Re: [R] help with matrix
On 19/07/2013 10:17 AM, Kátia Emidio wrote: Hi, I am a beginner at R and I would like to get information from a matrix, where the numbers are larger than or equal to 50m . This means that plants need to be apart at least 50m. I need to get the selection as in a way that I can identify which pair of species are 50m or more apart. here a peace of my matrix: Nauc_calo_1 Nauc_calo_2 Nauc_calo_3 Nauc_calo_1 0 22 36 Nauc_calo_2 22 0 165.8 Nauc_calo_3 36 165.8 0 The which() function with arr.ind=TRUE will identify which matrix entries meet a condition. For example, M - matrix(rnorm(20), 4, 5) M [,1] [,2][,3][,4] [,5] [1,] -0.6947070 1.2079620 0.77996512 -0.04287046 -1.5487528 [2,] -0.2079173 -1.1231086 -0.08336907 1.36860228 0.5846137 [3,] -1.2653964 -0.4028848 0.25331851 -0.22577099 0.1238542 [4,] 2.1689560 -0.4666554 -0.02854676 1.51647060 0.2159416 which(M 1.5, arr.ind=TRUE) row col [1,] 4 1 [2,] 4 4 Duncan Murdoch __ 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 with matrix
Hi, I am a beginner at R and I would like to get information from a matrix, where the numbers are larger than or equal to 50m . This means that plants need to be apart at least 50m. I need to get the selection as in a way that I can identify which pair of species are 50m or more apart. here a peace of my matrix: Nauc_calo_1 Nauc_calo_2 Nauc_calo_3 Nauc_calo_1 0 22 36 Nauc_calo_2 22 0 165.8 Nauc_calo_3 36 165.8 0 Thanks -- Kátia Emídio da Silva DSc Eng. Florestal Manaus/AM Forestry Engineer Manaus/AM-Brazil [[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] help on matrix column removal based on another matrix results
Hello,
Try removing the square parenthesis from the left hand side,
Vsim - Vsim[, NSErr 0.6]
Hope this helps,
Rui Barradas
Em 20-11-2012 04:07, Irucka Embry escreveu:
Hi Rui, how are you?
I want to thank you for your assistance again.
I'm sorry, but the code that you provided for me did not work this time.
The code and the warning message are both below:
NSErr - t(matrix(NSEr)) # which is a 1x1000 double matrix
Vsim[] - Vsim[ , NSErr 0.6]
Warning message:
In Vsim[] - Vsim[, NSErr 0.6] :
number of items to replace is not a multiple of replacement length
Thanks again.
Irucka
-Original Message-
From: Rui Barradas [ruipbarra...@sapo.pt]
Sent: 11/19/2012 10:35:00 AM
To: iruc...@mail2world.com
Cc: r-help@r-project.org
Subject: Re: [R] help on matrix column removal based on another matrix
results
Sorry, the comma is in the wrong place, it should be
Vsim[] - Vsim[ , NSErr 0.6]
Rui Barradas
Em 19-11-2012 16:18, Rui Barradas escreveu:
Hello,
Try
Vsim[] - Vsim[NSErr 0.6, ]
Hope this helps,
Rui Barradas
Em 19-11-2012 14:30, iembry escreveu:
Hi everyone, now I am trying to finish writing the code (I had asked
for
assistance on subtracting arrays)
This is what I what I am running in R:
source(/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Exampl
e/NSEr.R)
NSEr - function (obs, sim)
{
{jjh - (as.vector(obs) - sim)^2
Xjjhs - apply(Xjjh, 2, sum)
Yii - (obs - mean(obs))^2
Yiis - apply(Yii, 2, sum)
NSEr - 1 - (Xjjhs/Yiis)
}
NSEr}
Vsim - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,)
Vsim - as.matrix(Vsim[,-1]) # remove column 1 from analysis
Vobs - read.csv(Observed_Flow.csv, header = TRUE, sep =,)
Vobs - as.matrix(Vobs[,-1]) # remove column 1 from analysis
NSEr - NSEr(Vobs,Vsim);
write.table(NSEr, NSEr.csv, sep =,)
NSErr - t(matrix(NSEr))
## select the behavioural simulations and discard the rest
Vsim - Vsim[NSErr 0.6]
write.table(Vsim, Vsim.csv, sep =,)
**Vsim becomes numeric[42016] rather than a double matrix of
101x416.
What is the proper way to remove the columns in Vsim where the NSEr
for that
column is less than 0.6? I am trying to make Vsim a double matrix of
101x416.
Thank-you again.
Below is the rest of the code in R:
## normalise Qsim and compute the quantiles
NSEr - NSEr[NSEr 0.6]
write.table(NSEr, NSEr_great_0.6.csv, sep =,)
NSEr - NSEr - 0.6
write.table(NSEr, NSEr_minus0.6.csv, sep =,)
NSEr - NSEr/sum(NSEr)
write.table(NSEr, NSEr_normalized.csv, sep =,)
#NSEr = sum(NSEr)
limits - apply(Vsim, 1, wtd.quantile, weights = NSEr, probs =
c(0.05,0.95), normwt=F)
--
View this message in context:
http://r.789695.n4.nabble.com/help-on-matrix-column-removal-based-on-an
other-matrix-results-tp4650043.html
Sent from the R help mailing list archive at Nabble.com.
__
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Re: [R] help on matrix column removal based on another matrix results
Hi Rui, thank-you very much!
That worked perfectly.
Irucka
-Original Message-
From: Rui Barradas [ruipbarra...@sapo.pt]
Sent: 11/20/2012 5:42:36 AM
To: iruc...@mail2world.com
Cc: r-help@r-project.org
Subject: Re: [R] help on matrix column removal based on another matrix
results
Hello,
Try removing the square parenthesis from the left hand side,
Vsim - Vsim[, NSErr 0.6]
Hope this helps,
Rui Barradas
Em 20-11-2012 04:07, Irucka Embry escreveu:
Hi Rui, how are you?
