similar to: help to add a new column filled with value 1

Dear All, I have two data like this : $cat main.csv name,id,memory,storage mohan,1,100.20,1.10 ram,1,200,100 kumar,1,400,50 xxx,1,100,40 aaa,1,800,45 mount,1,200,80 > main <- read.csv(file='main.csv',sep=',' , header=TRUE) > main name id memory storage 1 mohan 1 100.2 10 2 ram 1 200.0 100 3 kumar 1 400.0 50 4 xxx 1

Dear All, I have a data frame like this: name ip Bsent Breceived a 1 0.00 0.00 a 2 1.43 19.83 a 1 0.00 0.00 a 2 1.00 1.00 b 1 0.00 2.00 b 3 0.00 2.00 b 2 2.00 0.00 b 2 2.00 0.00 b 1 24.40 22.72 c

Dear All, I have .csv file it looks like this : rawdata <- read.csv(file='/home/Mohan/Rworks/tmp/VMList_User.txt',sep='\t' , header=FALSE) > head(rawdata,n=5) Tenant Domain Owner Current State 1 \\N ROOT admin Running 2 \\N ROOT admin Stopped 3 \\N ROOT admin Running 4 \\N

Dear R users: Is there a built-in and simple way to insert new columns after other columns in a dataframe? I.e. currently I have: V1 V2 V3 V4 [1,] [2,] Etc. But I want V1 V5 V2 V3 V4 [1,] [2,] Etc. Can this be done in one line? Jon Minton [[alternative HTML version deleted]]

Dear All, I am using this query it returns id : id <- dbGetQuery(con1,"SELECT id FROM tenants WHERE name LIKE '%consim%'") But In my case the string "consim" is there in another variable(it is coming from configuration file); > str <- "consim" I am trying to replace the string some this like, but it not working: > id <-

Hii, I will create boxplots from matrices. I have the following data sets: 5.0 1.78 2.99 2.019 0 10.0 1.79 3.00 1.744 0 15.0 1.78 2.98 1.936 0 20.0 1.78 2.99 1.975 0 25.0 1.73 2.91 3.591 0 30.0 1.79 3.00 1.966 0 35.0 1.79 3.00 2.451 0 40.0 1.79 3.00

Hello List, My goal is to apply a user-defined function on several columns of a data frame. When testing the code on a reproducible example below, I get the following error message. > #now Write a new function using the above cut ()/quantile function to apply on different columns of the data frame > > CutQuintiles <- function(x) { + cut (test1$x,quantile (test1$x,

I am recoding some data. Many values that should be 1.5 are recorded as 1-2. Some example data and my solution is below. I am curious about better approaches or any other suggestions. Thanks! # example input data myData <- read.table(textConnection("id, v1, v2, v3 a,1,2,3 b,1-2,,3-4 c,,3,4"),header=TRUE,sep=",") closeAllConnections() # the first column is IDs so remove

Full_Name: Steven McKinney Version: 2.9.0 OS: Mac OS X 10.5.6 Submission from: (NULL) (142.103.207.10) A corrupt data frame can be constructed as follows: foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4]

Hi all .., I have a list of 7000 dataframes with similar column headers and I wanted to add a new column to each dataframe based on a certain condition which is the same for all dataframes. When I extract one dataframe and apply my code it works very well as follows :- First suppose this is my first dataframe in the list > OneDF <- Mylist[[1]] > OneDF ID Pdate

Folks: I consider my reply below rather clumsy: One has to keep track of index numbers other than that which is inserted and must separately change column names. Is there as "essentially better" way to do this, either via base R or via an R package. I leave it to you to define "essentially better." Thanks. Cheers, Bert On Mon, Aug 1, 2011 at 10:17 AM, Bert Gunter

Hi useRs, A recent coding infelicity along these lines yielded a corrupt data frame. foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4] > bar X1 X2 X3 X4 NewCol 1 1 4 7 10 <NA> 2 2 5 8 11

Dear list, I have two vectors: x <- c("one","two") y <- paste(rep(x,2),"blah") I want to replace all occurrences of each element of x in y with something else, so that y looks like this: y [1] "something else blah" "something else blah" "something else blah" [4] "something else blah" I can do this using a loop: for (

Dear list, I have data like this: dat1 <- data.frame(subject=rep(1:10,2), cond1=rep(c("A","B"),each=5), cond2=rep(c("C","D"),each=10), choice=sample(0:1,10,replace=TRUE)) I would like to compare subjects' "choice" for (cond1=="A" & cond2=="C") vs

Hi, I'm trying to concatenate values from two columns in a data frame. For example, I have the following data.frame: C1 C2 C3 C4 C5 A B *F C* Q G H *I J* T K D *R S* E P L *M N* O I'd like to concatenate text from columns C3 and C4, to yield either a list or vector, like so: NewCol FC IJ RS MN Is this feasible in R? Thanks!

Dear friends, when i use glm() to fit my data, i use glm(formula = snail ~ vegtype + mhveg + humidity + elevation + soiltem, *family = Gamma(link = inverse),* data =a,)) It shows: error in eval(expr, envir, enclos) : *gamma distribution don't allow negative value*. But i use result<-glm(formula = snail ~ vegtype + mhveg + humidity + elevation + soiltem, family = poisson, data =a) #this

Dear Rers, I have a function to barplot() a matrix, eg myfun <- function(x, ...) { barplot(x , ... )} (The real function is more complicated, it does things to the matrix first.) So I can do: m1 <- matrix(1:20,4) myfun(m1) myfun(m1, main="My title") I'd like to be able to add the number of rows of the matrix passed to the function to the "..." argument, eg

Hi list, I have data in a dataframe t1, with a column for different amounts spent, a column what it was spent on, and a column with dates, from which I create a new column with months. Example: amount <- rep(c(10,20,30),3) what <- rep(c("food","books","cycling"),3) when <- c(rep("2008-09-05",5),rep("2008-10-07",4)) t1 <-

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Dear list, I would like to start some R workshops at King's College London, and to do so, I would like to use the "Use R!" logo at http://www.agrocampus-ouest.fr/math/useR-2009//useR%21%202008_fichiers/useR-middle.png Since it seems to be difficult to get a shell account at KCL, I also went ahead and registered use-r.org.uk and am starting to put together a website at