Exporting marginal means results after conjoint command into table in word

[Katharina Stuckradt]

Dear all,

I've been using the conjoint command to estimate marginal means (MMs) on my conjoint data and created graphs as id neatly explained in the documentation of conjoint.
In a next step, I want to export the results into a table for publication in word. Usually, I would use estout to report regression results. This does not work here. The only command that works is asdoc. However, most of the asdoc options do not seem to work and the final table is messy.

Ideally, I would like to end up with a table with four columns, reporting MMs of four different subgroup estimations, with SE in parentheses and p-values reported as * p<0.001, p<0.01, * p<0.05.

Can anyone help me out with this?
Thank you in advance, Katharina

** Forced Choice Outcome
* women
conjoint choice values appl train equota wquota if female == 1 , est(mm) id(id) 
matrix female_fc = e(results)

* men
conjoint choice values appl train equota wquota if female != 1 , est(mm) id(id) 
matrix male_fc = e(results)

** Rating Outcome 
* women
conjoint support_1 values appl train equota wquota if female == 1 , est(mm) id(id) 
matrix female_r = e(results)

* men
conjoint support_1 values appl train equota wquota if female != 1 , est(mm) id(id)
matrix male_r = e(results)

// results stored in matrices for creating graphs

[Roman Mostazir]

Can you post some data examples using -dataex- ? See FAQ on how to use -dataex-. Not everyone, I suppose, will be familiar with the -conjoint- here (I am not). But looking at your codes, I assume that you can obtain the e(results) matrix in a separate 'k' matrix with command: - mat k = e(results) - and then write a loop to obtain the desired values from the -k matrix-.

[Katharina Stuckradt]

Dear Roman, absolutely, thank you for pointing that out.

* Example generated by -dataex-. For more info, type help dataex
clear
input float choice double(values appl train equota wquota) float(female id)
0 1 3 2 1 2 0 1
1 2 3 2 1 1 0 1
0 1 1 2 3 1 0 1
1 1 2 1 2 1 0 1
0 2 2 1 1 3 0 1
1 2 3 2 1 3 0 1
0 1 2 2 3 3 0 1
1 2 1 1 2 3 0 1
1 2 1 1 2 2 0 1
0 1 3 1 2 3 0 1
0 1 2 2 3 2 0 1
1 2 3 1 2 3 0 1
0 2 2 1 3 3 0 1
1 2 1 1 1 3 0 1
1 1 3 1 2 1 0 1
0 1 1 1 3 2 0 1
1 2 2 2 1 1 0 1
0 1 2 2 1 1 0 1
0 1 3 1 2 3 0 1
1 1 2 1 2 2 0 1
1 1 1 1 1 2 0 1
0 2 2 2 3 2 0 1
0 1 2 1 3 1 0 1
1 1 3 1 2 3 0 1
1 2 3 2 2 1 0 1
0 2 1 2 3 3 0 1
0 2 2 2 3 2 0 1
1 1 3 1 1 2 0 1
1 2 1 1 2 3 0 1
0 1 1 2 3 3 0 1
0 1 2 1 3 3 0 1
1 1 2 2 1 1 0 1
end
label values values values
label def values 1 "diversity", modify
label def values 2 "performance", modify
label values appl formalization
label def formalization 1 "unstandardized", modify
label def formalization 2 "standardized", modify
label def formalization 3 "anonymous", modify
label values train training
label def training 1 "diversity training", modify
label def training 2 "hiring training", modify
label values equota quotas
label values wquota quotas
label def quotas 1 "none", modify
label def quotas 2 "tiebreaker", modify
label def quotas 3 "strict quota", modify
label values female female
label def female 0 "male", modify

[Roman Mostazir]

I am sorry, I am not being able to run your commands posted in #1 with the data example you posted in #3. Make sure your commands work with the data example you provide. As I mentioned before that I do not know or ever used 'conjoint'; it is not an official Stata command and I don't have time to look further why the commands are not running. One reason definitiely is not having enough variation in the data example provided. See below:

conjoint choice values appl train equota wquota if female == 1 , est(mm) id(id)
no observations
r(2000);

end of do-file

r(2000);


conjoint choice values appl train equota wquota if female != 1 , est(mm) id(id)
.\ not found
r(111);

end of do-file

r(111);


. conjoint support_1 values appl train equota wquota if female == 1 , est(mm) id(id)
variable support_1 not found
r(111);

end of do-file

r(111);


. conjoint support_1 values appl train equota wquota if female != 1 , est(mm) id(id)
variable support_1 not found
r(111);

[Andrew Musau]

You can do this to show the structure of the matrices created by this command:

** Forced Choice Outcome
* women
conjoint choice values appl train equota wquota if female == 1 , est(mm) id(id)
matrix f = e(results)
clear
svmat f, names(matcol)
dataex in 1/20

Copy and paste the dataex output.