I want to thank you for your assistance again.
I'm sorry, but the code that you provided for me did not work this
time.
The code and the warning message are both below:
NSErr - t(matrix(NSEr)) # which is a 1x1000 double matrix
Vsim[] - Vsim[ , NSErr 0.6]
Warning message:
In Vsim[] - Vsim[, NSErr 0.6] :
number of items to replace is not a multiple of replacement length
Thanks again.
Irucka
-Original Message-
From: Rui Barradas [ruipbarra...@sapo.pt]
Sent: 11/19/2012 10:35:00 AM
To: iruc...@mail2world.com
Cc: r-help@r-project.org
Subject: Re: [R] help on matrix column removal based on another
matrix
results
Sorry, the comma is in the wrong place, it should be
Vsim[] - Vsim[ , NSErr 0.6]
Rui Barradas
Em 19-11-2012 16:18, Rui Barradas escreveu:
Hello,
Try
Vsim[] - Vsim[NSErr 0.6, ]
Hope this helps,
Rui Barradas
Em 19-11-2012 14:30, iembry escreveu:
Hi everyone, now I am trying to finish writing the code (I had
asked
for
assistance on subtracting arrays)
This is what I what I am running in R:
source(/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Exampl
e/NSEr.R)
NSEr - function (obs, sim)
{
{jjh - (as.vector(obs) - sim)^2
Xjjhs - apply(Xjjh, 2, sum)
Yii - (obs - mean(obs))^2
Yiis - apply(Yii, 2, sum)
NSEr - 1 - (Xjjhs/Yiis)
}
NSEr}
Vsim - read.csv(1000Samples_Vsim.csv, header = TRUE,
sep =,)
Vsim - as.matrix(Vsim[,-1]) # remove column 1 from analysis
Vobs - read.csv(Observed_Flow.csv, header = TRUE, sep
=,)
Vobs - as.matrix(Vobs[,-1]) # remove column 1 from analysis
NSEr - NSEr(Vobs,Vsim);
write.table(NSEr, NSEr.csv, sep =,)
NSErr - t(matrix(NSEr))
## select the behavioural simulations and discard the rest
Vsim - Vsim[NSErr 0.6]
write.table(Vsim, Vsim.csv, sep =,)
**Vsim becomes numeric[42016] rather than a double matrix of
101x416.
What is the proper way to remove the columns in Vsim where the
NSEr
for that
column is less than 0.6? I am trying to make Vsim a double
matrix of
101x416.
Thank-you again.
Below is the rest of the code in R:
## normalise Qsim and compute the quantiles
NSEr - NSEr[NSEr 0.6]
write.table(NSEr, NSEr_great_0.6.csv, sep =,)
NSEr - NSEr - 0.6
write.table(NSEr, NSEr_minus0.6.csv, sep =,)
NSEr - NSEr/sum(NSEr)
write.table(NSEr, NSEr_normalized.csv, sep =,)
#NSEr = sum(NSEr)
limits - apply(Vsim, 1, wtd.quantile, weights = NSEr,
probs =
c(0.05,0.95), normwt=F)
--
View this message in context:
http://r.789695.n4.nabble.com/help-on-matrix-column-removal-based-on-an
other-matrix-results-tp4650043.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.
__
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.
.
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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.
[R] help on matrix column removal based on another matrix results
Hi everyone, now I am trying to finish writing the code (I had asked for
assistance on subtracting arrays)
This is what I what I am running in R:
source(/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Example/NSEr.R)
NSEr - function (obs, sim)
{
{jjh - (as.vector(obs) - sim)^2
Xjjhs - apply(Xjjh, 2, sum)
Yii - (obs - mean(obs))^2
Yiis - apply(Yii, 2, sum)
NSEr - 1 - (Xjjhs/Yiis)
}
NSEr}
Vsim - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,)
Vsim - as.matrix(Vsim[,-1]) # remove column 1 from analysis
Vobs - read.csv(Observed_Flow.csv, header = TRUE, sep =,)
Vobs - as.matrix(Vobs[,-1]) # remove column 1 from analysis
NSEr - NSEr(Vobs,Vsim);
write.table(NSEr, NSEr.csv, sep =,)
NSErr - t(matrix(NSEr))
## select the behavioural simulations and discard the rest
Vsim - Vsim[NSErr 0.6]
write.table(Vsim, Vsim.csv, sep =,)
**Vsim becomes numeric[42016] rather than a double matrix of 101x416.
What is the proper way to remove the columns in Vsim where the NSEr for that
column is less than 0.6? I am trying to make Vsim a double matrix of
101x416.
Thank-you again.
