Dear Martin:
your proposal will result in the 3-year-mean of means; but I want to get the 3-year-mean of (yearly) medians (across observations). This is why I cannot use meanonly; I would need "medianonly" for your solution, which unfortunately doesn't exist.
Please correct me if I'm wrong.
This is why I want to do it like this, but I don't kow what is wrong (Stata says: if not allowed r(101))
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'
}
}
Hope someone could tell me..Thanks.
-------- Original-Nachricht --------
> Datum: Fri, 3 Oct 2008 14:52:12 +0200
> Von: "Martin Weiss" <[email protected]>
> An: [email protected]
> Betreff: Re: st: Re: creating average over past three years within groups
> True, consecutive -summarize-s will overwrite the rclass results so you
> have
> to store the r(mean) them in -local-s. Apart from that, I do not know what
> is wrong with my proposal...
>
> HTH
> Martin
> _______________________
> ----- Original Message -----
> From: "Fabian Brenner" <[email protected]>
> To: <[email protected]>
> Sent: Friday, October 03, 2008 2:40 PM
> Subject: Re: st: Re: creating average over past three years within groups
>
>
> > Dear Martin,
> >
> > I don't know if I got what you mean, but as far as I know your proposal
> > using -summarize, meanonly- would lead to the average (across the most
> > recent three years) of the mean (across observations within each year).
> >
> > But I want to get the average (across the most recent three years) of
> the
> > median (across observations within each year), i.e.
> > average1997=(median1997+median1996+median1995)/3
> >
> > Hope someone could help me.
> > Fabian
> >
> >
> > [I guess your proposal would look like the following:
> > quietly forvalues k = 1/10 {
> > quietly forvalues y = 1974/2000 {
> > summarize age if year == `= `y' - 3' & group == `k', meanonly
> > local median3 = r(mean)
> > summarize roe if year == `= `y' - 2' & group == `k', meanonly
> > local median2 = r(mean)
> > summarize roe if year == `= `y' - 1' & industry == `k', meanonly
> > replace fiveyearmedian = (`median3' + `median2' + r(mean1 ) / 3 if year
> ==
> > `y' & group == `k'
> > }
> > }
> >
> >
> >
> > -------- Original-Nachricht --------
> >> Datum: Fri, 3 Oct 2008 14:22:16 +0200
> >> Von: "Martin Weiss" <[email protected]>
> >> An: [email protected]
> >> Betreff: st: Re: creating average over past three years within groups
> >
> >> Not sure what you are doing with all those -locals-. Just -summarize,
> >> meanonly- with your -if- statements and type -ret li- afterwards. There
> >> you
> >> will find the average as r(mean) to plug into your -replace- operation.
> >>
> >> HTH
> >
> >> Martin
> >> _______________________
> >> ----- Original Message -----
> >> From: "Fabian Brenner" <[email protected]>
> >> To: <[email protected]>
> >> Sent: Friday, October 03, 2008 2:17 PM
> >> Subject: st: creating average over past three years within groups
> >>
> >>
> >> > 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
> >> >
> >> >
> >> > --
> >> > GMX startet ShortView.de. Hier findest Du Leute mit Deinen
> Interessen!
> >> > Jetzt dabei sein:
> >> > http://www.shortview.de/wasistshortview.php?mc=sv_ext_mf@gmx
> >> > *
> >> > * For searches and help try:
> >> > * http://www.stata.com/help.cgi?search
> >> > * http://www.stata.com/support/statalist/faq
> >> > * http://www.ats.ucla.edu/stat/stata/
> >> >
> >>
> >>
> >> *
> >> * For searches and help try:
> >> * http://www.stata.com/help.cgi?search
> >> * http://www.stata.com/support/statalist/faq
> >> * http://www.ats.ucla.edu/stat/stata/
> >
> > --
> > GMX startet ShortView.de. Hier findest Du Leute mit Deinen Interessen!
> > Jetzt dabei sein:
> > http://www.shortview.de/wasistshortview.php?mc=sv_ext_mf@gmx
> > *
> > * For searches and help try:
> > * http://www.stata.com/help.cgi?search
> > * http://www.stata.com/support/statalist/faq
> > * http://www.ats.ucla.edu/stat/stata/
> >
>
>
> *
> * For searches and help try:
> * http://www.stata.com/help.cgi?search
> * http://www.stata.com/support/statalist/faq
> * http://www.ats.ucla.edu/stat/stata/
--
GMX Kostenlose Spiele: Einfach online spielen und Spa� haben mit Pastry Passion!
http://games.entertainment.gmx.net/de/entertainment/games/free/puzzle/6169196
*
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/
