similar to: Pairwise Comparrisons

There was a BioConductor thread today where the poster wanted to find pairwise difference between columns of a matrix. I suggested the slow solution below, hoping that someone might suggest a faster and/or more elegant solution, but no other response. I tried unsuccessfully with the apply() family. Searching the mailing list was not very fruitful either. The closest I got to was a cryptic chunk

Dear R users, I have a data generated as the following, dat <- data.frame(matrix(sample(24), nrow=4)) dimnames(dat) <-list(rownames=c('g1','g2','g3','g4'), colnames=c("A_CH1","A_CH2","B_CH1","B_CH2","C_CH3","C_CH3")) » dat A_CH1 A_CH2 B_CH1 B_CH2 C_CH3 C_CH3 g1 16 24 7 9 14

Hi I am using the following script to average a set of data 0f 62 columns into 31 colums. The data consists of values of ln(0.01) or -4.60517 instead of NA's. These need to be averaged for each row (i.e 2 values being averaged). What I would I need to change for me to meet the conditions: 1. If each run of the sample has a value, the average is given 2. If only one run of the sample has a

?s 17:39 de 25/07/2024, Yuan Chun Ding via R-help escreveu: > Hi R users, > > I generated a square correlation matrix for the dat dataframe below; > dat<-data.frame(g1=c(1,0,0,1,1,1,0,0,0), > g2=c(0,1,0,1,0,1,1,0,0), > g3=c(1,1,0,0,0,1,0,0,0), > g4=c(0,1,0,1,1,1,1,1,0)) > library("Hmisc") > dat.rcorr =

HI Rui, Thank you for the help! You did not remove a row if zero values exist in both column pair, right? Ding From: Rui Barradas <ruipbarradas at sapo.pt> Sent: Thursday, July 25, 2024 11:15 AM To: Yuan Chun Ding <ycding at coh.org>; r-help at r-project.org Subject: Re: [R] please help generate a square correlation matrix ?s 17:?39 de 25/07/2024, Yuan Chun Ding via R-help

Hi all, I'd like to write a piece of code which will remove columns from a matrix, if the column contains only one value, say every value in the column is a "3": Matrix <- matrix(NA, nrow=5, ncol=4) Matrix[,1] <- c(1,2,3,4,5) Matrix[,2] <- c(3,3,3,3,3) Matrix[,3] <- c(5,4,3,2,1) Matrix[,4] <- c(5,1,4,3,2) [,1] [,2] [,3] [,4] [1,] 1 3 5 5 [2,] 2

On how to even search for this "feature" as I have no idea on what it can be. I've got a meridian linked to * (by EuroISDN) which is linked to a ISDN30. I can make calls from the meridian, and receive calls into the meridian. Great stuff. However, if someone dials an invalid number, then instead of hearing a "three tone", the line just drops and goes dead. The console

Hi Rui, You are always very helpful!! Thank you, I just modified your R codes to remove a row with zero values in both column pair as below for my real data. Ding dat<-gene22mut.coded r <- P <- matrix(NA, nrow = 22L, ncol = 22L, dimnames = list(names(dat), names(dat))) for(i in 1:22) { #i=1 x <- dat[[i]] for(j in (1:22)) { #j=2 if(i == j) { #

?s 20:47 de 25/07/2024, Yuan Chun Ding escreveu: > Hi Rui, > > You are always very helpful!! Thank you, > > I just modified your R codes to remove a row with zero values in both column pair as below for my real data. > > Ding > > dat<-gene22mut.coded > r <- P <- matrix(NA, nrow = 22L, ncol = 22L, > dimnames = list(names(dat),

Hola Javier, Intenta los siguientes cambios g4 <- function(codigo_llega, n_caracteres){ codigo_llega <- as.character(codigo_llega) if(n_caracteres == 6) res <- substr(codigo_llega, start=0, stop=4) else res <- codigo_llega res } g4 <- Vectorize(g4) x <- c('Jorge Velez','Javier Marcuzzi','Daniel Merino','Susana deus Alvarez', 'Carlos