Below is the rest of the code in R:
## normalise Qsim and compute the quantiles
NSEr - NSEr[NSEr 0.6]
write.table(NSEr, NSEr_great_0.6.csv, sep =,)
NSEr - NSEr - 0.6
write.table(NSEr, NSEr_minus0.6.csv, sep =,)
NSEr - NSEr/sum(NSEr)
write.table(NSEr, NSEr_normalized.csv, sep =,)
#NSEr = sum(NSEr)
limits - apply(Vsim, 1, wtd.quantile, weights = NSEr, probs =
c(0.05,0.95), normwt=F)
--
View this message in context:
http://r.789695.n4.nabble.com/help-on-matrix-column-removal-based-on-another-matrix-results-tp4650043.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] help on matrix column removal based on another matrix results
Hello,
Try
Vsim[] - Vsim[NSErr 0.6, ]
Hope this helps,
Rui Barradas
Em 19-11-2012 14:30, iembry escreveu:
Hi everyone, now I am trying to finish writing the code (I had asked for
assistance on subtracting arrays)
This is what I what I am running in R:
source(/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Example/NSEr.R)
NSEr - function (obs, sim)
{
{jjh - (as.vector(obs) - sim)^2
Xjjhs - apply(Xjjh, 2, sum)
Yii - (obs - mean(obs))^2
Yiis - apply(Yii, 2, sum)
NSEr - 1 - (Xjjhs/Yiis)
}
NSEr}
Vsim - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,)
Vsim - as.matrix(Vsim[,-1]) # remove column 1 from analysis
Vobs - read.csv(Observed_Flow.csv, header = TRUE, sep =,)
Vobs - as.matrix(Vobs[,-1]) # remove column 1 from analysis
NSEr - NSEr(Vobs,Vsim);
write.table(NSEr, NSEr.csv, sep =,)
NSErr - t(matrix(NSEr))
## select the behavioural simulations and discard the rest
Vsim - Vsim[NSErr 0.6]
write.table(Vsim, Vsim.csv, sep =,)
**Vsim becomes numeric[42016] rather than a double matrix of 101x416.
What is the proper way to remove the columns in Vsim where the NSEr for that
column is less than 0.6? I am trying to make Vsim a double matrix of
101x416.
Thank-you again.
Below is the rest of the code in R:
## normalise Qsim and compute the quantiles
NSEr - NSEr[NSEr 0.6]
write.table(NSEr, NSEr_great_0.6.csv, sep =,)
NSEr - NSEr - 0.6
write.table(NSEr, NSEr_minus0.6.csv, sep =,)
NSEr - NSEr/sum(NSEr)
write.table(NSEr, NSEr_normalized.csv, sep =,)
#NSEr = sum(NSEr)
limits - apply(Vsim, 1, wtd.quantile, weights = NSEr, probs =
c(0.05,0.95), normwt=F)
--
View this message in context:
http://r.789695.n4.nabble.com/help-on-matrix-column-removal-based-on-another-matrix-results-tp4650043.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.
__
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] help on matrix column removal based on another matrix results
Sorry, the comma is in the wrong place, it should be
Vsim[] - Vsim[ , NSErr 0.6]
Rui Barradas
Em 19-11-2012 16:18, Rui Barradas escreveu:
Hello,
Try
Vsim[] - Vsim[NSErr 0.6, ]
Hope this helps,
Rui Barradas
Em 19-11-2012 14:30, iembry escreveu:
Hi everyone, now I am trying to finish writing the code (I had asked for
assistance on subtracting arrays)
This is what I what I am running in R:
source(/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Example/NSEr.R)
NSEr - function (obs, sim)
{
{jjh - (as.vector(obs) - sim)^2
Xjjhs - apply(Xjjh, 2, sum)
Yii - (obs - mean(obs))^2
Yiis - apply(Yii, 2, sum)
NSEr - 1 - (Xjjhs/Yiis)
}
NSEr}
Vsim - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,)
Vsim - as.matrix(Vsim[,-1]) # remove column 1 from analysis
Vobs - read.csv(Observed_Flow.csv, header = TRUE, sep =,)
Vobs - as.matrix(Vobs[,-1]) # remove column 1 from analysis
NSEr - NSEr(Vobs,Vsim);
write.table(NSEr, NSEr.csv, sep =,)
NSErr - t(matrix(NSEr))
## select the behavioural simulations and discard the rest
Vsim - Vsim[NSErr 0.6]
write.table(Vsim, Vsim.csv, sep =,)
**Vsim becomes numeric[42016] rather than a double matrix of 101x416.
What is the proper way to remove the columns in Vsim where the NSEr
for that
column is less than 0.6? I am trying to make Vsim a double matrix of
101x416.
Thank-you again.
Below is the rest of the code in R:
## normalise Qsim and compute the quantiles
NSEr - NSEr[NSEr 0.6]
write.table(NSEr, NSEr_great_0.6.csv, sep =,)
NSEr - NSEr - 0.6
write.table(NSEr, NSEr_minus0.6.csv, sep =,)
NSEr - NSEr/sum(NSEr)
write.table(NSEr, NSEr_normalized.csv, sep =,)
#NSEr = sum(NSEr)
limits - apply(Vsim, 1, wtd.quantile, weights = NSEr, probs =
c(0.05,0.95), normwt=F)
--
View this message in context:
http://r.789695.n4.nabble.com/help-on-matrix-column-removal-based-on-another-matrix-results-tp4650043.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.
__
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.
Re: [R] help on matrix column removal based on another matrix results
Hi Rui, how are you?
I want to thank you for your assistance again.
I'm sorry, but the code that you provided for me did not work this time.
The code and the warning message are both below:
NSErr - t(matrix(NSEr)) # which is a 1x1000 double matrix
Vsim[] - Vsim[ , NSErr 0.6]
Warning message:
In Vsim[] - Vsim[, NSErr 0.6] :
number of items to replace is not a multiple of replacement length
Thanks again.
