[R] Is there a faster way to do this?
Hi guys, I am still learning R, and not well familiar with all the apply
functions.
I am trying to find faster alternatives to replace the for cycle.
Is there a faster way to do the example below?
nm - 1000
b - matrix (rnorm (5000, 0, 1), nrow = 500, ncol = nm)
a - matrix (0, nm, nm)
for (i in 1 : nm) {
for (j in 1 : nm) {
if ( j == i) {
next }
a[i, j] - t (b [, i]) %*% b[, j]
}
}
thanks
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Re: [R] Is there a faster way to do this?
Tena koe Marcio
Seems like you are simply multiplying transpose b by b and replacing the
diagonal with 0. If this is correct, then use
a - t(b) %*% b
diag(a) - 0
If this is not a correct interpretation of what you are trying to do, could you
show us with a small reproducible example.
HTH .
Peter Alspach
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Márcio Resende
Sent: Friday, 26 March 2010 2:15 p.m.
To: r-help@r-project.org
Subject: [R] Is there a faster way to do this?
Hi guys, I am still learning R, and not well familiar with all the
apply
functions.
I am trying to find faster alternatives to replace the for cycle.
Is there a faster way to do the example below?
nm - 1000
b - matrix (rnorm (5000, 0, 1), nrow = 500, ncol = nm)
a - matrix (0, nm, nm)
for (i in 1 : nm) {
for (j in 1 : nm) {
if ( j == i) {
next }
a[i, j] - t (b [, i]) %*% b[, j]
}
}
thanks
--
View this message in context: http://n4.nabble.com/Is-there-a-faster-
way-to-do-this-tp1691601p1691601.html
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__
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PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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Re: [R] Is there a faster way to do this?
Márcio, Think matrix! Do you really want b to be 100 copies of the same numbers? You are asking for a strange crossproduct with the main diagonal zeroed out. a2 - crossprod(a) a2[cbind(1:1000, 1:1000)] - 0 all.equal(a, a2) Rich [[alternative HTML version deleted]] __ 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.
Re: [R] Is there a faster way to do it?
You might also want to consider using na.string=9 in the scan().
jim holtman wrote:
Here is a faster way of doing the replacement: (provide reproducible
data next time)
x - matrix(sample(6:9, 64, TRUE), 8)
x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]87767879
[2,]77867677
[3,]77769667
[4,]99768766
[5,]69988989
[6,]97697867
[7,]79897978
[8,]99699886
x.f - 1:8 # replacement values based on column
x.ind - which(x == 9, arr.ind=TRUE)
x.ind
row col
[1,] 4 1
[2,] 6 1
[3,] 8 1
[4,] 4 2
[5,] 5 2
[6,] 7 2
[7,] 8 2
[8,] 5 3
[9,] 6 4
[10,] 7 4
[11,] 8 4
[12,] 3 5
[13,] 8 5
[14,] 5 6
[15,] 7 6
[16,] 1 8
[17,] 5 8
x[x.ind] - x.f[x.ind[,'col']]
x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]87767878
[2,]77867677
[3,]77765667
[4,]12768766
[5,]62388688
[6,]17647867
[7,]72847678
[8,]12645886
On Wed, Oct 28, 2009 at 12:55 PM, Marcio Resende
mresende...@yahoo.com.br wrote:
#Mdarts is a matrix 2343x788
#frequencia is a vector 2343x1
# 9 in Mdarts[fri,frj] stands for my missing values which i want to replace
by the value in the vector frequencia
Mdarts-t(matrix(scan(C:/GWS/CNB/dartg.txt),ncol=nindT,nrow=nm, byrow=T))
frequencia - matrix(scan(C:/GWS/CNB/freq.txt),ncol=1)
for (fri in 1:nindT){
for (frj in 1:nm){
Mdarts[fri,frj] - if (Mdarts[fri,frj] == 9) frequencia[frj] else
Mdarts[fri,frj]
Mdarts[fri,frj] - Mdarts[fri,frj]/1-(frequencia[frj]^2)
}
}
Is there a faster way to it?
Maybe using any apply function?
Thanks in advance
--
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__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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and provide commented, minimal, self-contained, reproducible code.
[R] Is there a faster way to do it?
#Mdarts is a matrix 2343x788
#frequencia is a vector 2343x1
# 9 in Mdarts[fri,frj] stands for my missing values which i want to replace
by the value in the vector frequencia
Mdarts-t(matrix(scan(C:/GWS/CNB/dartg.txt),ncol=nindT,nrow=nm, byrow=T))
frequencia - matrix(scan(C:/GWS/CNB/freq.txt),ncol=1)
for (fri in 1:nindT){
for (frj in 1:nm){
Mdarts[fri,frj] - if (Mdarts[fri,frj] == 9) frequencia[frj] else
Mdarts[fri,frj]
Mdarts[fri,frj] - Mdarts[fri,frj]/1-(frequencia[frj]^2)
}
}
Is there a faster way to it?
Maybe using any apply function?
Thanks in advance
--
View this message in context:
http://www.nabble.com/Is-there-a-faster-way-to-do-it--tp26098223p26098223.html
Sent from the R help mailing list archive at Nabble.com.
__
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.
Re: [R] Is there a faster way to do it?
Here is a faster way of doing the replacement: (provide reproducible
data next time)
x - matrix(sample(6:9, 64, TRUE), 8)
x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]87767879
[2,]77867677
[3,]77769667
[4,]99768766
[5,]69988989
[6,]97697867
[7,]79897978
[8,]99699886
x.f - 1:8 # replacement values based on column
x.ind - which(x == 9, arr.ind=TRUE)
x.ind
row col
[1,] 4 1
[2,] 6 1
[3,] 8 1
[4,] 4 2
[5,] 5 2
[6,] 7 2
[7,] 8 2
[8,] 5 3
[9,] 6 4
[10,] 7 4
[11,] 8 4
[12,] 3 5
[13,] 8 5
[14,] 5 6
[15,] 7 6
[16,] 1 8
[17,] 5 8
x[x.ind] - x.f[x.ind[,'col']]
x
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]87767878
[2,]77867677
[3,]77765667
[4,]12768766
[5,]62388688
[6,]17647867
[7,]72847678
[8,]12645886
On Wed, Oct 28, 2009 at 12:55 PM, Marcio Resende
mresende...@yahoo.com.br wrote:
#Mdarts is a matrix 2343x788
#frequencia is a vector 2343x1
# 9 in Mdarts[fri,frj] stands for my missing values which i want to replace
by the value in the vector frequencia
Mdarts-t(matrix(scan(C:/GWS/CNB/dartg.txt),ncol=nindT,nrow=nm, byrow=T))
frequencia - matrix(scan(C:/GWS/CNB/freq.txt),ncol=1)
for (fri in 1:nindT){
for (frj in 1:nm){
Mdarts[fri,frj] - if (Mdarts[fri,frj] == 9) frequencia[frj] else
Mdarts[fri,frj]
Mdarts[fri,frj] - Mdarts[fri,frj]/1-(frequencia[frj]^2)
}
}
Is there a faster way to it?
Maybe using any apply function?
Thanks in advance
--
View this message in context:
http://www.nabble.com/Is-there-a-faster-way-to-do-it--tp26098223p26098223.html
Sent from the R help mailing list archive at Nabble.com.
__
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.
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
__
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and provide commented, minimal, self-contained, reproducible code.
