Re: [Gretl-users] loop question
El 30/11/11 13:06, Giuseppe Vittucci escribió:
Clearly, you can follow Jack's suggestion.
In my suggestion, i.e.:
loop foreach i 48 60 100 250 400 780
...
endloop
i is taken as a string.
But you can always recall it inside the loop using $i:
- defining a new scalar. E.g:
loop foreach i 48 60 100 250 400 780
scalar j = $i
...
endloop
Thanks, Giuseppe, that is a rather smart solution, much more simple than
what I was trying todo.
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