Specifying and calling a mata function inside stata loop

[Johannes Eigner]

Dear Statalisters,

I want to call a mata function inside a stata-loop since calling mata inside the stata loop will not be feasible.
I understand that I have to define the mata function outside the loop. The function should take up 25 variables with 25 observations each, create a Matrix A, and form an inverse L of that matrix.
Then I would like to store the columns of matrix L either as new variables in Stata or make them accessible with "getmata". Does the latter even work?

My code looks like this:

mata
function inversion() // The variables in the loop will always have the same name but different values. Therefore I tried to "hardcode" the function without arguments
{
    A = st_data(.,("var1","var2","var3") // Instead of 3 variables I specify 25 here.
    L = invsym(I(25)-A) // This creates the inversion matrix L
    return L  // Here the code throws an error
}
end

When I want to implement the function I get the error

'return' found where almost anything else expected

Inside the loop I want to call the function as follows

forvalues z=1(1)10000 {

// Some code

putmata var1 var2 var3, replace // all 25 variables

inversion() // calling the function


getmata (new_var1 new_var2 new_var3)=L, replace // storing the columns of L in 25 new variables.

I am new to Mata programming and grateful for any hints.

Best wishes,
Johannes Eigner

[Joseph Coveney]

Well, I believe return L would be return(L). And then, it will be problematic calling directly from Stata a function that returns something. You'd probably want to wrap your Mata code in a Stata program (in an ADO-file, probably, and one that would deliver everything neat-and-tidy back to Stata when the program exits) and call that in your Stata loop.

By the way, doesn't Stata, itself, have a invsym(),as well as a svmat?

[Johannes Eigner]

Thank you very much Joseph! I out-sourced the mata bit of the loop in a do-file which I called from within the loop. This works perfectly fine. I haven't tried to do it all in Stata but I will try it next time.

[Christophe Kolodziejczyk]

You don't have to define the mata function inside the loop. You define it once and then call it from the Stata loop. Below some suggestions of modification to your code. I think it would be more efficicent to store the whole data in Mata (not just 25 observations) and then make the loop in Mata, where you select 25 observations, instead of using putmata and getmata back and forth (10000 times). But I can't figure out exactly what you are trying to achieve, so I might well be mistaken .

clear mata
mata
real matrix function inversion() // The variables in the loop will always have the same name but different values. Therefore I tried to "hardcode" the function without arguments
{
    A = st_data(.,("var1","var2","var3") // Instead of 3 variables I specify 25 here.
    L = invsym(I(25)-A) // This creates the inversion matrix L
    return(L)  // Here the code throws an error
}
end

forvalues z=1(1)10000 {

// Some code

putmata var1 var2 var3, replace // all 25 variables

mata: L=inversion() // calling the function


getmata (new_var1 new_var2 new_var3)=L, replace // storing the columns of L in 25 new variables.
}

[Joseph Coveney]

. . .
forvalues z=1(1)10000 {

. . .

mata: L=inversion() // calling the function

. . .
}

I don't think that you can do that. I believe that you'll get an error if you try to call from Stata a Mata function that returns a matrix and also try to assign it to a Stata name like that.

You can do stuff like this

mata: inversion()

but that will just display the returned matrix.

I think that trying to assign a returned result (string or real scalar or matrix) from a Mata function to a token in Stata with a statement like

mata: my_Stata_name = my_Mata_function()

will give you an error.

[Christophe Kolodziejczyk]

Joseph
But L is not a Stata name, but a Mata object

the following code works fine

clear mata
mata:
real matrix foo()
{

return(1)
}

end

forval i = 1/10 {
mata: L = foo()
getmata (new_var1)=L , replace
}

[Joseph Coveney]

I'm mistaken. Thanks for setting me straight on this, Christophe.