Sorry, I didn't have time to check the speed, indeed.
However - isn't apply the same as a loop, just hidden?
D.
On Fri, Apr 6, 2012 at 6:59 PM, ilai ke...@math.montana.edu wrote:
On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
This works great:
Thank you very much, Rui.
This definitely produces the result needed.
Again, I have not checked the speed yet.
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
f - function(x){
nr - nrow(x)
result - matrix(0, nrow=nr, ncol=ncol(x))
colnames(result) - colnames(x)
Preoccupation with speed of execution is typically (certainly not
always) misplaced. Provided you have used sensible basic
vectorization, loops in whatever form work adequately. First get
working code. Then, if necessary, parallelization, byte compilation,
or complex vectorization strategies can
On Sun, Apr 8, 2012 at 8:26 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Sorry, I didn't have time to check the speed, indeed.
However - isn't apply the same as a loop, just hidden?
D.
Yes ?apply is a loop but not the same as ?for, see Intro to R. As
Bert Gunter pointed out
Agreed, Rui provided a very elegant vectorized solution - and I am
very thankful to him. Unfortunately (for myself), I am not as
proficient in vectorization - otherwise, I would not have asked the
question.
Why I did not follow on the original hint to use apply? For this reason (quote):
However -
Hello,
Oops!
What happened to the function 'f'?
Forgot to copy and pasted only the rest, now complete.
f - function(x){
nr - nrow(x)
result - matrix(0, nrow=nr, ncol=ncol(x))
colnames(result) - colnames(x)
inp.ord - order(x)[1:nr] - 1 # Keep only one per row,
Hello,
I maybe missing something but this seems like an indexing problem
which doesn't require a loop at all.
Yes, but with 'order'.
# Original example
input - as.matrix(data.frame(a=c(5,1,3,7), b=c(2,6,4,8)))
(input)
desired.result - as.matrix(data.frame(a=c(100,0,100,0),
Hello, everybody!
I have a matrix input (see example below) - with all unique entries
that are actually unique ranks (i.e., start with 1, no ties).
I want to assign a value of 100 to the first row of the column that
contains the minimum (i.e., value of 1).
Then, I want to assign a value of 100 to
I maybe missing something but this seems like an indexing problem
which doesn't require a loop at all. Something like this maybe?
(input-matrix(c(5,1,3,7,2,6,4,8),nc=2))
output - matrix(0,max(input),2)
output[input[,1],1] - 100
output[input[,2],2] - 100
output
Cheers
On Fri, Apr 6, 2012 at
I think the OP wants to fill values in an arbitrarily large matrix.
Now, first of all, I'd like to know what his real problem is, since this
seems like a very tedious and unproductive matrix to produce. But in
the meantime, since he also left out important information, let's
assume the input
Ok, how's this:
Rgames foo
[,1] [,2] [,3] [,4]
[1,]361 16
[2,] 10 14 125
[3,] 117 159
[4,]84 132
Rgames sapply(1:4,FUN=function(k){
foo[k,which(foo==k,arr.ind=T)[2]]-100;return(foo)})-bar
Rgames bar
[,1] [,2] [,3] [,4]
[1,]3
Yes, that's correct - my matrix has N rows.
Thank you very much, Carl. This works great:
input-as.matrix(data.frame(a=c(5,1,3,7),b=c(2,6,4,8)))
result-input
N-nrow(input)
for (k in 1:N){
foo - which (input == k,arr.ind=T)
result[k,foo[2]] -100
}
result[result !=100]-0
Dimitri
On Fri, Apr
On Fri, Apr 6, 2012 at 4:02 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
This works great:
Really ? surprising given it is the EXACT same for-loop as in your
original problem with counter i replaced by k and reorder to
matrix[!100]- 0 instead of matrix(0)[i]- 100
You didn't
13 matches
Mail list logo