[Simon Pfaff]

Katharina,

I wanted to take a look (although I am not familiar with conjoint at all). As Roman mentioned: The dataset is not sufficient, there are no "female" cases, variable support_1 is missing. Maybe you can fix this?

I modified the data and created an example for the first 2 models.

* Example generated by -dataex-. For more info, type help dataex
collect clear
clear
input float choice double(values appl train equota wquota) float(female id)
0 1 3 2 1 2 1 1
1 2 3 2 1 1 0 1
0 1 1 2 3 1 0 1
1 1 2 1 2 1 0 1
0 2 2 1 1 3 1 1
1 2 3 2 1 3 0 1
0 1 2 2 3 3 0 1
1 2 1 1 2 3 0 1
1 2 1 1 2 2 0 1
0 1 3 1 2 3 0 1
0 1 2 2 3 2 0 1
1 2 3 1 2 3 0 1
0 2 2 1 3 3 0 1
1 2 1 1 1 3 0 1
1 1 3 1 2 1 1 1
0 1 1 1 3 2 0 1
1 2 2 2 1 1 0 1
0 1 2 2 1 1 0 1
0 1 3 1 2 3 0 1
1 1 2 1 2 2 0 1
1 1 1 1 1 2 0 1
0 2 2 2 3 2 0 1
0 1 2 1 3 1 0 1
1 1 3 1 2 3 0 1
1 2 3 2 2 1 0 1
0 2 1 2 3 3 1 1
0 2 2 2 3 2 0 1
1 1 3 1 1 2 0 1
1 2 1 1 2 3 0 1
0 1 1 2 3 3 0 1
0 1 2 1 3 3 0 1
1 1 2 2 1 1 0 1

1 1 3 2 1 2 1 0
0 2 3 2 1 1 0 0
0 1 1 2 3 1 0 0
1 1 2 1 2 1 0 0
1 2 2 1 1 3 1 0
1 2 3 2 1 3 0 0
1 1 2 2 3 3 0 0
1 2 1 1 2 3 0 0
1 2 1 1 2 2 0 0
0 1 3 1 2 3 0 0
1 1 2 2 3 2 0 0
1 2 3 1 2 3 0 0
1 2 2 1 3 3 0 0
1 2 1 1 1 3 0 1
1 1 3 1 2 1 1 1
0 1 1 1 3 2 0 1
1 2 2 2 1 1 0 1
0 1 2 2 1 1 0 1
0 1 3 1 2 3 0 1
1 1 2 1 2 2 0 1
1 1 1 1 1 2 0 1
0 2 2 2 3 2 0 1
0 1 2 1 3 1 0 1
1 1 3 1 2 3 0 1
1 2 3 2 2 1 0 1
0 2 1 2 3 3 1 1
0 2 2 2 3 2 0 1
1 1 3 1 1 2 0 1
1 2 1 1 2 3 0 1
0 1 1 2 3 3 0 1
0 1 2 1 3 3 0 1
1 1 2 2 1 1 0 1

end
label values values values
label def values 1 "diversity", modify
label def values 2 "performance", modify
label values appl formalization
label def formalization 1 "unstandardized", modify
label def formalization 2 "standardized", modify
label def formalization 3 "anonymous", modify
label values train training
label def training 1 "diversity training", modify
label def training 2 "hiring training", modify
label values equota quotas
label values wquota quotas
label def quotas 1 "none", modify
label def quotas 2 "tiebreaker", modify
label def quotas 3 "strict quota", modify
label values female female
label def female 0 "male", modify
label def female 1 "female", modify

Unfortunately I cannot figure out how to include/attach the significance stars. But maybe the example is helpful anyways and somebody else has an idea?!

collect clear

* women
collect: conjoint choice values if female == 1 , est(mm) id(id)
matrix female_fc = e(results)

* men
collect: conjoint choice values appl train if female != 1 , est(mm) id(id)
matrix male_fc = e(results)

collect remap colname["P>|t|"] = colname[p]

collect style cell colname[SE], sformat((%s))
collect style cell colname[SE], nformat(%4.3f)
collect style cell colname[p], nformat(%4.2f)

*collect stars p 0.01 "***" 0.05 "**" 0.9 "*"        // not working
*collect remap result[stars] = colname[stars]

collect layout (rowname[diversity]#colname[Est.]#colname[SE]#colname[p] rowname[performance]#colname[Est.]#colname[SE]#colname[p]) (cmdset)