Irucka
-Original Message-
From: Rui Barradas [ruipbarra...@sapo.pt]
Sent: 11/19/2012 10:35:00 AM
To: iruc...@mail2world.com
Cc: r-help@r-project.org
Subject: Re: [R] help on matrix column removal based on another matrix
results
Sorry, the comma is in the wrong place, it should be
Vsim[] - Vsim[ , NSErr 0.6]
Rui Barradas
Em 19-11-2012 16:18, Rui Barradas escreveu:
Hello,
Try
Vsim[] - Vsim[NSErr 0.6, ]
Hope this helps,
Rui Barradas
Em 19-11-2012 14:30, iembry escreveu:
Hi everyone, now I am trying to finish writing the code (I had asked
for
assistance on subtracting arrays)
This is what I what I am running in R:
source(/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Exampl
e/NSEr.R)
NSEr - function (obs, sim)
{
{jjh - (as.vector(obs) - sim)^2
Xjjhs - apply(Xjjh, 2, sum)
Yii - (obs - mean(obs))^2
Yiis - apply(Yii, 2, sum)
NSEr - 1 - (Xjjhs/Yiis)
}
NSEr}
Vsim - read.csv(1000Samples_Vsim.csv, header = TRUE, sep =,)
Vsim - as.matrix(Vsim[,-1]) # remove column 1 from analysis
Vobs - read.csv(Observed_Flow.csv, header = TRUE, sep =,)
Vobs - as.matrix(Vobs[,-1]) # remove column 1 from analysis
NSEr - NSEr(Vobs,Vsim);
write.table(NSEr, NSEr.csv, sep =,)
NSErr - t(matrix(NSEr))
## select the behavioural simulations and discard the rest
Vsim - Vsim[NSErr 0.6]
write.table(Vsim, Vsim.csv, sep =,)
**Vsim becomes numeric[42016] rather than a double matrix of
101x416.
What is the proper way to remove the columns in Vsim where the NSEr
for that
column is less than 0.6? I am trying to make Vsim a double matrix of
101x416.
Thank-you again.
Below is the rest of the code in R:
## normalise Qsim and compute the quantiles
NSEr - NSEr[NSEr 0.6]
write.table(NSEr, NSEr_great_0.6.csv, sep =,)
NSEr - NSEr - 0.6
write.table(NSEr, NSEr_minus0.6.csv, sep =,)
NSEr - NSEr/sum(NSEr)
write.table(NSEr, NSEr_normalized.csv, sep =,)
#NSEr = sum(NSEr)
limits - apply(Vsim, 1, wtd.quantile, weights = NSEr, probs =
c(0.05,0.95), normwt=F)
--
View this message in context:
http://r.789695.n4.nabble.com/help-on-matrix-column-removal-based-on-an
other-matrix-results-tp4650043.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.
__
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.
.
span id=m2wTlpfont face=Arial, Helvetica, sans-serif size=2
style=font-size:13.5px___BRGet
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Email Storage #150; POP3 #150; Calendar #150; SMS #150; Translator #150;
Much More!/font/font/span
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Re: [R] Help: a matrix exponential
This is a matrix algebra problem, not an R problem. ej fails because betaligne and ZIcolone are not conformable matrices. dim(betaligne) 1 2 dim(ZIcolone) 76 1 Let a matrix be m x n, then you can only matrix multiply it with a matrix that is n x k (where m and k are arbitrary positive integers, but n must be the same). The resulting matrix will be m x k. So you can postmultiply t(betaligne) with t(ZIcolone) or ZIcolone with betaligne. Everything else is not conformable. HTH, Daniel Hi, I have a problem with the final results fj ; fj must firstly be a vector indexed by j = 8, and the same avoire Chifre (9.5), beta0=64.91151 beta1=17.70393 beta=c(beta0, beta1) j=1:8 n=76 i=1:n zi-c(26, 26, 26, 28, 28, 28, 28, 28, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 32, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 36, 38, 38, 38, 38, 38, 38, 38, 38) Xi-c(5.79, 1579.52, 2323.7, 68.85, 426.07, 110.29, 108.29, 1067.6, 17.05, 22.66, 21.02, 175.88, 139.07, 144.12, 20.46, 43.4, 194.9, 47.3, 7.74, 0.4, 82.85, 9.88, 89.29, 215.1, 2.75, 0.79, 15.93, 3.91, 0.27, 0.69, 100.58, 27.8, 13.95, 53.24, 0.96, 4.15, 0.19, 0.78, 8.01, 31.75, 7.35, 6.5, 8.27, 33.91, 32.52, 3.16, 4.85, 2.78, 4.67, 1.31, 12.06, 36.71, 72.89, 1.97, 0.59, 2.58, 1.69, 2.71, 25.5, 0.35, 0.99, 3.99, 3.67, 2.07, 0.96, 5.35, 2.9, 13.77, 0.47, 0.73, 1.4, 0.74, 0.39, 1.13, 0.09, 2.38) ZI-log(zi) betaligne-t(as.matrix(-beta)) ZIcolone-as.matrix(ZI) k=8 ej-1/k*sum(exp(betaligne %*% ZIcolone)*Xi) Error in betaligne %*% ZIcolone : non-conformable arguments PLEASE HELP ME THANKS Goual -- View this message in context: http://r.789695.n4.nabble.com/Help-a-matrix-exponential-tp4642585p4642589.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] help, please! matrix operations inside 3 nested loops
thank you for your help.
my input data looks like this (tab separated):
Ind.nr. Pop.nr. scm266 rms1280 scm247 rms1107
1 101 305 318 222 135
1 101 305 318 231 135
2 101 305 313 999 96
2 101 305 321 999 130
3 101 305 324 231 135
3 101 305 324 231 135
4 101 305 313 230 126
4 101 305 313 230 135
6 101 305 313 231 135
6 101 305 321 231 135
it is a dataset with genetic marker alleles for single individuals.
the first row is the header, all following rows are individuals. 2 rows
count for 1 individual.
first colum is the individual's number, second colum is the number for the
population the individual comes from, and all following colums are different
genetic markers.
what i want to do with this data in R, is to compare one individual with
each of the other individuals, allele-wise. there are five possibilities:
the two compared individuals share 4,3,2,1,0 alleles of the currently
examined marker (=colum). for each shared allele this pair of individuals
shall get 1 scoring point. for each pair of individuals, all scoring points
shall be summarized over all markers.
