similar to: maintaining variable types in data frames

Dear All, I'm not sure if I understand the parameter stringsAsFactors correctly. I'm trying to convert the string columns in aframe1 to factors. But it seems stringsAsFactors=T in as.data.frame() doesn't do anything. Could anybody let know what is the correct way to converting strings to factors? > aframe1=data.frame(x=LETTERS[1:10], y=LETTERS[1:10], stringsAsFactors=F) >

> On Jul 7, 2017, at 6:03 AM, John Kane via R-help <r-help at r-project.org> wrote: > > This is not serious problem but I just wonder if someone can explain what is happening. > The same command within a dataframe is giving me a factor and as a plain vector is giving me a character. It's probably something simple that I have read and forgotten but I thought I'd ask.

Thanks Marc. It never occurred to me that I would need a ""stringsAsFactors" expression in a data.frame.? I could have sworn I never did before when mocking up some data but clearly I was wrong or there has been a change in R v. 3.4.1 which seems unlikely. On Friday, July 7, 2017, 10:37:29 AM EDT, Marc Schwartz <marc_schwartz at me.com> wrote: > On Jul 7, 2017, at 6:03

> On Jul 7, 2017, at 7:03 PM, John Kane <jrkrideau at yahoo.ca> wrote: > > Thanks Marc. > It never occurred to me that I would need a ""stringsAsFactors" expression in a data.frame. I could have sworn I never did before when mocking up some data but clearly I was wrong or there has been a change in R v. 3.4.1 which seems unlikely. Welcome John. Going back to

Hi, Suppose you created a dataframe like this: set.seed(28) ?dat1<-as.data.frame(simplify2array(list(letters[1:5],sample(1:20,5,replace=TRUE),6:10)),stringsAsFactors=FALSE) ?str(dat1) #'data.frame':??? 5 obs. of? 3 variables: # $ V1: chr? "a" "b" "c" "d" ... # $ V2: chr? "1" "2" "10" "18" ... # $ V3: chr?

Dear Elisa, Try this: vec1<-c(33,18,13,47,30,10,6,21,39,25,40,29,14,16,44,1,41,4,15,20,46,32,38,5,31,12,48,27,36,24,34,2,35,11,42,9,8,7,26,22,43,17,19,28,23,3,49,37,50,45) vec2<-vec1[1:26] names(vec2)<-LETTERS[1:26] label1<-unlist(lapply(mapply(c,lapply(seq(0,45,5),function(x) x),lapply(seq(5,50,5),function(x) x),SIMPLIFY=FALSE),function(i)

Hello everyone, I've come up with a problem with using paste() inside apply() that I can't seem to solve. Briefly, if I'm using paste to collapse the rows of a data frame, AND the data frame contains strings with spaces, AND there are NA values in subsequent columns, then paste() introduces spaces. This only happens with that particular combination of data values and commands. I have

Hello R helpers, I would like to automate this code for many files of the same type. But I don´t know how to make it. In particular, i don´t know how to read many files each one as an r object with the name of the file. Then a for loop would be sufficient, right? Many thanks and a happy new year. Dominic datos <- read.table('global2001.csv',head=T,sep=';',stringsAsFactors=F)

I have a data frame with columns which draw on the same underlying universe, so I want them to be factors with the same level set: --8<---------------cut here---------------start------------->8--- > z <- data.frame(a=c("a","b","c"),b=c("b","c","d"),stringsAsFactors=FALSE) > str(z) 'data.frame': 3 obs. of 2

Hi, Try this: el<- read.csv("el.csv",header=TRUE,sep="\t",stringsAsFactors=FALSE) ?elsplit<- split(el,el$st) ? datetrial<-data.frame(date1=seq.Date(as.Date("1930.1.1",format="%Y.%m.%d"),as.Date("2010.12.31",format="%Y.%m.%d"),by="day")) elsplit1<- lapply(elsplit,function(x)

I have a solution (actually a few) to this problem, but none are computationally efficient enough to be useful. I'm hoping someone can enlighten me to a better solution. I have data frame of chromosome/position pairs (along with other data for the location). For each pair I need to determine if it is with in a given data frame of ranges. I need to keep only the pairs that are within any of

Hi all, I was trying to convert a data frame to a zoo object so I can use some time series functions like lag(). But it seems then everything became a factor, so I have to convert it back to numeric to run the correct regressions. Is there a way to avoid it? Here is an example: ############################# a <- data.frame(nn =as.character(c("a", "b", "c",

Hola usuarios de R, Antes de nada presentarme, soy nuevo en R (antiguo... y bueno actual usuario de SAS), llevo menos de un mes y estoy intentando empezar con pequeñas cosas. Hay un tema que me está llevando bastante tiempo y por más que investigo y hago pruebas no logro saber como se hace; veréis tengo un data.frame el cuál tiene bastante variables y todas ellas clasificadas como char, sin

Hi Adam, I hope this is what you wanted: dat1<- read.csv("example.csv",sep="\t",stringsAsFactors=FALSE) ?str(dat1) #'data.frame':??? 102 obs. of? 5 variables: # $ species? : chr? "B. barbastrellus" "E. nilssonii" "H. savii" "M. alcathoe" ... # $ period?? : chr? "dusk" "dusk" "dusk"

I use the following code to create two data.frames d1 and d2 from a list: types <- c("integer", "character", "double") nlines <- 10 d1 <- as.data.frame(lapply(types, do.call, list(nlines)), stringsAsFactor=FALSE) l2 <- lapply(types, do.call, list(nlines)) d2 <- as.data.frame(l2, stringsAsFactors=FALSE) I would expect d1 and d2 to be the

Hola, Una manera muy sencilla de hacerlo es esta: #------------------------------- whatcol <- 0 for(i in 1:ncol(d)){ whatcol[i] <-class(d[,i]) } whatcol #-------------------------------- > whatcol <- 0 > for(i in 1:ncol(d)){ + whatcol[i] <-class(d[,i]) + } > whatcol [1] "character" "character" "factor" "factor"

Hi all, I have a date issue and would appreciate any help. I am reading a field data and n one of the columns I am expecting a date but has non date values such as character and empty. space. Here is a sample of my data. KL <- read.table(header=TRUE, text='ID date 711 Dead 712 Uknown 713 20-11-08 714 11-28-07 301 302 09-02-02 303 09-21-02',stringsAsFactors = FALSE, fill =T)

Comments interspersed, and some code at the end. On Mon, 5 Mar 2018, Sariya, Sanjeev wrote: > Thanks. I think nabble is good for programming questions. Bear with me > if I'm incorrect. You may have found R-help archives at Nabble, but R-help has nothing to do with Nabble. > > Data: Genomics SNP information I know almost nothing about using R for genomics. > Goal: I need

Hello All, This works, results <- read.table("plink.txt",T) while this doesn't. results <- read.table("plink.txt") Make sure your data frame contains columns CHR, BP, and P What does adding the "T" in read.table do? Which argument does this correspond to? I tried searching for it but didn't find the answer in: read.table(file, header = FALSE,

This is not? serious problem but I just wonder if someone can explain what is happening. The same command within a dataframe is giving me a factor and as a plain vector is giving me a character.? It's probably something simple that I have read and forgotten but I thought I'd ask. Thanks #================================================ dat1 <- data.frame(aa = letters[1:10]) str(dat1)