If I have understood the request, I'm not sure that omitting all 0 pairs for each pair of columns makes much sense, but be that as it may, here's another way to do it by using the 'FUN' argument of combn to encapsulate any calculations that you do. I just use cor() as the calculation -- you can use anything you like that takes two vectors of 0's and 1's and produces fixed

I have a really simple question I have a data frame of 8 variables (the first column is the subjects' id): SubID G1 G2 G3 G4 W1 W2 W3 W4 1 6 5 6 2 6 2 2 4 2 6 4 7 2 6 6 2 3 3 5 5 5 5 5 5 4 5

Chris Mauritz wrote: > > The 32bit PPC release is for the older G3/G4 macs, no? I've got a > pile of unused G4 400-800mhz powermacs just sitting around The ppc32 distro will work on any NewWorld G3, G4 based machine ( and 32bit ibm power, like the older rs6k's ) - it _also_ works fine on the MacMini and G4 Xserve's. With some common sense tweaks, a G4 450Mhz is a very

Carlos: Te agradezco mucho tu rápida respuesta y mucho me apena haber planteado tan mal el problema. Porque la matriz en realidad es: g1 g2 g3 g4 g5 g6 g7 g1 0 18 13 16 11 12 15 g2 18 0 25 13 22 16 10 g3 13 25 0 28 23 13 25 g4 16 13 28 0 6 7 3 g5 11 22 23 6 0 18 17 g6 12 16 13 7 18 0 8 g7 15 10 25 3 17 8 0 Entonces cada cantidad debe conservar la pertenencia al grupo

Let's go back to the original posting. > > > >> in each column, less than 10% values are 1, most of them are 0; > > > > > > > >> so I want to remove a row with value of zero in both columns when calculate correlation between two columns. > > So we're talking about correlations between binary variables. Suppose we have two 0-1-valued

Hi All, How do I add these axis labels? ############################################### p=seq(0,1,length.out=500) p=p[-c(1,length(p))] g1=log(p/(1-p)) g2=qnorm(p) g3=log(-log(1-p)) g4=-log(-log(p)) plot(p,g1, 'n',ylim=c(-5,5),las=1, bty='n', xaxt='n',yaxt='n', xlab="",ylab="" ) lines(p,g1,lty=1,col=1) lines(p,g2,lty=1,col=2)

Dear R users, I have noted a difference in the merge distances given by hclust using centroid method. For the following data: x<-c(1009.9,1012.5,1011.1,1011.8,1009.3,1010.6) and using Euclidean distance, hclust using centroid method gives the following results: > x.dist<-dist(x) > x.aah<-hclust(x.dist,method="centroid") > x.aah$merge [,1] [,2] [1,] -3 -6

Dear R helpers,   Following is a R script I am using to run the Fast Fourier Transform. The csv files has 10 columns with titles m1, m2, m3 .....m10.     When I use the following commands, I am getting the required results. The probelm is if there are 100 columns, it is not wise to define 100 commands as fk <- ONS$mk and so on. Thus, I need some guidance to write the loop for the STEP A and

Hi We deployed centos 4.8 on HP DL380 G2, G3 and G4 servers (same number of cpus). The same software is running on all servers. Only on G2 servers we see a low cpu utilization combined with high load-average. For example, during the same time of day and same workload, on a G2 I see a l-avg of 11.31 and on a G4 I see 2.68. The server response time seems unchanged and we don't have any problem

Hi, I'm new to R. I was wondering if anyone has an OS X native installation (not Carbon) of R on a G3 Macintosh that is able to run linear regression? I installed Jan de Leeuw's R binary for OS 10.2 Jaguar on my laptop, which is still a G3 model. When I try to run a simple regression, using lm, I get an "Illegal Instruction" message and R quits. Jan thinks it may be the case