my code again, modified according to your suggestions:
#1) read in data:
daten-read.table('K:/Analysen/STRUCTURE/test.txt', header=TRUE, sep=\t)
daten-as.data.frame(daten)
#2) create empty matrix:
indxind-matrix(0,nrow=617, ncol=617)
indxind[1:20,1:19]
#3) compare cells to each other, score:
#for the whole dataset: s in 3:34, z1 in 1:617, z2 in 1:617
for (s in 3:6) { #walks though the matrix colum by colum, starting at
colum 3
for (z1 in 1:6) { #for each current colum, take one row (z1)...
for (z2 in 1:6) { #...and compare it to another row (z2) of the current
colum
if (z1!=z2) {topf-indxind[z1,z2]
if (daten[2*z1-1,s]==daten[2*z2-1,s]) topf-topf+1
#actually, 2 rows make up 1 individual,
if (daten[2*z1-1,s]==daten[2*z2,s]) topf-topf+1
#therefore i compare 2 rows
if (daten[2*z1,s]==daten[2*z2-1,s]) topf-topf+1
#with another 2 rows
if (daten[2*z1,s]==daten[2*z2,s]) topf-topf+1
indxind[z1,z2]-topf
indxind[z2,z1]-topf
}
#print(c(s,z1,z2,indxind[1,2])) ##counts s, z1 and z2 properly, but
gives always 8 for indxind[1,2]
}
#indxind[1:5,1:5] #empty matrix
}
#indxind[1:5,1:5] #empty matrix
}
#4) check:
indxind[1:5,1:5]
@ Michael Weylandt: i've done my best with regard to the big picture of my
algorithm and the small reproducible example. i hope both is sufficient.
@ Petr Pikal-3: in this case, there are only numerical values, but it's a
useful hint for my other codes.
@ Petr Pikal-3 and Berend Hasselman: initializing indxind with 0's instead
of NAs helps, it fills something in indxind now. but it does the calculation
only for the first marker (colum 3), afterwards i get an error:
Fehler in if (daten[2 * z1 - 1, s] == daten[2 * z2 - 1, s]) topf - topf +
:
Fehlender Wert, wo TRUE/FALSE nötig ist
Error in if (daten[2 * z1 - 1, s] == daten[2 * z2 - 1, s]) topf - topf + :
Missing value, where TRUE/FAlse is required
Has this something to do with the changing to daten-as.data.frame(daten) in
line 3 (instead of as.matrix before)?
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Re: [R] help, please! matrix operations inside 3 nested loops
SORRY it should be:
Fridolin wrote
for (s in 3:6) { #walks though the matrix colum by colum, starting at
colum 3
for (z1 in 1:5) { #for each current colum, take one row (z1)...
for (z2 in 1:5) { #...and compare it to another row (z2) of the
current colum
error is gone now SORRY!!!
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Re: [R] help, please! matrix operations inside 3 nested loops
all problems solved. thank you for your help!
for the sake of completeness, here my solution:
#1) read in data:
daten-read.table('K:/Analysen/STRUCTURE/test.txt', header=TRUE, sep=\t)
daten-as.data.frame(daten)
#2) create empty matrix:
indxind-matrix(0,nrow=617, ncol=617)
#indxind[1:20,1:19]
#3) compare cells to each other, score:
#for the whole dataset: s in 3:34, z1 in 1:617, z2 in 1:617
z1-1 #running variable for rows in daten
z2-1 #running variable for rows in daten
l1-1 #running variable for rows in indxind
l2-1 #running variable for rows in indxind
for (s in 3:6) { #walks though the matrix colum by colum, starting at
colum 3
while (z111) { #for each current colum, take one row
(z1)...
while (z211) { #...and compare it to
another row (z2) of the current colum
if (z1!=z2) {
l1
topf-indxind[l1,l2]
if
(daten[z1,s]==daten[z2,s]) topf-topf+1 #actually, 2 rows make up 1
individual,
if
(daten[z1,s]==daten[z2+1,s]) topf-topf+1 #therefore i compare 2 rows
if
(daten[z1+1,s]==daten[z2,s]) topf-topf+1 #with another 2 rows
if
(daten[z1+1,s]==daten[z2+1,s]) topf-topf+1
indxind[l1,l2]-topf
}
z2-z2+2
l2-l2+1
}
z2-1
l2-1
z1-z1+2
l1-l1+1
}
z1-1
l1-1
}
#4) check:
indxind[1:5,1:5]
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Re: [R] help, please! matrix operations inside 3 nested loops
Hi
thank you for your help.
my input data looks like this (tab separated):
Ind.nr. Pop.nr. scm266 rms1280 scm247 rms1107
1 101 305 318 222 135
1 101 305 318 231 135
2 101 305 313 999 96
2 101 305 321 999 130
3 101 305 324 231 135
3 101 305 324 231 135
4 101 305 313 230 126
4 101 305 313 230 135
6 101 305 313 231 135
6 101 305 321 231 135
Better to use dput(your.data) for sharing data. Anyway I am still confused
but you probably are able to clarify things further.
it is a dataset with genetic marker alleles for single individuals.
the first row is the header, all following rows are individuals. 2 rows
count for 1 individual.
first colum is the individual's number, second colum is the number for
the
population the individual comes from, and all following colums are
different
genetic markers.
what i want to do with this data in R, is to compare one individual with
In those 2 rows for one individual sometimes the genetic marker differs
test[1:2, scm247]
[1] 222 231
What do you want to do with them?
each of the other individuals, allele-wise. there are five
possibilities:
the two compared individuals share 4,3,2,1,0 alleles of the currently
examined marker (=colum). for each shared allele this pair of
individuals
shall get 1 scoring point. for each pair of individuals, all scoring
points
shall be summarized over all markers.
