similar to: the problem about normalize the data frame

Hello, I have a data frame >str(dat)'data.frame': 20000 obs. of 30 variables it contains two information-two types of cancers:stage A(A1 to A10) and stage B(B1 to B20) ##totally 30 patients-20000 sets of gene expression I'd like to find the lists for top 20 differentially expressed genes using t-test (by P-value). Here is my code, unfortunately it doesn't work...I need the

Dear all, Hi, I have the problems about converting the matrix to adjacency matrix.Here's my example, a b c d e fa 1.0000000 0.4048823682 0.1228531 0.49046991 0.4945158868 0.307443317b 0.4048824 1.0000000000 0.4367475 0.96949219 0.0007378596 0.560747765c 0.1228531 0.4367474719 1.0000000 0.40037341 0.3157538204

Hi, I'm trying to build R packages. My package name is TEST. (/Users/apple/Documents/R/TEST)I type something in Terminal.But it showed an error... applede-macbook-pro:~ apple$ R CMD BUILD TESTError: cannot change to directory 'anRpackage1' I don't know where is wrong...Hope that you could help me!! Thank you!! Best Regards,Gina

Hi, I'm new to R language. There is a problem I couldn't understand. Hope you can answer my question. when i type >for (i in 1:10){ + print(sample(9,4,replace=T)) +} and it shows ten of four numbers and how do I do to calculate the frequencies in each list? I know there is a hint; list10<-vector(mode="list",length=4) But I don't know how to

Sarah's solutions are good, and here's another, even more basic: tmp1 <- unique(dat1$B) tmp2 <- seq_along(tmp1) dat1$C <- tmp2[ match( dat1$B, tmp1) ] > dat1 N B C 1 1 29_log 1 2 2 29_log 1 3 3 29_log 1 4 4 27_cat 2 5 5 27_cat 2 6 6 1_log 3 7 7 1_log 3 8 8 1_log 3 9 9 1_log 3 10 10 1_log 3 11 11 3_cat 4 12 12 3_cat 4 As a single line

I need to split a given matrix in a sequential order. Let my matrix is : > dat <- cbind(sample(c(100,200), 10, T), sample(c(50,100, 150, 180), 10, > T), sample(seq(20, 200, by=20), 10, T)); dat [,1] [,2] [,3] [1,] 200 100 80 [2,] 100 180 80 [3,] 200 150 180 [4,] 200 50 140 [5,] 100 150 60 [6,] 100 50 60 [7,] 100 100 100 [8,] 200 150 100 [9,]

Buenas tardes a todos, tengo un problema con R: ejecuto el mismo script en el ordenador del trabajo y en mi portátil con los mismos datos y obtengo resultados diferentes (siendo los correctos, los obtenidos en el trabajo): rm(list=ls()) directorio<-"C:\\Documents and Settings\\BDs\\" library(RODBC) library(car) library(gdata) ### DATOS A ###

Hi, Here's one way to approach it, using the coercion of factor to numeric. Note that I changed your data.frame() statement to avoid coercing strings to factors, just to make it simpler to set the levels. dat1 <-data.frame(N=seq(1, 12,1), B=c("29_log","29_log", "29_log", "27_cat", "27_cat", "1_log", "1_log",

Hi Sarah, Thank you so much!! I got your good ideas. Ding -----Original Message----- From: Sarah Goslee [mailto:sarah.goslee at gmail.com] Sent: Friday, May 11, 2018 11:40 AM To: Ding, Yuan Chun Cc: r-help mailing list Subject: Re: [R] add one variable to a data frame [Attention: This email came from an external source. Do not open attachments or click on links from unknown senders or

Dear all, let suppose I have following vector:   > dat1 <- c(rep("asd", 5), rep("xyz", 12), rep("erd", 17)) > dat1 <- dat1[sample(1:length(dat1), length(dat1), replace=F)] > dat1  [1] "erd" "xyz" "erd" "asd" "asd" "erd" "xyz" "asd" "erd" "erd"

