similar to: show plot

Dear All I can not download R for Linux. I do not know which file I should install? Best Regards, Soheila

Dear All Is there any function for "median test" in R? Best Regards, Soheila [[alternative HTML version deleted]]

Dear All How can I transform 1 column to 2 columns in R? 211217_s_at GO:0005249 211217_s_at GO:0005251 211217_s_at GO:0005515 211217_s_at GO:0015271 211217_s_at GO:0030955 211217_s_at GO:0005249 211217_s_at GO:0005251 211217_s_at GO:0005515 211217_s_at GO:0015271 211217_s_at GO:0030955 Best Wishes, Soheila [[alternative HTML version deleted]]

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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.

Dear All I want to chose just spacial columns in R. (read table) data1<- read.table("/home/Documents/data.txt",header=F,sep = "\t", as.is =F) data.2<-data1[sub.data[,1],] The dimension of data.2 is correct but my data are transformed NA head(data.2) V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20 V21 V22 V23 V24 NA <NA> NA NA NA

Dear All I want to arrange our output in 3 columns in R. Example: N 04320 03010 00514 00604 00563 03430 03022 V 22.000000 0.000000e+00 15.0000000 15.0000 13.0000 1.300000e+01 51.0000000 p 0.640625 2.910383e-11 0.4257812 0.4375 0.6875 3.356934e-04 0.2434998 N 03060 04140 03050 00562 00565 V 20.0000000 62.00000000 273.0000000

Dear ALL I have this function in R: func_LN <- function(data){ med_ge <- matrix(c(rep(NA,nrow(data)*ncol(data))), nrow = nrow(data), ncol=ncol(data), byrow=TRUE) T <- matrix(c(rep(NA,length(n)*ncol(data))), nrow = length(n), ncol=ncol(data), byrow=TRUE) Tdiff<- matrix(c(rep(NA,length(n)*ncol(data))), nrow = length(n), ncol=ncol(data), byrow=TRUE) T1<- c(rep(NA,ncol(data)))

Unable to resolve error in seq.default(which(text == "")[1] + 1, length(text), 1): wrong sign in 'by' argument I am trying to run this code get.msg <- function(path) { con <- file(path, open = "rt", encoding = "latin1") text <- readLines(con) # The message always begins after the first full line break msg <- text[seq(which(text ==

Hello Everyone, My name is Thomas and I have been using R for one week. I recently found your site and have been able to search the archives of posts. This has given me some great information that has allowed me to craft an initial design to an inquiry I would like to make into the breakdown of McNemar's test. I have read an intro to R manual and the posting guides and hope I am not violating

Dear R users I searched some sources but i did not find an answer.Please give me some hint to following problem. I would like to compute a summary statistic for some vector for different factor levels. I know I can use tapply or aggregate but I do not know if there is a way how to use function with several (two) variable input (like weighted.mean). I wrote a simple a function for factor

On 8 September 2017 at 11:25, PIKAL Petr <petr.pikal at precheza.cz> wrote: > > Moller Skarbiniks Pedersen > My program which is slow looks like this: > > > > filename <- "digits.txt" > > lines <- readLines(filename) > > why you do not read a file as a whole e.g. by > > lines<-read.table("digits.txt") > Good idea.

Hi, with the code below I try to organize sections of a data.frame. The dataframe has 9 data columns. I'd like to order the columns of the data.frame by the sum of the columns' entries. The sections of the columns are identified by "i". The order of the columns is given by the vector "z". Each section of the data.frame is ordered differently. This can be seen when

Hi dear, I want to estimate d.f for Chi-squared distribution: est.chi[i,]<-c( fitdistr(as.numeric(data2[,i]),"chi-squared",start=list(df=1))$estimate)Warning message:In optim(x = c(7.86755, 7.50852, 7.86342, 7.70589, 7.70153, 7.58272, : one-diml optimization by Nelder-Mead is unreliable: use "Brent" or optimize() directly Who can help me to solve this problem? Best

Sorry to hijack this thread. I have a similar but slightly different situation. Using the original poster's example, how to elegantly get the mean of column V2 when column V1 is either A or C and F1 is 0? Thanks, Gang On Nov 13, 2007, at 5:30 AM, Petr PIKAL wrote: > Hi > > r-help-bounces@r-project.org napsal dne 13.11.2007 10:59:09: > >> Dear list, >> >> I

Dear sirs, I have tried to find a command to fit(plot) a regression line but I haven't found any command on the html.help. My question is if I am using a regression for instance reg<-lm(y~x) and I would like to make a regression line, to see how well my data is fit. Can you please help me with this?? Yours Sincerely Soheila Zad

Thank you Pikal and Bert. My apology for posting parts of my previous email in HTML. Bert's suggestion will work but i am wondering if there is an alternative especially in the case where the data frames are big; that is the difference in lengths among them is large. Below is a list of sample date frames and desired result. EK

Dear R community, I am not experienced in writing functions and need your help to understand the strategy to face the following problem: I have a group of independent numerical variables, let's say a<-c(1,3,2,6,9,2,2,3,4,1) b<-c(2,3,3,4,5,6,2,1,1,6) c<-c(0,2,4,4,7,6,7,1,2,2) d<-c(0,0,0,1,0,2,1,3,1,2) e<-c(9,2,3,1,1,1,0,2,5,6) and a grouping variable,

Building on Petr's suggestion, you could modify his code to get all 10 samples at once in a compact format: > Samples <- lapply(lll, function(x) replicate(10, sample(x, rep=TRUE))) # Samples is a list containing 3 matrices, one for each group # Each column gives the index (row) numbers for a particular sample > str(Samples) List of 3 $ A: int [1:3, 1:10] 2 1 3 1 1 3 2 3 1 2 ... $

Hi R User, I am new in R and trying to create tables with selecting rows randomly (but with replacement) for each group but each group should have same number as original. Is it possible to create it using the following example data set? Your help is highly appreciated. dat1<-structure(list(RegionA = structure(c(1L, 1L, 2L, 3L, 3L, 4L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L, 6L),