Based on your example,
dput(test)
structure(list(Ind.nr. = c(1L, 1L, 2L, 2L, 3L, 3L, 4L, 4L, 6L,
6L), Pop.nr. = c(101L, 101L, 101L, 101L, 101L, 101L, 101L, 101L,
101L, 101L), scm266 = c(305L, 305L, 305L, 305L, 305L, 305L, 305L,
305L, 305L, 305L), rms1280 = c(318L, 318L, 313L, 321L, 324L,
324L, 313L, 313L, 313L, 321L), scm247 = c(222L, 231L, 999L, 999L,
231L, 231L, 230L, 230L, 231L, 231L), rms1107 = c(135L, 135L,
96L, 130L, 135L, 135L, 126L, 135L, 135L, 135L)), .Names = c(Ind.nr.,
Pop.nr., scm266, rms1280, scm247, rms1107), class =
data.frame, row.names = c(NA,
-10L))
what is your desired result?
Regards
Petr
my code again, modified according to your suggestions:
#1) read in data:
daten-read.table('K:/Analysen/STRUCTURE/test.txt', header=TRUE,
sep=\t)
daten-as.data.frame(daten)
#2) create empty matrix:
indxind-matrix(0,nrow=617, ncol=617)
indxind[1:20,1:19]
#3) compare cells to each other, score:
#for the whole dataset: s in 3:34, z1 in 1:617, z2 in 1:617
for (s in 3:6) { #walks though the matrix colum by colum, starting at
colum 3
for (z1 in 1:6) { #for each current colum, take one row (z1)...
for (z2 in 1:6) { #...and compare it to another row (z2) of the
current
colum
if (z1!=z2) {topf-indxind[z1,z2]
if (daten[2*z1-1,s]==daten[2*z2-1,s]) topf-topf+1
#actually, 2 rows make up 1 individual,
if (daten[2*z1-1,s]==daten[2*z2,s]) topf-topf+1
#therefore i compare 2 rows
if (daten[2*z1,s]==daten[2*z2-1,s]) topf-topf+1
#with another 2 rows
if (daten[2*z1,s]==daten[2*z2,s]) topf-topf+1
indxind[z1,z2]-topf
indxind[z2,z1]-topf
}
#print(c(s,z1,z2,indxind[1,2])) ##counts s, z1 and z2 properly,
but
gives always 8 for indxind[1,2]
}
#indxind[1:5,1:5] #empty matrix
}
#indxind[1:5,1:5] #empty matrix
}
#4) check:
indxind[1:5,1:5]
@ Michael Weylandt: i've done my best with regard to the big picture
of my
algorithm and the small reproducible example. i hope both is sufficient.
@ Petr Pikal-3: in this case, there are only numerical values, but it's
a
useful hint for my other codes.
@ Petr Pikal-3 and Berend Hasselman: initializing indxind with 0's
instead
of NAs helps, it fills something in indxind now. but it does the
calculation
only for the first marker (colum 3), afterwards i get an error:
Fehler in if (daten[2 * z1 - 1, s] == daten[2 * z2 - 1, s]) topf - topf
+
:
Fehlender Wert, wo TRUE/FALSE nötig ist
Error in if (daten[2 * z1 - 1, s] == daten[2 * z2 - 1, s]) topf - topf
+ :
Missing value, where TRUE/FAlse is required
Has this something to do with the changing to
daten-as.data.frame(daten) in
line 3 (instead of as.matrix before)?
--
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matrix-operations-inside-3-nested-loops-tp4639592p4639730.html
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Re: [R] help, please! matrix operations inside 3 nested loops
Hi
all problems solved. thank you for your help!
for the sake of completeness, here my solution:
#1) read in data:
daten-read.table('K:/Analysen/STRUCTURE/test.txt', header=TRUE,
sep=\t)
daten-as.data.frame(daten)
not needed, daten is already data frame
#2) create empty matrix:
indxind-matrix(0,nrow=617, ncol=617)
#indxind[1:20,1:19]
#3) compare cells to each other, score:
#for the whole dataset: s in 3:34, z1 in 1:617, z2 in 1:617
z1-1 #running variable for rows in daten
z2-1 #running variable for rows in daten
l1-1 #running variable for rows in indxind
l2-1 #running variable for rows in indxind
for (s in 3:6) { #walks though the matrix colum by colum, starting at
colum 3
while (z111) { #for each current colum, take one row
(z1)...
while (z211) { #...and compare it to
another row (z2) of the current colum
if (z1!=z2) {
l1
topf-indxind[l1,l2]
if
(daten[z1,s]==daten[z2,s]) topf-topf+1 #actually, 2 rows make up 1
individual,
if
(daten[z1,s]==daten[z2+1,s]) topf-topf+1 #therefore i compare 2
rows
if
(daten[z1+1,s]==daten[z2,s]) topf-topf+1 #with another 2 rows
if
(daten[z1+1,s]==daten[z2+1,s]) topf-topf+1
indxind[l1,l2]-topf
}
z2-z2+2
l2-l2+1
}
z2-1
l2-1
z1-z1+2
l1-l1+1
}
z1-1
l1-1
}
#4) check:
indxind[1:5,1:5]
I believe that above cycles can be simplified, maybe by changing your
daten to three dimensional array or some clever **ply construction but if
your loops works it is not probably worth en effort.