-----------------------------
            |       1       2
------------+----------------
diversity   |                
  Est.      |     .75   .4375
  SE        | (0.135) (0.051)
  p         |    0.32    0.44
performance |                
  Est.      |     .25     .75
  SE        | (0.405) (0.042)
  p         |    0.65    0.11
-----------------------------

[Roman Mostazir]

Based on Simon's data above, here is mine that will export the results in excel with significance. But, I am sure people here will write more efficient codes than mine:

lab var choice "Choice" //labelling the 'choice' variable

putexcel set test.xlsx, sheet(test) modify

loc row = 4

foreach v of varlist choice { //add your other variables here
    
    forval i = 0/1 {
        loc gender_lab : lab female `i'
  
        conjoint `v' values if female == `i', est(mm) id(id)
        matrix k = e(results)

        loc rownames : rownames k
        loc wc : word count `rownames'
        
        forval j = 1/`wc' {
            
            loc rowlab : word `j' of `rownames'
            loc b     : di %3.2f k[`j', 1]
            loc se     : di %3.2f k[`j', 2]
            loc p    : di %4.3f k[`j', 4]
        
            if `p' <= .001 {
                loc sig = "***"
            }
            else if `p' <= .01 {
                loc sig = "**"
            }
            else if `p' <= .05 {
                loc sig = "*"
            }
            else  {
                loc sig = ""
            }
            
            putexcel A`row' = "`:var lab `v''"
            putexcel B`row' = "`gender_lab'"
            putexcel C`row' = "`rowlab'"
            putexcel D`row' = "`b' (`se')`sig'"
            
            loc ++row
        }
    }
}

[Katharina Stuckradt]

Dear Roman, absolutely, thank you for pointing that out.

