similar to: if else statement

Dear All, I have a matrix with data that is not organised. I would like to go through this and extract it. Each feature has 2 vectors which express the data. I also have an index of the places where the data should be cut. eg. >class(cc) "matrix" >cc [,1] [,2] [1,] 1 26 [2,] 2 27 [3,] 3 28 [4,] 4 29 [5,] 5 30 [6,] 6 31 [7,] 7 32

Hello, Sorry to all for the lack of communication, and thanks to those with the suggestions. So I have two matrixes which are different sizes, same column number but different row number. What I would like to do is to compare A to B. In the 6 and 9th column there are ref numbers (molecular masses to be exact). I would like to check A and B by these columns. If I find that the number in A[9,x]

Hello, I hope this doesn't turn into a statistics question but here I go. I am using the nls function with a Gaussian distribution, see coding below. When I run the nls I get an error back saying that I have a linear gradient. I then, of course am unable to do anything else. The data that I am using are intensity values from some mass spectrometry data. Is there something I can

Hello, I apologise if someone has already answered this but I searched and googled but didn't find anything. I have a matrix which gives me the similarity of each item to each other. I would like to turn this matrix into something like what they have in the graph package with the nodes and edges. http://cran.r-project.org/doc/packages/graph.pdf . However I cannot find a method to

Dear all, Thank you for any help. I have a data.frame and would like to plot it in 3D. I have tried wireframe() and cloud(), I got scatterplot3d(xs) Error: could not find function "scatterplot3d" > wireframe(xs) Error in wireframe(xs) : no applicable method for "wireframe" > persp(x=x, y=y, z=xs) Error in persp.default(x = x, y = y, z = xs) : (list)

Hello all and thank you, I will try to make this as clear as I can. I have a matrix, at the top of the matrix are 4 time points, the first col describes the sample, then under each time point I have the intensity values of that sample. So : Samples 0 10 30 120 <- time A 200 300 400 1000 <- data B 100 400 200 100 C 500 400 200 1000 ^ | samples I would like to plot these however, all I

Hello all, I know this is a pretty easy question but I can't find it in S poetry or R help. How can I make a negative number positive. Such as -5 to be +5 I tried +(-5), but that didn't work. So no, I don't mean taking a -5^2 just to get a positive number. This is in a function so it's not just -5 it's x. :) Thanks, Paul -- Research Technician Mass Spectrometry o The

hello all, Quick question, how do I get the p value out of the anova? Thanks, Paul > pb<-aov(as.numeric(diff[5,16:33]) ~ grF) > summary(pb) Df Sum Sq Mean Sq F value Pr(>F) grF 3 2.7860e+10 9.2867e+09 4.2236 0.02534 * Residuals 14 3.0783e+10 2.1988e+09 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 '

Thanks again everyone, I have a problem trying to install RMySQL on Suse 10.0. I have used both the install.packages() and the 'R CMD INSTALL RMySQL..." neither worked. I get the error message telling me that it cannot find mysql. So I have set the flags using export PKG_CPPFLAGS="-I</usr/bin/>" export PKG_LIBS="-L</usr/lib/mysql>" After that I run R

Hello all, I consider myself pretty new to the whole OO based programming so I'm sorry if I'm doing something stupid. > xml<-read.metlin(url) Error in function (classes, fdef, mtable) : unable to find an inherited method for function "read.metlin", for signature "url" >read.metlin standardGeneric for "read.metlin" defined from package

Umm sorry to bother everyone again but I'm having trouble with my if statement. I come from a perl background so that's probably my problem! :) So here is my code: if (any(lgAB>4) | any(lgAB<-4)){ freq_AB<-hist(lgAB, type="o", plot=F) else freq_AB<-hist(lgAB, breaks=br,type ="o", plot=F) } And I get > source("E:/R/GMDA-1.1.R") Error in

Hello all, I'm looking for a way to be able to read a text file into R. It's a csv file but when I do "txt <-read.table("F00.csv", header=T, sep=",")" It doesn't read the file properly, and I only get 2 columns. If I open it up in OOc or Excel it open right with 7 columns. What I would really like to do is read the file as text and then split it and

Dear all, Just a quick one, hopefully. I have a histogram made from the method 'hist()'. How do I get labels on the bars? Such that the bars will have the x axis on the bar, not the frequency of the point but the number of the point itself. To make a quick summary, I want the the numbers from 'losses' (below) to be on the bar's. Thanks, Paul

So thanks for the help, I have a matrix (AB) which in the first column has my bin numbers so -4 - +4 in 0.1 bin units. Then I have in the second column the frequency from some data. I have plotted them and they look roughly Gaussian. So I want to fit them/ find/optimize "mu", "sigma", and "A". So I call the nls function : nls_AB <- nls(x ~ (A/sig*sqrt(2*pi))*

Hello, This should be a fairly simple question. I have 2 matrix's and I want to get the difference between them and then remove the difference from one of them. So A vs B, anything I see in B I want to remove from A, and then print out the final matrix. I have some code here but at the moment I'm getting stuck on the creating a matrix, cbind problems. clean <- function (A, B,

Hello everyone, I have an overlay plot it's nice but you can't see all the data. I would like to know if there is a way to get a plot that gives a side by side plot so that each plot would be next to each other. The two plots have the same data are of different species. At the moment this is the code I'm using: exp<-cbind(abs(round(rnorm(10),2)*10), seq(100, 200, by=10))

Hi, I need to convert numbers to characters in order to index an array when I encountered the number 100000 which turned into "1e+05". Does anyone know how to get around this problem? Lana Schaffer Biostatistics/Informatics The Scripps Research Institute DNA Array Core Facility La Jolla, CA 92037 (858) 784-2263 (858) 784-2994 schaffer at scripps.edu

Hi, I am using this subset statement and it works outside a function. LIS[[i]]<- lapply(LI, subset, select=cov[[i]]) However, wrapped inside a function this statement produces the same values for every LIS[[1]] which is only the first subset of LI. Does anyone know why is not working correctly inside a function? ff = factor(covariate) nLev <- nlevels(ff) cov <-

Hello all, Let me try this again with some code. I am trying to make a new object and it's not working. I get : Error in rematchDefinition(definition, fdef, mnames, fnames, signature) : methods can add arguments to the generic only if '...' is an argument to the generic as my error. I have included my code. It is from the xcms package. Any help would be great even if

Hello all, I would like to compare two sets of count data which form Poisson distributions. I'd like to generate some sort of p-value of the likely-hood that the distributions are the same. Thanks in advance for your advice. Cheers, Mark Mark Gosink, Ph.D. Head of Computational Biology Scripps Florida 5353 Parkside Drive - RFA Jupiter, FL 33458 tel: 561-799-8921 fax: