Dear all,
I have a matrix X with 47 lines and say 500 columns - values are in {0,1}.
I'd like to compare lines.
For that, I first did:
for (i in 1:(dim(X)[1]-1))
for (j in (i+1):dim(X)[1]) {
Y <- X[i,]+Y[j,]
etc.
but, since it takes a long time, I would prefer avoding loops;
for that, my first idea was to add this matrix:
X1=X[,rep(1:46,46:1)]
to this one:
res=NULL
for (i in (2:47)) res=c(res,i:47)
X2=X[,res]
(Is it a nice alternative way ?)
Is there a way to create the second matrix X2 without a loop, such as for X1
?
Thanks in advance,
Jacques VESLOT
Please give a simple example of the input data and output that you desire. It is difficult to understand from you partial codes what you mean. For example what is Y ? Are you trying to find add values from pairs of rows ? If so, please see my posting "pairwise difference operator" where I wanted to find the differences between pairs of columns. http://tolstoy.newcastle.edu.au/R/help/04/07/1633.html Otherwise, please send a sample input and output. Thank you. Regards, Adai On Wed, 2005-01-26 at 11:40 +0400, Jacques VESLOT wrote:> Dear all, > > I have a matrix X with 47 lines and say 500 columns - values are in {0,1}. > I'd like to compare lines. > > For that, I first did: > > for (i in 1:(dim(X)[1]-1)) > for (j in (i+1):dim(X)[1]) { > Y <- X[i,]+Y[j,] > etc. > > but, since it takes a long time, I would prefer avoding loops; > for that, my first idea was to add this matrix: > > X1=X[,rep(1:46,46:1)] > > to this one: > > res=NULL > for (i in (2:47)) res=c(res,i:47) > > X2=X[,res] > > (Is it a nice alternative way ?) > Is there a way to create the second matrix X2 without a loop, such as for X1 > ? > > Thanks in advance, > > Jacques VESLOT > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html >
See if this does what you want:> m <- matrix(round(runif(24)), 4, 6) # simulate some data > m[,1] [,2] [,3] [,4] [,5] [,6] [1,] 0 1 0 1 0 1 [2,] 0 1 1 1 0 0 [3,] 1 1 1 0 0 0 [4,] 1 0 0 0 1 0> library(gtools) # Install the gregmisc package if you don't have it. > idx <- combinations(nrow(m), 2) > res <- m[idx[,1],] + m[idx[,2]] > rownames(res) <- paste(idx[,1], idx[,2], sep="+") > res[,1] [,2] [,3] [,4] [,5] [,6] 1+2 0 1 0 1 0 1 1+3 1 2 1 2 1 2 1+4 1 2 1 2 1 2 2+3 1 2 2 2 1 1 2+4 1 2 2 2 1 1 3+4 2 2 2 1 1 1 Andy> From: Jacques VESLOT > > It is part of a function to determine Dice's index in the > framewok of AFLP > analysis. > > X is a binary matrix which value for each strain (lines) and > each base pair > (columns) is 1 where there is a peak and 0 where there is no peak as > biologists explained to me. > > The first step is to compare each strain with one another by > counting the > number of 0, 1 and 2, respectively where there is no peak, > one peak or 2 > peaks for each base pair. > > In that respect, I want to add together each pair of X's lines. > > For the moment, there is a double loop calculating, at each > step, the sum of > two lines as a vector Y and counting the number of 0, 1 and 2 > in it for > inclusion in other operations. > > I read your posting... > > Thanks for helping, > > Jacques VESLOT > > > -----Message d'origine----- > De : Adaikalavan Ramasamy [mailto:ramasamy at cancer.org.uk] > Envoy? : mercredi 26 janvier 2005 15:57 > ? : jacques.veslot at cirad.fr > Cc : R-help > Objet : Re: [R] Still avoiding loops > > > Please give a simple example of the input data and output that you > desire. It is difficult to understand from you partial codes what you > mean. For example what is Y ? > > Are you trying to find add values from pairs of rows ? If so, > please see > my posting "pairwise difference operator" where I wanted to find the > differences between pairs of columns. > http://tolstoy.newcastle.edu.au/R/help/04/07/1633.html > > Otherwise, please send a sample input and output. Thank you. > > Regards, Adai > > > On Wed, 2005-01-26 at 11:40 +0400, Jacques VESLOT wrote: > > Dear all, > > > > I have a matrix X with 47 lines and say 500 columns - > values are in {0,1}. > > I'd like to compare lines. > > > > For that, I first did: > > > > for (i in 1:(dim(X)[1]-1)) > > for (j in (i+1):dim(X)[1]) { > > Y <- X[i,]+Y[j,] > > etc. > > > > but, since it takes a long time, I would prefer avoding loops; > > for that, my first idea was to add this matrix: > > > > X1=X[,rep(1:46,46:1)] > > > > to this one: > > > > res=NULL > > for (i in (2:47)) res=c(res,i:47) > > > > X2=X[,res] > > > > (Is it a nice alternative way ?) > > Is there a way to create the second matrix X2 without a > loop, such as for > X1 > > ? > > > > Thanks in advance, > > > > Jacques VESLOT > > > > ______________________________________________ > > R-help at stat.math.ethz.ch mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html > > > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html > >
