[R] Still avoiding loops

[Robert Kruus]

Slight edit?

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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]
> > Envoyé : mercredi 26 janvier 2005 15:57
> > À : 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
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> 


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-- 
robert.kruus at utoronto.ca
"There are ten church members by inheritance for every one by
conviction."                           [Anonymous]
t