Thanks Nick, David, and Euler, for your responses. Both of Nicks solutions
solve my problem so to close.
My problem I have 3 variables
+--------------------------------+
p1 p2 p3
--------------------------------
.0193647 .7684672 .2121681
.0110648 .0962676 .8926677
.00574 .8762589 .1180012
.0045997 .1808127 .8145875
.0024693 .3957042 .6018265
--------------------------------
.0067998 .3438436 .6493566
.0113164 .0771485 .9115351
.0123369 .133595 .854068
.0103629 .2671735 .7224636
.0085421 .1814262 .8100317
+--------------------------------+
I need to generate a 4th variable that takes on the values 1 2 and 3 if p1
p2 p3 are the largest in the group respectively (I am not concerned about
ties at the moment because they have not occured)
A specific fix (Nick1)
gen pred = 1
replace pred = 2 if p2 == max(p1,p2,p3)
replace pred = 3 if p3 == max(p1,p2,p3)
A more general fix for pn (where n is the p1 p2 p3 ....pn) (Nick2)
gen pred = .
forval i = 1/n {
replace pred = `i' if p`i' == max(p1,p2,..pn)
}
Thanks (I will add that they are both neat fixes)
Ronnie
-----Original Message-----
From: [email protected]
[mailto:[email protected]]On Behalf Of Nick Cox
Sent: 8. april 2003 14:53
To: [email protected]
Subject: st: RE: max value
Ronnie Babigumira
>
> Hi (stata8, win2k)
> I have 3 variables p1 p2 p3. I need to generate a new variable pred
>
> "pred" ==
> 1 if p1 is the largest of the 3,
> 2 if p2 is the largest
> and 3 if p3 is the largest
>
> Any elegant ideas (thats all I seem to get from the list
> and for that I am
> forever grateful)
None of these is very elegant.
gen pred = 1
replace pred = 2 if p2 == max(p1,p2,p3)
replace pred = 3 if p3 == max(p1,p2,p3)
gen pred = .
forval i = 1/3 {
replace pred = `i' if p`i' == max(p1,p2,p3)
}
gen id = _n
reshape long p, i(id)
bysort id (p) : gen pred = _j[_N]
reshape wide
Are you worried about ties?
Nick
[email protected]
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