similar to: String frequencies in rows

Dear all, I''m trying to write a function, that will take as an argument, some aligned genome sequences, and using a sliding window, do pairwise comparisons of sequence similarity. Coding the sliding window I think I can manage but what I''m trying to get to grips with is getting it so as every pairwise comparison is made, no matter how many genomes are added, from 3 to N. So if

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

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

?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 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 = rcorr(as.matrix(dat)) dat.r <-round(dat.rcorr$r,2) however, I want to modify this correlation calculation;

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

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

We were looking at conducting a pilot program with Dell to virtualize our lab environments. We are specifically looking at Dell/Citrix provisioning servers. The OS in this setup is streamed from the server on every boot. I was curious if the R Project had any licensing stipulations in this type of setup. This is in the beginning phases and we wanted to answer some questions about licensing.

Hello, Were having issues with macs writing to our gluster system. Gluster vol info at end. On a mac, if I make a file in the shell I get the following message: smoke:hunter david$ echo hello > test -bash: test: Operation not permitted And the file is made but is zero size. smoke:hunter david$ ls -l test -rw-r--r-- 1 david realise 0 Mar 3 08:44 test glusterfs/nfslog logs thus:

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) { #

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

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

?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),

I'm trying to sort a DATAFRAME by a column "ID" that contains alphanumeric data. Specifically,"ID" contains integers all preceeded by the character "g" as in: g1, g6, g3, g19, g100, g2, g39 I am using the following code: DATAFRAME=DATAFRAME[order(DATAFRAME1$ID),] and was hoping it would sort the dataframe by ID in the following manner g1, g2, g3, g6, g19,

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

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

Does the "legend" command work in the windows version of R? Here is the command that I'm using: legend(1,1,legend=c("G1","G2","G3","G4","G5","G6"),pch=array(1:6,dim=c (1,6))) When I use this on my existing graph in a window I get a solid black box at the lower left. If I use this with the postscript driver my output