* Example generated by -dataex-. For more info, type help dataex
clear
input float choice double(values appl train equota wquota) float(female id)
0 1 3 2 1 2 0 1
1 2 3 2 1 1 0 1
0 1 1 2 3 1 0 1
1 1 2 1 2 1 0 1
0 2 2 1 1 3 0 1
1 2 3 2 1 3 0 1
0 1 2 2 3 3 0 1
1 2 1 1 2 3 0 1
1 2 1 1 2 2 0 1
0 1 3 1 2 3 0 1
0 1 2 2 3 2 0 1
1 2 3 1 2 3 0 1
0 2 2 1 3 3 0 1
1 2 1 1 1 3 0 1
1 1 3 1 2 1 0 1
0 1 1 1 3 2 0 1
1 2 2 2 1 1 0 1
0 1 2 2 1 1 0 1
0 1 3 1 2 3 0 1
1 1 2 1 2 2 0 1
1 1 1 1 1 2 0 1
0 2 2 2 3 2 0 1
0 1 2 1 3 1 0 1
1 1 3 1 2 3 0 1
1 2 3 2 2 1 0 1
0 2 1 2 3 3 0 1
0 2 2 2 3 2 0 1
1 1 3 1 1 2 0 1
1 2 1 1 2 3 0 1
0 1 1 2 3 3 0 1
0 1 2 1 3 3 0 1
1 1 2 2 1 1 0 1
0 2 1 1 1 3 1 2
1 1 2 2 3 1 1 2
0 1 1 2 3 2 1 2
1 1 1 1 1 2 1 2
0 2 1 1 2 2 1 2
1 2 1 1 1 1 1 2
0 1 2 1 2 1 1 2
1 2 1 2 3 3 1 2
0 1 3 1 2 2 1 2
1 1 3 2 2 2 1 2
0 1 3 1 3 1 1 2
1 2 1 1 3 3 1 2
0 1 1 1 2 1 1 2
1 2 2 1 1 1 1 2
0 2 3 1 3 3 1 2
1 2 2 1 2 1 1 2
0 1 1 1 2 1 1 2
1 1 1 1 1 1 1 2
1 2 2 2 1 2 1 2
0 1 1 1 3 2 1 2
0 2 1 1 2 3 1 2
1 2 3 1 1 3 1 2
1 2 1 1 1 2 1 2
0 1 2 1 2 1 1 2
0 2 3 1 1 2 1 2
1 1 2 2 3 1 1 2
1 2 1 1 3 1 1 2
0 2 3 1 3 3 1 2
0 1 2 1 2 3 1 2
1 1 2 2 2 3 1 2
0 1 3 1 3 1 1 2
1 2 1 2 2 1 1 2
0 2 1 2 2 3 0 3
1 2 1 2 2 2 0 3
1 1 1 1 2 1 0 3
0 2 3 2 3 3 0 3
0 1 3 2 2 3 0 3
1 2 3 1 2 2 0 3
1 2 3 1 2 1 0 3
0 1 1 2 3 2 0 3
0 2 1 2 1 1 0 3
1 2 1 2 1 3 0 3
1 1 3 1 1 2 0 3
0 1 1 1 2 1 0 3
0 2 2 2 3 1 0 3
1 2 3 1 2 1 0 3
1 2 3 1 1 1 0 3
0 2 1 2 1 2 0 3
0 2 2 1 3 1 0 3
1 2 3 1 1 3 0 3
0 2 2 1 1 3 0 3
1 2 3 1 1 2 0 3
0 2 3 2 3 3 0 3
1 1 1 1 2 1 0 3
1 2 1 2 2 1 0 3
0 1 2 2 1 1 0 3
0 2 2 2 2 3 0 3
1 2 1 2 3 3 0 3
0 2 3 2 3 1 0 3
1 1 3 2 3 1 0 3
1 1 3 2 2 2 0 3
0 2 3 2 2 3 0 3
1 1 3 2 3 1 0 3
0 2 3 2 1 3 0 3
0 1 1 1 3 1 1 4
1 2 1 1 3 1 1 4
1 2 2 1 2 2 1 4
0 2 2 2 1 3 1 4
1 2 2 2 2 3 1 4
0 2 2 1 3 3 1 4
0 1 2 2 3 3 1 4
1 2 1 2 3 2 1 4
1 2 1 1 1 2 1 4
0 2 2 1 3 2 1 4
0 1 2 2 3 1 1 4
1 2 2 1 3 3 1 4
1 1 3 1 2 1 1 4
0 1 3 2 3 1 1 4
0 2 3 1 3 1 1 4
1 1 2 2 2 3 1 4
0 1 2 1 1 1 1 4
1 2 2 2 1 2 1 4
1 1 1 1 3 1 1 4
0 2 3 1 3 2 1 4
0 1 2 2 2 3 1 4
1 2 1 2 3 1 1 4
0 1 2 2 3 2 1 4
1 1 1 2 2 3 1 4
0 1 2 1 1 1 1 4
1 2 2 2 2 2 1 4
1 2 1 1 3 1 1 4
0 1 3 1 2 3 1 4
0 1 1 2 3 3 1 4
1 2 1 2 3 3 1 4
1 1 1 1 2 1 1 4
0 2 2 2 1 3 1 4
1 2 1 2 2 3 0 5
0 1 3 2 3 1 0 5
0 1 1 2 3 3 0 5
1 1 1 2 2 1 0 5
0 1 3 2 2 2 0 5
1 2 2 1 1 2 0 5
0 1 3 1 1 1 0 5
1 2 2 2 3 1 0 5
0 2 2 1 3 3 0 5
1 2 2 1 1 2 0 5
1 2 1 2 2 2 0 5
0 2 1 2 2 3 0 5
0 2 1 2 3 2 0 5
1 2 1 2 1 1 0 5
1 2 2 2 2 2 0 5
0 1 3 1 1 3 0 5
0 1 2 1 1 3 0 5
1 2 2 1 3 2 0 5
1 2 3 2 3 1 0 5
0 1 3 2 1 3 0 5
0 1 2 2 3 1 0 5
1 2 2 2 1 1 0 5
1 2 3 2 1 1 0 5
0 2 1 2 3 1 0 5
1 1 1 2 1 1 0 5
0 1 2 1 1 3 0 5
0 1 1 2 3 2 0 5
1 1 2 2 2 1 0 5
0 1 1 2 2 3 0 5
1 1 1 2 1 2 0 5
0 1 1 1 1 1 0 5
1 2 3 1 2 2 0 5
end
label values values values
label def values 1 "diversity", modify
label def values 2 "performance", modify
label values appl formalization
label def formalization 1 "unstandardized", modify
label def formalization 2 "standardized", modify
label def formalization 3 "anonymous", modify
label values train training
label def training 1 "diversity training", modify
label def training 2 "hiring training", modify
label values equota quotas
label values wquota quotas
label def quotas 1 "none", modify
label def quotas 2 "tiebreaker", modify
label def quotas 3 "strict quota", modify
label values female female
label def female 0 "male", modify
label def female 1 "female", modify

Apologies, I included a larger number of cases in the example data. The command I provided above does now work with the example data.

[Katharina Stuckradt]

Thank you both very much, Roman and Simon. Let me dive into both of your approaches and see with which one I can go further.