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Re: st: nested loop for mean and median


From   Nick Cox <njcoxstata@gmail.com>
To   "statalist@hsphsun2.harvard.edu" <statalist@hsphsun2.harvard.edu>
Subject   Re: st: nested loop for mean and median
Date   Mon, 29 Apr 2013 15:26:19 +0100

Lots of problems here. Some arise from just not reading the documentation.

1. What is throwing you out is a type mismatch. At a guess, one or
more of -fyear- -sic- -country- is actually a string variable. I can't
tell you which: look at -describe-.

2. -summarize- does not calculate the median except with the -detail-
option. When it does it saves it as r(p50). Why did you type
r(median)?

3. For each variable, second time round the second loop

gen mean_Assets = ...

for example will fail because the variable already exists.

Something like this is much easier:

foreach v in  Assets Liab Income {
        egen mean_`v' = mean(`v'), by(country year sic)
        egen median_`v' = median(`v'), by(country year sic)
}

Nick
njcoxstata@gmail.com


On 29 April 2013 15:15, Nahla Betelmal <nahlaib@gmail.com> wrote:
> Dear Statalis,
>
> I need to get the mean and median for some firms in a certain year
> (2000-2012) and certain industry (73 industries). e.g. the mean and
> median for assets for the firms operate in industry 6 in year 2004,
> and so on. I tried the following nested loop but I got error (109).
>
> p.s. I need the mean and median as new variables as they will be used
> in further calculations.
>
>
>
>
> foreach v of var  Assets Liab Income {
>         forval y=2000/2012 {
>                          forval  i= 1/73 {
>                                 qui su `v' if  fyear==`y'& sic==`i'& country==1
>                                 gen mean_`v' = r(mean)
>                                 gen median_`v' = r(median)
> }
> }
> }
>
> type mismatch
> r(109);
>
>
> Many thanks
>
> Nahla Betelmal
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