On Sun, 28 Mar 2010, Economics Guy wrote:
It is really the same thing. One of many possibilities:
theFrame <- data.frame(theValues=runif(150,-10,10))
exact <- diag(15)[1+ (rank(theFrame$theValues)-1)%/%10,]
not.so.exact <- diag(15)[1+
(rank(theFrame$theValues+runif(150,0,3))-1)%/%10,]
If what you actually wanted was one factor with fifteen levels, just wrap
the subscript in the last assignment in factor() instead.
HTH,
Chuck
Thanks Chuck, this does what I asked for:
theValues <- runif(150,-10,10)
exact <- factor(1+(rank(theValues)-1)%/%100)
Unfortunately, it looks like my example may have been too contrived
for my actual program.
In the solution that Chuck proposed he antisipated that I may not want
the assignment of factor levels to be exact:
notSoExact <- factor(1+(rank(theValues+runif(150,0,3)-1)%/%100)
This is close to what I need. However, in the real program what I need
to be able to do is precisely vary the degree of exactness in the
assignment. So I need to be able to have the assignment range from
completely random assignment to the exact assignment that "exact"
above provides.
Anyone think of a simple way to do this?
That's div 10, not div 100 (check your typing).
The range of the runif() call determines the 'exactness'.
Alternatively, apply weights of a and 1-a, for 0<=a<=1, to the components
in rank().
Chuck
Thanks Agan,
That Economics Guy
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Charles C. Berry (858) 534-2098
Dept of Family/Preventive Medicine
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