Regards
Petr
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[R] help, please! matrix operations inside 3 nested loops
hello, this is my script:
#1) read in data:
daten-read.table('K:/Analysen/STRUCTURE/input_STRUCTURE_tab_excl_5_282_559.txt',
header=TRUE, sep=\t)
daten-as.matrix(daten)
#2) create empty matrix:
indxind-matrix(nrow=617, ncol=617)
indxind[1:20,1:19]
#3) compare cells to each other, score:
for (s in 3:34) { #walks though the matrix colum by colum, starting at
colum 3
for (z1 in 1:617) { #for each current colum, take one row (z1)...
for (z2 in 1:617) { #...and compare it to another row (z2) of the
current colum
if (z1!=z2) {topf-indxind[z1,z2]
if (daten[2*z1-1,s]==daten[2*z2-1,s]) topf-topf+1
#actually, 2 rows make up 1 individual,
if (daten[2*z1-1,s]==daten[2*z2,s]) topf-topf+1
#therefore i compare 2 rows
if (daten[2*z1,s]==daten[2*z2-1,s]) topf-topf+1
#with another 2 rows
if (daten[2*z1,s]==daten[2*z2,s]) topf-topf+1
indxind[z1,z2]-topf
indxind[z2,z1]-topf
}
#print(c(s,z1,z2,indxind[1,2])) ##counts s, z1 and z2 properly, but
gives NA for indxind[1,2]
}
#indxind[1:5,1:5] #empty matrix
}
#indxind[1:5,1:5] #empty matrix
}
#4) check:
indxind[1:5,1:5]
this results no errors, but my matrix indxind remains empty (only NAs).
though all columns and rows are counted properly. R needs quite a while to
get through all this (there are probably smarter and faster ways to
calculate this but i am not too deep into R and bioinformatics, and i need
to calculate this only once). could the 3 for-loops already be too
computationally intense for adding matrix operations?
any help would be much appreciated!
thx, frido
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Re: [R] help, please! matrix operations inside 3 nested loops
On Wed, Aug 8, 2012 at 9:06 AM, Fridolin smells_like_r...@gmx.net wrote:
hello, this is my script:
#1) read in data:
daten-read.table('K:/Analysen/STRUCTURE/input_STRUCTURE_tab_excl_5_282_559.txt',
header=TRUE, sep=\t)
daten-as.matrix(daten)
#2) create empty matrix:
indxind-matrix(nrow=617, ncol=617)
indxind[1:20,1:19]
#3) compare cells to each other, score:
for (s in 3:34) { #walks though the matrix colum by colum, starting at
colum 3
for (z1 in 1:617) { #for each current colum, take one row (z1)...
for (z2 in 1:617) { #...and compare it to another row (z2) of the
current colum
if (z1!=z2) {topf-indxind[z1,z2]
if (daten[2*z1-1,s]==daten[2*z2-1,s]) topf-topf+1
#actually, 2 rows make up 1 individual,
if (daten[2*z1-1,s]==daten[2*z2,s]) topf-topf+1
#therefore i compare 2 rows
if (daten[2*z1,s]==daten[2*z2-1,s]) topf-topf+1
#with another 2 rows
if (daten[2*z1,s]==daten[2*z2,s]) topf-topf+1
indxind[z1,z2]-topf
indxind[z2,z1]-topf
}
#print(c(s,z1,z2,indxind[1,2])) ##counts s, z1 and z2 properly, but
gives NA for indxind[1,2]
}
#indxind[1:5,1:5] #empty matrix
}
#indxind[1:5,1:5] #empty matrix
}
#4) check:
indxind[1:5,1:5]
this results no errors, but my matrix indxind remains empty (only NAs).
though all columns and rows are counted properly. R needs quite a while to
get through all this (there are probably smarter and faster ways to
calculate this but i am not too deep into R and bioinformatics, and i need
to calculate this only once). could the 3 for-loops already be too
computationally intense for adding matrix operations?
any help would be much appreciated!
thx, frido
Hi Frido,
I'm afraid I get a little lost in your code, but I'd be willing to bet
we can cut the loops out entirely and speed things up.
Can you give us a big picture description of the algorithm you're
implementing as well as (if it's not too hard) a small reproducible
example [1]?
Note also that most of us don't use Nabble so you'll need to
explicitly quote any relevant context.
Thanks,
Michael
[1]
http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
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Re: [R] help, please! matrix operations inside 3 nested loops
Fridolin wrote
hello, this is my script:
#1) read in data:
daten-read.table('K:/Analysen/STRUCTURE/input_STRUCTURE_tab_excl_5_282_559.txt',
header=TRUE, sep=\t)
daten-as.matrix(daten)
#2) create empty matrix:
indxind-matrix(nrow=617, ncol=617)
indxind[1:20,1:19]
You should at least initialize indxind to 0 with
indxind-matrix(0,nrow=617, ncol=617)
because the default for matrix is to use NA for data.
Berend
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and provide commented, minimal, self-contained, reproducible code.
[R] Help in matrix
Hi I have a matrix a = 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 b = 5 6 7 8 9 10 I want output as c = 5 2 3 8 2 3 1 6 3 1 9 3 1 2 7 1 2 10 Please any on help to avoid reduncy of my code - Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/Help-in-matrix-tp4449622p4449622.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] Help in matrix
On Mar 6, 2012, at 13:02 , arunkumar wrote: Hi I have a matrix a = 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 b = 5 6 7 8 9 10 I want output as c = 5 2 3 8 2 3 1 6 3 1 9 3 1 2 7 1 2 10 Please any on help to avoid reduncy of my code Here's one way: a - matrix(rep(1:3,each=6),6) a [,1] [,2] [,3] [1,]123 [2,]123 [3,]123 [4,]123 [5,]123 [6,]123 b - matrix(5:10,2,byrow=T) ix - cbind(1:6,rep(1:3,each=2)) a[ix] - b a [,1] [,2] [,3] [1,]523 [2,]823 [3,]163 [4,]193 [5,]127 [6,]12 10 -- Peter Dalgaard, Professor Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.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] Help in matrix
Thanks so much it worked very well - Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/Help-in-matrix-tp4449622p4450560.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.