> From: Robert Kruus > > Slight edit?Yes. Thank you. Jacques caught that as well... Andy> -------- > > It is rumored that on Wed, 26 Jan 2005 09:24:44 -0500 > "Liaw, Andy" <andy_liaw at merck.com> wrote: > > > See if this does what you want: > > > > > m <- matrix(round(runif(24)), 4, 6) # simulate some data > > > m > > [,1] [,2] [,3] [,4] [,5] [,6] > > [1,] 0 1 0 1 0 1 > > [2,] 0 1 1 1 0 0 > > [3,] 1 1 1 0 0 0 > > [4,] 1 0 0 0 1 0 > > > library(gtools) # Install the gregmisc package if you don't have > > > it. idx <- combinations(nrow(m), 2) > > > res <- m[idx[,1],] + m[idx[,2]] > > I think you missed a "," (if you want pairwise row sums) > res <- m[idx[,1],] + m[idx[,2],] > > > > rownames(res) <- paste(idx[,1], idx[,2], sep="+") > > > res > > [,1] [,2] [,3] [,4] [,5] [,6] > > 1+2 0 1 0 1 0 1 > > 1+3 1 2 1 2 1 2 > > 1+4 1 2 1 2 1 2 > > 2+3 1 2 2 2 1 1 > > 2+4 1 2 2 2 1 1 > > 3+4 2 2 2 1 1 1 > > > > Andy > > > > > > > From: Jacques VESLOT > > > > > > It is part of a function to determine Dice's index in the > > > framewok of AFLP > > > analysis. > > > > > > X is a binary matrix which value for each strain (lines) and > > > each base pair > > > (columns) is 1 where there is a peak and 0 where there is > no peak as > > > biologists explained to me. > > > > > > The first step is to compare each strain with one another by > > > counting the > > > number of 0, 1 and 2, respectively where there is no peak, > > > one peak or 2 > > > peaks for each base pair. > > > > > > In that respect, I want to add together each pair of X's lines. > > > > > > For the moment, there is a double loop calculating, at each > > > step, the sum of > > > two lines as a vector Y and counting the number of 0, 1 and 2 > > > in it for > > > inclusion in other operations. > > > > > > I read your posting... > > > > > > Thanks for helping, > > > > > > Jacques VESLOT > > > > > > > > > -----Message d'origine----- > > > De : Adaikalavan Ramasamy [mailto:ramasamy at cancer.org.uk] > > > Envoy? : mercredi 26 janvier 2005 15:57 > > > ? : jacques.veslot at cirad.fr > > > Cc : R-help > > > Objet : Re: [R] Still avoiding loops > > > > > > > > > Please give a simple example of the input data and output that you > > > desire. It is difficult to understand from you partial codes what > > > you mean. For example what is Y ? > > > > > > Are you trying to find add values from pairs of rows ? If so, > > > please see > > > my posting "pairwise difference operator" where I wanted > to find the > > > differences between pairs of columns. > > > http://tolstoy.newcastle.edu.au/R/help/04/07/1633.html > > > > > > Otherwise, please send a sample input and output. Thank you. > > > > > > Regards, Adai > > > > > > > > > On Wed, 2005-01-26 at 11:40 +0400, Jacques VESLOT wrote: > > > > Dear all, > > > > > > > > I have a matrix X with 47 lines and say 500 columns - > > > values are in {0,1}. > > > > I'd like to compare lines. > > > > > > > > For that, I first did: > > > > > > > > for (i in 1:(dim(X)[1]-1)) > > > > for (j in (i+1):dim(X)[1]) { > > > > Y <- X[i,]+Y[j,] > > > > etc. > > > > > > > > but, since it takes a long time, I would prefer avoding loops; > > > > for that, my first idea was to add this matrix: > > > > > > > > X1=X[,rep(1:46,46:1)] > > > > > > > > to this one: > > > > > > > > res=NULL > > > > for (i in (2:47)) res=c(res,i:47) > > > > > > > > X2=X[,res] > > > > > > > > (Is it a nice alternative way ?) > > > > Is there a way to create the second matrix X2 without a > > > loop, such as for > > > X1 > > > > ? > > > > > > > > Thanks in advance, > > > > > > > > Jacques VESLOT > > > > > > > > ______________________________________________ > > > > R-help at stat.math.ethz.ch mailing list > > > > https://stat.ethz.ch/mailman/listinfo/r-help > > > > PLEASE do read the posting guide! > > > http://www.R-project.org/posting-guide.html > > > > > > > > > > ______________________________________________ > > > R-help at stat.math.ethz.ch mailing list > > > https://stat.ethz.ch/mailman/listinfo/r-help > > > PLEASE do read the posting guide! > > > http://www.R-project.org/posting-guide.html > > > > > > > > > > ______________________________________________ > > R-help at stat.math.ethz.ch mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide! > > http://www.R-project.org/posting-guide.html > > > > > -------- > > -- > robert.kruus at utoronto.ca > "There are ten church members by inheritance for every one by > conviction." [Anonymous] > t > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html > >