Fabian,
the error message you get relates to the -local- statements. -local-
does not allow -if-. A local macro acts like a placeholder and not
like a variable.
Take a look at your code:
local median3 = oneyearmedian if year == `= `y' - 3' & industry == `k',
First, you don't need the `=`y'-3' construct here, and the comma at
the end is illegal. More importantly, you ask Stata to perform
something in one step for which it needs two steps:
- find out what the median is for industry k three years ago,
- store this in a local macro.
You can do it like this:
sum oneyearmedian if year==`y'-3 & industry == `k'
local median3 = r(mean)
However, you could get all of this a lot easier by calculating the
three year average median for one observation per year and industry,
and then propagating it along. I assume you have some panel identifier
called id.
sort year industry id
by year industry: gen first = _n==1
sort first industry year
by first industry: gen threeyearmean = (oneyearmedian[_n-1] +
oneyearmedian[_n-2] + oneyearmedian[_n-3])/3 if first==1
bysort year industry (first) : replace threeyearmean = threeyearmean[_N]
(Untested.)
Hope this helps,
Eva
2008/10/3 Fabian Brenner <[email protected]>:
> Dear all:
>
> I used -bysort year group : egen oneyearmedian = median(age)- to create the median within each group (1-10) for each year (1974-2000).
>
> Now I want to get the three year average of the three median for each group for the most recent three years:
>
> generate threeyearaverage = .
> sort year group
>
> quietly forvalues k = 1/10 {
> quietly forvalues y = 1974/2000 {
> local median3 = oneyearmedian if year == `= `y' - 3' & industry == `k',
> local median2 = oneyearmedian if year == `= `y' - 2' & industry == `k',
> local median1 = oneyearmedian if year == `= `y' - 1' & industry == `k',
> replace fiveyearmedian = (`median3' + `median2' + `median1' ) / 3 if year == `y' & group == `k'
> }
> }
>
> But stata always says: if not allowed r(101);
>
> What do I wrong? Hope someone can help me.
> Fabian Brenner
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