On 2014/8/11 14:50, William Dunlap wrote:
You can use
m[m > 0 & m <= 1.0] <- 1
m[m > 1 ] <- 2
or, if you have lots of intervals, something based on findInterval(). E.g.,
m[] <- findInterval(m, c(-Inf, 0, 1, Inf)) - 1
(What do you want to do with non-positive numbers?)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
Thank you very much.
I think findInterval() is what I want.
Regards,
Jinsong
On Mon, Aug 11, 2014 at 2:40 PM, Jinsong Zhao <jsz...@yeah.net> wrote:
Hi there,
I hope to replace a range of numeric in a matrix with a integer. For
example, in the following matrix, I want to use 1 to replace the elements
range from 0.0 to 1.0, and all larger than 1. with 2.
(m <- matrix(runif(16, 0, 2), nrow = 4))
[,1] [,2] [,3] [,4]
[1,] 0.7115088 0.55370418 0.1586146 1.882931
[2,] 0.9068198 0.38081423 0.9172629 1.713592
[3,] 1.5210150 0.93900649 1.2609942 1.744456
[4,] 0.3779058 0.03130103 0.1893477 1.601181
so I want to get something like:
[,1] [,2] [,3] [,4]
[1,] 1 1 1 2
[2,] 1 1 1 2
[3,] 2 1 2 2
[4,] 1 1 1 2
I wrote a function to do such thing:
fun <- function(x) {
if (is.na(x)) {
NA
} else if (x > 0.0 && x <= 1.0) {
1
} else if (x > 1.0) {
2
} else {
x
}
}
Then run it as:
apply(m,2,function(i) sapply(i, fun))
However, it seems that this method is not efficient when the dimension is
large, e.g., 5000x5000 matrix.
Any suggestions? Thanks in advance!
Best regards,
Jinsong
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
