Hi:
I had Harold's idea (matrix indexing), but I was curious to see which of
these ran fastest. I simulated
1000 rows and three columns of binary data, along with a fourth column that
sampled the values 1:3
1000 times. Here are the timings:
> f <- as.data.frame(matrix(rbinom(3000, 1, 0.4), nrow = 1000))
> names(f) <- LETTERS[1:3]
> f$D <- sample(1:3, 1000, replace = TRUE)
> system.time(E1 <- f[cbind(1:nrow(f), f$D)])
user system elapsed
0 0 0
> system.time(E2 <- apply(f, 1, function(x) x[eval(x)["D"]]))
user system elapsed
0.03 0.00 0.03
> system.time(E3 <- diag(as.matrix(f[f$D])))
user system elapsed
0.26 0.03 0.30
> identical(E1, E2)
[1] TRUE
> identical(E2, E3)
[1] TRUE
HTH,
Dennis
On Wed, Jun 9, 2010 at 7:03 AM, Malcolm Fairbrother <
m.fairbrot...@bristol.ac.uk> wrote:
> Dear all,
>
> I have a data frame f, with four variables:
>
> f <- data.frame(A=c(0,0,1,1), B=c(0,1,0,1), C=c(1,1,0,1), D=c(3,1,2,3))
> f
> A B C D
> 1 0 0 1 3
> 2 0 1 1 1
> 3 1 0 0 2
> 4 1 1 1 3
>
> I want to create a new variable (f$E), such that each of its elements is
> drawn from either f$A, f$B, or f$C, according to the value (for each row) of
> f$D (values of which range from 1 to 3).
>
> In the first row, D is 3, so I want the value from the third variable (C),
> which for the first row is 1. In the second row, D is 1, so I want the value
> from the first variable (A), which for the second row is 0. And so forth,
> such that in the end my new data frame looks like:
>
> A B C D E
> 1 0 0 1 3 1
> 2 0 1 1 1 0
> 3 1 0 0 2 0
> 4 1 1 1 3 1
>
> My question is: How do I do this for a much larger dataset, where my "index
> variable" (f$D in this example) actually indexes a much larger number of
> variables (not just three)?
>
> I know that in principle I could do this with a long series of nested
> ifelse statements (as below), but I assume there is some less cumbersome
> option, and I'd like to know what it is. Any help would be much appreciated.
> Apologies if I'm missing something obvious.
>
> f$E <- ifelse(f$D==3, f$C, ifelse(f$D==2, f$B, f$A))
>
> Thanks,
> Malcolm
>
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