similar to: New class: data.table

Greetings all, I am getting an error message that is stifling me. Any ideas? > ## Define Directories ## > load_from <- "/home/mcc/Dropbox/abrodsky/kegg_combine_data/" > save_to <- "/home/mcc/Dropbox/abrodsky/ttest_results/" > > ############################### > ## Define Columns To Compare ## > compareA <- "log_b_rich" > compareB

Hi, I am trying to use the nls() function to closely approximate a vector of values, colC and I'm running into trouble. I am not sure how if I am asking the program to do what I think its doing, because the same minimization in Excel's Solver does not run into problems. If anyone can tell me what is going wrong, and why I'm getting a singular convergence(7) error, please tell me. I

Hi, I am making my way down the learning curve of R, and find it a great language with so many helpful users! Below is an example of what I'm trying to do, but can't quite figure out the right path to go down. Here's what I have: Main is a time series of data with columns Cola and Colb Cola Colb 1 1 1 2 1

Hello every body, I am quite a new user of R so I beg your pardon for this naive question and the lake of syntax with wich I ask it: I have a data frames like this: cola colb 1 c 1 i 1 i 1 c 2 i 2 c 2 c 2 i 2 i ... 10000 I would like ,for each level of cola and for x in colb: -if colb[x]=="i" and colb[x-1] does not exist (first row in dataframes), then replace colb[x] by

Hi, I have two dataframes: The first, df1, contains some missing data: cola colb colc cold cole 1 NA 5 9 NA 17 2 NA 6 NA 14 NA 3 3 NA 11 15 19 4 4 8 12 NA 20 The second, df2, contains the following: cola colb colc cold cole 1 1.4 0.8 0.02 1.6 0.6 I'm wanting all missing data in df1$cola to be replaced by the value of df2$cola.

Hi, May be this helps: funcName<- function(df1, x){ ?whatCol=df1[[x]] ?print("Got it") ?print(whatCol) ?} ? funcName(df,"ColA") #[1] "Got it" #[1] 1 2 3 4 5 ? funcName(df,"ColB") #[1] "Got it" #[1] A B C D E #Levels: A B C D E A.K. >I am trying to extract the 2nd column from a dataframe using a function called funcName. Note this is an

Greetings all, I am curious to know if either of these two sets of code is more efficient? Example1: ## t-test ## colA <- temp [ , j ] colB <- temp [ , k ] ttr <- t.test ( colA, colB, var.equal=TRUE) tt_pvalue [ i ] <- ttr$p.value or Example2: tt_pvalue [ i ] <- t.test ( temp[ , j ], temp[ , k ], var.equal=TRUE) ------------- I have three loops, i, j, k. One to test the all of

Dear all, I have a problem with the "subset()" function. I spent all day yesterday with a collegue to solve it and we did not find a satisfying solution (even in the archived mails), so I ask for your help. Let's say (for a simple example) a matrix mat: R> mat cola colb colc [1,] 1 4 7 [2,] 2 5 8 [3,] 3 6 9 My goal is to select the lines of the matrix on the basis of the

Hello all! I am currently trying to sort a data frame in a particular way, but I am having some difficulties with this. Specifically I want to sort the below dataset in such a way that there is only one line per ProteinID and if there are multiple GeneID or GeneName entries for a single proteinID, that they be concatenated with a comma separating them. The way I have done it earlier worked fine

colnames(dd) #[1] "col1" "colb" null_vector<- colnames(dd) sapply(null_vector,makeNull,dd) #???? col1 colb #[1,]?? NA??? 4 #[2,]??? 2?? NA #[3,]??? 3??? 2 #[4,]??? 4?? NA #[5,]??? 1??? 4 #[6,]?? NA??? 5 #[7,]??? 1??? 6 A.K. >I am trying to make a column value in a dataframe = NA if there is a 0 or high value in that column. I need to do this process repeatedly, hence

Hello, I have a general formatting question. I have two columns of data: ColA <- c("m", "m", "m", "m") ColB<- c("d","d","d","d") And I would like to reorder them into a new column that looks like this: ColC<-

I am trying to read some data off the zillow site. Newbie to xml, html, parsing and the xml package. I've been able to load the web page I'm interested with the following code but I'm not sure of the next step to get the information I'm interested in into R : library(XML) url <- "http://www.zillow.com/homes/511 W Lafayette St, Norristown, PA_rb" doc <-doc <-

Why does this sapply code change df3 but not df1? Thanks df1 <- read.table(text=" cola colb colc cold cole 1 NA 5 9 NA 17 2 NA 6 NA 14 NA 3 3 NA 11 15 19 4 4 8 12 NA 20 ", header=TRUE) df2 <-df1*2 df1 df2 df3 <-sapply(names(df1),function(x) {df1[[x]]<- df2[[x]]}) df1 df3 [[alternative HTML version deleted]]

Hi, Please see below for post on r-help regarding data.frame() and the possibility of dropping rownames, for space and time reasons. I've made some changes, attached, and it seems to be working well. I see the expected space (90% saved) and time (10 times faster) savings. There are no doubt some bugs, and needs more work and testing, but I thought I would post first at this stage. Could some

Hi, The following function works, but is there a neater way to write it? f = function(x,...) { # return a character vector of the arguments passed in after 'x' gsub(" ","",unlist(strsplit(deparse(substitute(list(...))),"[(,)]")))[-1] } > f(x,a,b,c*d) [1] "a" "b" "c*d" > Thanks. [[alternative HTML

Hi all, Does anyone know a solution for this error ? > tkwidget(dlg, "iwidgets::spinint", range="{0 23}") Error in structure(.External("dotTclObjv", objv, PACKAGE = "tcltk"), class = "tclObj") : [tcl] wrong # args: should be ".31.1.19 configure -range {begin end}". Thanks, Matthew [[alternative HTML version

Dear all, The data.table package was released back in August 2008. This email is to publicise its existence in response to several suggestions to do so. It seems I didn't send a general announcement about it at the time and therefore perhaps, not surprisingly, not many people know about it. Glancing at some r-help threads recently supports the idea of sending a public announcement. The

Dear all, The data.table package was released back in August 2008. This email is to publicise its existence in response to several suggestions to do so. It seems I didn't send a general announcement about it at the time and therefore perhaps, not surprisingly, not many people know about it. Glancing at some r-help threads recently supports the idea of sending a public announcement. The

Gracias, Carlos. Habia pensado en algo similar usando sapply(): sapply(seq(1, ncol(vtmp), by = 2), function(i) c(rbind(as.character(vtmp[, i]), as.character(vtmp[, i+1])))) Dependiendo de la dimension de los datos, quizas mapply() sea mas eficiente que sapply(). Saludos cordiales, Jorge.- 2015-02-25 1:01 GMT+11:00 Carlos Ortega <cof en qualityexcellence.es>: > Hola, > > Este

Hi, In the example below why is d 10 times bigger than m, according to object.size ? It also takes around 10 times as long to create, which fits with object.size() being truthful. gcinfo(TRUE) also indicates a great deal more garbage collector activity caused by data.frame() than matrix(). $ R --vanilla .... > nr = 1000000 > system.time(m<<-matrix(integer(1), nrow=nr, ncol=2)) [1]