> From: "Gabor Grothendieck" <[EMAIL PROTECTED]>
> Date: Wed, 17 Dec 2003 15:02:49 -0500 (EST)
>
> Define function f to take a vector as input representing
> a single input row. f should (1) transform this to a vector
> representing the required row of output or else (2) produce
> NULL if no row is to be output for that input row.
>
> Then use this code where z is your input matrix:
>
> t( matrix( unlist( apply( z, 1, f ) ), 2) )
>
But as has been pointed out recently, apply really is still just a for
loop.
> > From: Adrian Dusa <[EMAIL PROTECTED]>
> > Date: Wed, 17 Dec 2003 21:28:05 +0200
> >
> > I have a (rather theoretical) programming problem for which I have found
> > a solution, but I feel it is a rather poor one. I wonder if there's some
> > other (more clever) solution, using (maybe?) vectorization or
> > subscripting.
Here is a subscripting solution, where (for consistency with above) z is
your data [from read.table(filename, header=T)]:
> z
rel1 rel2 rel3 age0 age1 age2 age3 sex0 sex1 sex2 sex3
1 1 3 NA 25 23 2 NA 1 2 1 NA
2 4 1 3 35 67 34 10 2 2 1 2
3 1 4 4 39 40 59 60 1 2 2 1
4 4 NA NA 45 70 NA NA 2 2 NA NA
> res <- matrix(NA, nrow=length(z[, 1]), ncol=2,
dimnames=list(rownames=rownames(z), colnames=c("ageh", "agew")))
> w <- w0 <- w1 <- w2 <- which(z[, c("rel1", "rel2", "rel3")] == 1, T)
# find spouse entries
> w0[, 2] <- z[, "sex0"][w[, 1]] # indices for respondent's age
> w1[, 2] <- 3 - w0[, 2] # indices for spouse's age
> w2[, 2] <- 4 + w[, 2] # indices of spouse's age
> res[w0] <- z[, "age0"][w[, 1]] # set respondent's age
> res[w1] <- z[w2] # set spouse's age
> res
colnames
rownames ageh agew
1 25 23
2 34 35
3 39 40
4 NA NA
>
Ray Brownrigg
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