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Re: st: e(wexp) versus e(wexp): different routines return different things


From   Stas Kolenikov <[email protected]>
To   [email protected]
Subject   Re: st: e(wexp) versus e(wexp): different routines return different things
Date   Sat, 27 Aug 2011 14:28:12 -0500

Oh those overqualified programmers working in Stata :). Regular
expressions are an overkill for situation like this. You can achieve
what you need with the string functions.

tempvar wvar
if "`e(wexp)'" == "" generate byte `wvar' = 1
else {
  local mywexp = subsinstr("`e(wexp)'","=","",1)
  generate double `wvar' = `mywexp'
}

and then you can pass `wvar'.

On Fri, Aug 26, 2011 at 8:49 PM, Rodini, Mark
<[email protected]> wrote:
> Greetings,
>
> I have a program which executes after I run an estimation procedure, and it does a collapse where weighting is an option.
> Within the program is the following line:
>
>
> if "`e(wexp)'" != "" {
>                        collapse (mean) `y_sample' `xb_sample' [fw `e(wexp)']
>                }
>
> I'm running an estimation using "reg" and one using "newey2".  Suppose the name of the weighting variable, should I opt to use it, is "mycount" so for example, I run:
>
> reg y high low cows [aw=mycount]
>
> and then execute the program.
>
>
> Both estimation routines create as output an estimation "variable" called e(wexp).  Note that this is what is passed to the program as indicated above.
>
> Here is the kicker: reg returns e(wexp) as "= mycount", but newey2 returns e(wexp) as "mycount" (without the equals sign!)
>
> The reg version properly executes the program above, but newey2 gives an error about an inability to weight, since the syntax requires an equals sign.
>
> I have tried within the program to create a tempname or tempvar, assign e(wexp) to it and then tried running
>
> scalar `wts'=regexr(`e(wexp)',"=","")
>
> I then replace the `e(wexp)' in the program with `wts' and add an equals sign explicitly in the program.  The idea I was hoping for is that it would replace the "=" with nothing in the macro variable, if one were there.
>
> No matter how I try it, it fails, usually with a type mismatch error.  I tried adding double quotes, etc.  I am assuming that because I'm trying to pass it as a scalar, that is what bombs it.  Any thoughts?  I'm guessing it's something pretty basic --I'm kind of new to writing complicated programs which pass lots of stuff.



-- 
Stas Kolenikov, also found at http://stas.kolenikov.name
Small print: I use this email account for mailing lists only.

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