similar to: converting the string columns in a data.frame to factors?

Suppose that I have the following list of lists of frames 'root' (let's call it a 'tree' of frames). I want to flatten it to be a list of frames. However, if I unlist(root), it will flatten the frames as well. Is there a simply way to flatten the tree to certain depth? aframe1=data.frame(x=1:3,y=1:3) aframe2=data.frame(u=7:9,v=11:13) aframe3=data.frame(p=3:5,q=6:8)

Hi, I have a list which looks like this... > str(y) List of 10 $ : chr [1:4] "ABCD" "5" "0" "1" $ : chr [1:4] "DEF" "15" "1" "16" $ : chr [1:4] "AAA" "2" "17" "8" $ : chr [1:4] "SSS" "15" "25" "1" $ : chr [1:4] "III"

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?

Suppose X and Y are two data frames with the same structures, variable names and dimensions but with different data and different patterns of missing. I want to replace missing values in Y with corresponding values from X. I'll construct a simple two-by-two case: > X <- as.data.frame(matrix(c("a","b",1,2),2,2), stringsAsFactors=FALSE) > X[,2] <-

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

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

> 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.

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)

> 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

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,

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",

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 everybody! I'm still quite new in R and I don't really understand this whole "vectorization" thing. For now, I use a for loop (and it works fine), but I think it would be useful to replace it. I want to export the result of a test statistic, which is stored as a list, into a csv file. I therefore have to export each element of the list separately. Here is the code: ----

If row numbers can be dispensed with, then tidyr makes this easy with the unnest function: ##### library(dplyr) #> #> Attaching package: 'dplyr' #> The following objects are masked from 'package:stats': #> #> filter, lag #> The following objects are masked from 'package:base': #> #> intersect, setdiff, setequal, union library(purrr)

Hola: A ver si me podéis ayudar que estoy atascado... Necesito contar los subcasos de la columna 2 de un dataframe respecto a los casos de la columna 1. Es decir, tengo un data.frame a<-c(rep(c('a','b','c','b','c'),3),'b') b<-c(rep(c('x','y','z','w'),4)) c<-c(rep(c(1,1,0,0),4))

Dear all, how can I perform a string operation like strsplit(x," ") on a column of a dataframe, and put the first or the second item of the split into a new dataframe column? (so that on each row it is consistent) Thanks Boris

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)

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

Hi, I am running the following script for a different (much larger data frame): DF = data.frame(read.table(textConnection(" A B C D E 1 1 a 1999 1 0 2 1 b 1999 0 1 3 1 c 1999 0 1 4 1 d 1999 1 0 5 2 c 2001 1 0 6 2 d 2001 0 1 7 3 a 2004 0 1 8 3 b 2004 0 1 9 3 d 2004 0 1 10 4 b 2001 1 0 11 4 c 2001 1 0 12 4 d 2001 0