HI R help, I was trying to get identical data frame from a list using two methods. #Suppose my list is: listdat1<-list(rnorm(10,20),rep(LETTERS[1:2],5),rep(1:5,2)) #Creating dataframe using cbind dat1<-data.frame(do.call("cbind",listdat1)) colnames(dat1)<-c("Var1","Var2","Var3") #Second dataframe conversion

Dear all, I have following two list object, both are basically collection of matrices : dat1 <- matrix(rnorm(25*6), ncol=6) dat1 <- split(dat1, seq(5,25,by=5)) dat1 <- lapply(dat1, matrix, ncol=6) dat2 <- matrix(rnorm(25*4), ncol=4) dat2 <- split(dat2, seq(5,25,by=5)) dat2 <- lapply(dat2, matrix, ncol=4) Now I want to replace last 4 columns of each matrix at "dat1"

Sarah et. al.: As a matter of aesthetics (i.e. my personal ocd-ness) I prefer using the public API of an object, i.e. *not* to makes use of the representation of a factor as essentially an integer vector with labels, but rather to use its documented behavior. (Feel free to ignore this remark!) Anyway, >cumsum(!duplicated(dat1$B)) [1] 1 1 1 2 2 3 3 3 3 3 4 4 will do it. This is very

Dear all, Please forgive me if there is a duplicate post; my previous mail perhaps didnt reach the list....... Let say I have following time series library(zoo) > dat1 <- zooreg(rnorm(10), start=as.Date("2010-01-01"), frequency=1) > dat1[c(3, 7,8)] = NA > dat1 2010-01-01 2010-01-02 2010-01-03 2010-01-04 2010-01-05 2010-01-06 2010-01-07 2010-01-08 2010-01-09

Um, maybe just dat1$C <- match(dat1$B, unique(dat1$B)) Indexing 1:k with numbers between 1 and k is a bit of a no-op... AFAICT, this even works without stringsAsFactors=FALSE -pd > On 11 May 2018, at 21:30 , MacQueen, Don <macqueen1 at llnl.gov> wrote: > > dat1$C <- seq(length(unique(dat1$B)))[ match( dat1$B, unique(dat1$B) )] -- Peter Dalgaard, Professor, Center for

I have the following code to write the output from auto.arima function. The issue is not in finding the model but to divert its out put fit to a file order_fit.txt. code runs but nothing is written to order_fit.txt where am I going wrong library(forecast) for (i in 1:2) { filen = paste("file",i,".txt",sep="") data <- read.table(filen) dat1 <- data[,1] xt <-

Dear R users, I came up to a problem dealing with percentiles in R. >From my previous questions: I do have a big data.frame, with lots of columns and rows. The following command enables me to calculate means for all data frame. dat1$newID<-rep(1:(nrow(dat1)/12),each=12) #if nrow(dat1)/12 is integer dat2<-with(dat1,aggregate(cbind(dat1[,1:71]),by=list(newID),mean)) What I need is to

Hello, Apologies for a similar earlier post. I didn't include enough details in that one. I am having a little trouble subsetting some data based on a grouping variable. I am using an instrument that does depth profiles of a water column. The instrument records on the way down as well as the way up. So thanks to an off-list reply I can subset the data so that all data collected at the

Hi, May be this helps: set.seed(25) dat1<- data.frame(ID=c("1001#01","1001#02","1001#03","1002#01","1002#02"),val=rnorm(5),stringsAsFactors=FALSE) ?dat1$ID<-as.numeric(gsub("#.*","",dat1$ID)) ?dat1 #??? ID??????? val #1 1001 -0.2118336 #2 1001 -1.0415911 #3 1001 -1.1533076 #4 1002? 0.3215315 #5 1002 -1.5001299 A.K.

Hello, I'm trying to use the bam() function in the R mgcv package for a large set of grouped binary data. However, I have found that this function does not take data in the format of cbind(numerator, denominator) on the left hand side of the formula. As an example, consider the following dat1 <- data.frame(id=rep(1:6, each=3), num=rbinom(18, size=10, prob=0.8), den=rbinom(18, size=5,