Thank you Maarten and Nick for the help and codes. And thanks Nick for
the detailed corrections. I appreciate it.
Thanks again
Nahla
On 29 April 2013 15:26, Nick Cox <njcoxstata@gmail.com> wrote:
> 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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