressing (2) and (3) would be helpful.
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    3. COI wasn't sure how your data were coded, so I went with how I would code it. With the coding that @StasK describes you will have a hard time addressing (2) and (3). I think your best best is to recode your data.
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