Thanks Nick. This works great. I didn't use the newvar2 since I really
need only the three outcomes to do further testing.
Tonya
From "Nick Cox" <[email protected]>
To <[email protected]>
Subject st: RE: Assigning value to highest rank
Date Mon, 24 May 2004 12:24:36 +0100
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There may be a neater way than this rather
awkward grinding through cases:
First you can generate a row maximum by
gen max = max(pl,pd,pu)
Then the number of values equal to the maximum is
gen nmax = (pd == max) + (pu == max) + (pl == max)
The case of a unique maximum is
gen newvar = (pd == max) + 0.5 * (pu == max) if nmax == 1
Then your ties between two are
replace newvar = cond(uniform() < 0.5, 1, 0.5) if pd == max & pu == max
replace newvar = cond(uniform() < 0.5, 0.5, 0) if pu == max & pl == max
replace newvar = cond(uniform() < 0.5, 1, 0) if pd == max & pl == max
Presumably the case of three ties needs addressing
gen random = uniform()
replace newvar = cond(random < 1/3, 1, cond(random < 2/3, 0.5, 0)) if nmax
== 3
Any interest in
gen newvar2 = ((pd == max) + 0.5 * (pl == max)) / nmax
as a simpler scoring system? That way,
pd is unique max => 1
pl is unique max => 0.5
pu is unique max => 0
pd and pl tie => 0.75
pd and pu tie => 0.5
pl and pu tie => 0.25
pd, pl, pu tie => 0.5
Nick
[email protected]
Tonya Cropper
>
> I have three variables within each observation that I am
> trying to assign a
> value for a new variable based on which of my three variables have the
> highest value. If the highest value is pl, I would like to
> assign newvar =
> 0, if the highest value is pd: newvar = 1, or if the highest
> value is pu:
> newvar = 0.5. However, if there's a tie for the highest
> value, I would like
> to randomly choose the winner and then assign the value as above.
>
>
>
> Here's an example of the data and what I would want newvar to
> indicate. I
> would greatly appreciate any assistance. Thank you.
>
>
>
> Id pd pl pu newvar
>
> 117 11 11 8 (1 or 0
> based on whether
> pd or pl win toss)
>
> 118 12 8 10 1
>
> 120 18 10 14 1
>
> 121 6 11 7 0
>
> 122 4 9 9 (0 or
> 0.5 based on
> whether pl or pu win toss)
>
> 123 6 8 10 0.5
>
> 126 9 9 12 0.5
>
> 127 10 7 10 (1 or
> 0.5 based on
> whether pd or pu win toss)
>
> 129 9 11 10 0
>
> 131 9 5 4 1
>
> 132 4 5 9 0.5
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