> From: Simon Cullen > On Mon, 26 Jan 2004 20:15:51 +0100, <[EMAIL PROTECTED]> 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 [EMAIL PROTECTED] ------------------------------------------------------------------------------ Notice: This e-mail message, together with any attachments,...{{dropped}} ______________________________________________ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html