From: Simon Cullen On Mon, 26 Jan 2004 20:15:51 +0100, <uaca at alumni.uv.es> wrote:
I want to conditionally operate on certain elements of a
matrix, let me
explain it with a simple vector example
z<- c(1, 2, 3) zz <- c(0,0,0) null <- (z > 2) & ( zz <- z) zz
[1] 1 2 3 why zz is not (0, 0, 3) ?????
<snip>
in the other hand, it curious that null has reasonable values
null
[1] FALSE FALSE TRUE
What you have done there is create a boolean vector, null, of the same length as z (and zz). For instance: (z > 2) & (zz <- z) =(F F T) & (T T T) (as assignment - presumably - returns T)
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Don't think so. zz <- z has the value z; i.e., c(1, 2, 3). When evaluated as logicals, non-zero values are treated as true (as in C), I believe. For example:
z <- rep(0, 3) ifelse(zz <- z, 1, 0)
[1] 0 0 0
(zz <- z) == TRUE
[1] FALSE FALSE FALSE However, what tripped me is the fact that even though non-zero is logically `true', it's not necessarily equal to TRUE (which is numerically equal to 1):
z <- 0:2 (zz <- z) == TRUE
[1] FALSE TRUE FALSE
ifelse(zz <- z, 1, 0)
[1] 0 1 1
if(3) TRUE else FALSE
[1] TRUE Andy
=(F F T). What will work is: z <- c(1, 2, 3) index <- z>2 zz <- z * index -- SC Simon Cullen Room 3030 Dept. Of Economics Trinity College Dublin Ph. (608)3477 Email cullens at tcd.ie
------------------------------------------------------------------------------
Notice: This e-mail message, together with any attachments,...{{dropped}}