On Tue, 29 Jun 2010, Sven Schreiber wrote:
> Am 29.06.2010 10:46, schrieb Marcus Marktanner:
> > Please apologize my ignorance, but I hope that someone can help me. I would
> > like to classify a country as either one with some kind of a green
> > revolution or not [...]
>
> Normally I would start by saying please start a new thread for a new
> topic, but here by coincidence the solution is probably similar to the
> one in the thread that you replied to.
>
> So something like:
>
> genr time
> matrix greencountries = {}
> numofunits = max($unit)
> loop for i=1..numofunits
> smpl $unit=i --restrict
> smpl 1961 1980 # not sure if this works here [...]
Sven's idea is fine, but that last "smpl" line will work only for
time-series data. Let's suppose you have a variable called "year"
in your data set which identifies the year. (If you don't already,
you can easily construct one, e.g. "year = time + 1960").
Here's a tested version of Sven's idea which fakes what you want
to do (as I understand it), using the panel dataset greene14_1.gdt
that's supplied with gretl:
<script>
open greene14_1.gdt
genr time
matrix greencountries = {}
numofunits = max($unit)
loop for i=1..numofunits
# first time sub-sample
smpl ($unit==i && year>=1970 && year<=1978) --restrict
ols Q const time --quiet
temp1 = $coeff(time)
# second time sub-sample
smpl ($unit==i && year>1978) --restrict --replace
ols Q const time --quiet
temp2 = $coeff(time)
green = (temp2 > temp1)
greencountries |= green
smpl full
endloop
print greencountries
</script>
Allin Cottrell