[R] Help with Matrix code optimization
The chol and solve methods for dpoMatrix (Matrix package) are much
faster than the default methods. But, the time required to coerce a
regular matrix to dpoMatrix swamps the advantage.
Hence, I have the following problem, where use of dpoMatrix is worse
than a regular matrix.
library(Matrix)
x - diag(10)
system.time(
for(r in seq(0.1, 0.9, length.out=1000)) {
m - r^abs(row(x)-col(x));
chol(m); solve(m);
})
system.time(
for(r in seq(0.1, 0.9, length.out=1000)) {
M - as(r^abs(row(x)-col(x)), 'dpoMatrix')
chol(M); solve(M);
})
Any ideas?
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[R] help with matrix column cleaning
Dear R Group I have a matrix object in R, where i want to retain only those columns which have each row from a different group.. see the example below.. team1-c(a1,a2,b1,b2,b3,c1,c2) teams-combn(team1, 3) I want a function which will delete all columns in the teams matrix that have more than 1 row having same letter starts.. For instance in the example object teams, I want to retain column 8,11,12,13,14,18,19,21,22,23,24 . I am looking at doing this in larger matrix objects using a function preferrably that i can apply to the matrix object to check for the beginning letter of each cell from each row and delete columns that have more than one row having the same start letter. Thanks for your help. Regards Vijayan Padmanabhan What is expressed without proof can be denied without proof - Euclide. Please visit us at www.itcportal.com ** This Communication is for the exclusive use of the intended recipient (s) and shall not attach any liability on the originator or ITC Ltd./its Subsidiaries/its Group Companies. If you are the addressee, the contents of this email are intended for your use only and it shall not be forwarded to any third party, without first obtaining written authorisation from the originator or ITC Ltd./its Subsidiaries/its Group Companies. It may contain information which is confidential and legally privileged and the same shall not be used or dealt with by any third party in any manner whatsoever without the specific consent of ITC Ltd./its Subsidiaries/its Group Companies. [[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] help with matrix column cleaning
On Feb 14, 2012, at 12:46 AM, Vijayan Padmanabhan wrote: Dear R Group I have a matrix object in R, where i want to retain only those columns which have each row from a different group.. see the example below.. team1-c(a1,a2,b1,b2,b3,c1,c2) teams-combn(team1, 3) I want a function which will delete all columns in the teams matrix that have more than 1 row having same letter starts.. For instance in the example object teams, I want to retain column 8,11,12,13,14,18,19,21,22,23,24 . This will give you a logical vector which you can use with [ in the column position to select those columns: apply(teams, 2, function(x) length( unique(substr(x, 1,1))) ==3 ) [1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE TRUE TRUE FALSE TRUE TRUE [13] TRUE TRUE FALSE FALSE FALSE TRUE TRUE FALSE TRUE TRUE TRUE TRUE [25] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE I am looking at doing this in larger matrix objects using a function preferrably that i can apply to the matrix object to check for the beginning letter of each cell from each row and delete columns that have more than one row having the same start letter. Thanks for your help. Regards Vijayan Padmanabhan -- David Winsemius, MD West Hartford, CT __ 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 with matrix
Hi I have 3 vectors, say g1, g2 and f, where g1 and g2 contain the row and column indices of a matrix, with corresponding elements in f. How can I create a matrix containing the values in f ? For example: g1 g2 f 3 4 10 2 1 30 4 4 50 How can then get a matrix that looks as follows: 0 0 0 0 30 0 0 0 0 0 0 10 0 0 0 50 I am still learning R. I think reshape will help, but it is not clear how that works here. Any help is much appreciated. Thank you. Jacob Jacob L van Wyk Department of Statistics University of Johannesburg, APK Box 524 Auckland Park 2006 South Africa Office Tel: +27 11 559 3080 Fax: +27 11 559 2832 [[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.
[R] help with matrix
Hi, dear all:
I am a beginner. I appreciate any help or hint from you.
I am trying to do calculation with matrices. I have 3 matrices. One is
matrixA, 2nd is matrixB, and last is matrixC.
Here is matrixA:
1.8511.40.0831.001
0.8771.30.1161.33
1.9021.21.1020.302
0.8640.1261.110.252
1.8230.2161.0020.307
Next is matrixB:
0.8761.770.1930.328
0.8911.0090.2381.004
0.8641.1150.2760.22
0.8871.3060.1660.239
0.8521.0011.0080.251
Last is matrixC:
0.5 1.0 0.5 1.0
1.0 1.0 1.0 0.5
1.0 1.0 1.0 0.5
1.0 1.0 1.0 0.5
0.5 0.5 1.0 0.5
What I want is to match both matrixA and matrixB with matrixC. Ignore the 1st
column in matrixA and matrixB. When the element im matrixC is 1.0, multilply
the corresponding element in matrixA by the corresponding element in matrixB.
Otherwise, ignore it. I wrote the following code with some error in the logic:
A-read.table('matrixA.txt', header=F)
B-read.table('matrixB.txt', header=F)
C-read.table('matrixC.txt', header=F)
a1-as.matrix(A[,1])
arest-as.matrix(A[,2:5])
b1-as.matrix(B[,1])
brest-as.matrix(B[,2:5])
C-as.matrix(C)
if(C= =1){
ra-(arest*brest)
}else {
throw-(arest*brest)
}
ra
I got a warning message:
Warning message:the condition has length 1 and only the first element will be
used in: if.
When I changed the if part to:
ifelse (C= =1,ra-(arest*brest),throw-(arest*brest))
The result is the multiplication of each element in matrixA and matrixB.
Anyone has an idea about what is wrong. Thanks
Riddle
Be a better friend, newshound, and
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