# Re: [R] What's the BEST way in R to adapt this vector?

From: <Jagat.K.Sheth_at_wellsfargo.com>
Date: Sun, 23 Nov 2008 20:00:00 -0600

bEST is up to you to define. Here is one simple way

y.new <- c(t(model.matrix(~factor(y)-1)))

-----Original Message-----
From: r-help-bounces_at_r-project.org [mailto:r-help-bounces_at_r-project.org] On Behalf Of zerfetzen
Sent: Saturday, November 22, 2008 12:00 PM To: r-help_at_r-project.org
Subject: [R] What's the BEST way in R to adapt this vector?

Goal:
Suppose you have a vector that is a discrete variable with values ranging from 1 to 3, and length of 10. We'll use this as the example:

y <- c(1,2,3,1,2,3,1,2,3,1)

...and suppose you want your new vector (y.new) to be equal in length to the possible discrete values (3) times the length (10), and formatted in such a way that if y == 1, then y.new[1:3] == c(1,0,0), and if y == 2, then y.new[4:6] == c(0,1,0). For example, the final goal should be:

y.new <- c(1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,1,0,0,0,1,0,0,0,1,1,0,0)

Note: I know how to do this with loops, but that's not taking advantage of R's capabilities with vectors and, I suspect, matrices.

So far, my best guess would be to start as follows:

```y1 <- ifelse(y == 1, 1, 0)
y2 <- ifelse(y == 2, 1, 0)
y3 <- ifelse(y == 3, 1, 0)

```

>into

y.new? Is that even the most efficient way? If it is, I'm sure I can get them into a matrix, but how do I read them out correctly? Thanks for any input.

```--
View this message in context:
p20638991p20638991.html
Sent from the R help mailing list archive at Nabble.com.

______________________________________________
R-help_at_r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.

______________________________________________
R-help_at_r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help