For the example of the function you gave, it is already 'vectorized':
myfunc - function(x1, x2) {
+ x1 + x2
+ }
myfunc(1:10, 1:10)
[1] 2 4 6 8 10 12 14 16 18 20
outer(1:10, 1:10, myfunc)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,]2345678
Hi All,
I want to do something along the lines of:
for (i in 1:n){
for (j in 1:n){
A[i,j]-myfunc(x[i], x[j])
}
}
The question is what would be the most efficient way of doing this. Would
using functions such as sapply be more efficient that using a for loop?
Note that n can be a
Take a look at 'outer' and vectorized your function. Also look at
'expand.grid'.
On Sunday, November 27, 2011, Sachinthaka Abeywardana
sachin.abeyward...@gmail.com wrote:
Hi All,
I want to do something along the lines of:
for (i in 1:n){
for (j in 1:n){
A[i,j]-myfunc(x[i],
Hi Sachin,
The technique you are suggesting is likely to be just as efficient as
any other if indeed myfunc must be called on each x[i] x[j] element
individually. I would take the additional steps of instantiating A as
a matrix (something like:
A - matrix(0, nrow = n, ncol = n)
Depending, you
Hi Jim,
What exactly do you mean by vectorized. I think outer looks like what I was
looking for. BUT there was a (weighted) distance matrix calculation that I
was trying to vectorize, which wasnt related to this post. Could you proved
a bit more details as to what you were referring to, and maybe
Here is an example, of course, this is predicated on how myfunc()
behaves---if it could not handle adding a constant to a vector, things
would choke:
## Current method
myfunc - function(x1, x2) {
x1 + x2
}
x - 1:10
n - length(x)
A - matrix(0, nrow = n, ncol = n)
for (i in 1:n){
for (j in
Thank you Phil and Bert!
I was sure there must be an efficient way using some kind of
indexing trick but totally did not see the as.matrix solution.
Thanks again
Philipp
On Wed, Nov 05, 2008 at 02:59:20PM -0800, Phil Spector wrote:
Philipp -
res = matrix(NA,5,3)
#reshape package should do it
library(reshape)
foo - data.frame(row=1:5, col=1:3, val=rnorm(15))
cast(foo, row~col)
On Wed, Nov 5, 2008 at 5:47 PM, Philipp Pagel [EMAIL PROTECTED] wrote:
Dear R experts,
Suppose I have a data frame of three variables:
foo - data.frame(row=1:5,
Philipp Pagel wrote:
Dear R experts,
Suppose I have a data frame of three variables:
foo - data.frame(row=1:5, col=1:3, val=rnorm(15))
foo
row col val
11 1 -1.00631642
22 2 0.77715344
33 3 0.17358793
44 1 -1.67226988
55 2 1.08218836
61
])] - foo[,3]
See ?[ for details.
-- Bert Gunter
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of stephen sefick
Sent: Wednesday, November 05, 2008 2:52 PM
To: Philipp Pagel
Cc: r-help@r-project.org
Subject: Re: [R] Efficient way to fill a matrix
#reshape
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