search for: name1

Hi R-Helpers, Sorry to bother you, but I have a simple task that I can't figure out how to do. For example, I have some names in two columns NamesWide<-data.frame(Name1=c("Tom","Dick"),Name2=c("Larry","Curly")) and I simply want to get a single column NamesLong<-data.frame(Names=c("Tom","Dick","Larry","Curly")) > NamesLong Names 1 Tom 2 Dick 3 Larry 4 Curly Stack produces...

Just to repeat: you have NamesWide<-data.frame(Name1=c("Tom","Dick"),Name2=c("Larry","Curly")) and you want NamesLong<-data.frame(Names=c("Tom","Dick","Larry","Curly")) There must be something I am missing, because NamesLong <- data.frame(Names = c(NamesWi...

Hi all, let say I have following matrix: > Dat <- matrix(1:30, 5, 6); colnames(Dat) <- rep(c("Name1", "Names2"), 3) > Dat Name1 Names2 Name1 Names2 Name1 Names2 [1,] 1 6 11 16 21 26 [2,] 2 7 12 17 22 27 [3,] 3 8 13 18 23 28 [4,] 4 9 14 19 24 29 [5,] 5 10 15 20...

...s=() ip=() while read -r LINE do NODENAME=` echo $LINE | cut -f 1 -d ','` IP=` echo $LINE | cut -f 2 -d ','` names[index]="$NODENAME" ip[index]="$IP" index=`expr index+1` total=`expr total+1` done <<< $(cat list.txt) simple file: more list.txt name1,ip1 name2,ip2 name3,ip3 output when running: sh -x ./test_bash.sh + index=0 + total=0 + names=() + ip=() ++ cat list.txt + read -r LINE ++ echo name1,ip1 name2,ip2 name3,ip3 ++ cut -f 1 -d , + NODENAME=name1 ++ echo name1,ip1 name2,ip2 name3,ip3 ++ cut -f 2 -d , + IP='ip1 name2' + names[i...

I have some questions about the RODBC package. library(RODBC) # required for those who want to repeat these lines 1st, I noticed that the following sequence does not work: channel <- odbcConnextExcel("test.xls") tables <- sqlTables(channel) name1 <- tables[1, "TABLE_NAME"] # this should be the name plan1 <- sqlFetch(channel, name1) # bang! odbcClose(channel) However, I can circumvent this with: channel <- odbcConnextExcel("test.xls") tables <- sqlTables(channel) name1 <- tables[1, "TABL...

...timeM1 statusM1 xM1 timeM2 statusM2 xM2 timeM3 statusM3 xM3 1 2 3 4 5 6 Where M1,M2, M3 hve no similarity except they have a max string length of 7. Examples are mcw0045, adl0003, lei0101. Now, what I want to do is Function(M1, M2, M3, datafile) { xnew <- readtable(datafile) name1 <- paste("time",M1",sep"") name2 <- paste(status",M1,sep"") name3 <- paste("x",M1,sep"") fit1 <- survfit(Surv(name1,name2)~name3,data=xnew) . . repeat for M2 and M3 . . par(mfcol=c(1,3))...

...-bounces at r-project.org> On Behalf Of Heinz Tuechler Sent: Monday, April 3, 2023 4:39 PM To: r-help at r-project.org Subject: Re: [R] Simple Stacking of Two Columns Jeff Newmiller wrote/hat geschrieben on/am 03.04.2023 18:26: > unname(unlist(NamesWide)) Why not: NamesWide <- data.frame(Name1=c("Tom","Dick"),Name2=c("Larry","Curly")) NamesLong <- data.frame(Names=with(NamesWide, c(Name1, Name2))) > > On April 3, 2023 8:08:59 AM PDT, "Sparks, John" <jspark4 at uic.edu> wrote: >> Hi R-Helpers, >> >> Sorry...

Hi, You were on the right track using stack(), but you just pass the entire data frame as a single object, not the separate columns: > stack(NamesWide) ? values ? ind 1 ? ?Tom Name1 2 ? Dick Name1 3 ?Larry Name2 4 ?Curly Name2 Note that stack also returns the index (second column of 'ind' values), which tells you which column in the source data frame the stacked values originated from. Thus, if you just want the actual data: > stack(NamesWide)$values [1] "To...

Hi, Let us suppose I have a list x = list () x $ name1 = 1 x $ name2 = 'a' in the work environment. Let us suppose that in the body of a function I want to acces to a component of x by using its name as argument of that function. How can this by done? For instance, I was expecting f = function ( name ) x $ name to output 1 ( that is, x $...

Hi! I suspect there must be an easy way to access components of a data frame by name, i.e. the input should look like "name1 name2 name3 ..." and the output be a data frame of those components with the corresponding names. I ´ve been trying for hours, but only found the long way to do it (which is not feasible, since I have lots of components to select): dframe[names(dframe)=="name1" | dframe=="na...

Hi, all How to format and join strings ? For example, like following short python examples. ********* name1 = 'sample-plot' filename = '%s.png' % name1 inputdir = '/path/to/dir' os.path.join(inputdir, filename) ********** Best, Hyunchul Kim [[alternative HTML version deleted]]

HI Elisa, You could use ?cut() 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) 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) paste(i[1],"<x<=",i[2],sep="")))

Say I have a file called exp.batch which contains 2 cols The first col contains names of R objects the user would like to use. The second col contains the file names which will be read in using read.table i.e. exp.batch may look like this..... name1 complex/filename/path1.txt name2 complex/filename/path2.txt name3 complex/filename/path3.txt name4 complex/filename/path4.txt I want to have a function which will read in the files and make them into the objects named in the 1st column - automatically (by just providing exp.batch) for(i in 1:...

Dear All, I need some help arranging data that was imported. The imported data frame looks something like this (the actual file is huge, so this is example data) DF: IDKey X1 Y1 X2 Y2 X3 Y3 X4 Y4 Name1 21 15 25 10 Name2 15 18 35 24 27 45 Name3 17 21 30 22 15 40 32 55 I would like to create a new data frame with the following NewDF: IDKey X Y Name1 21 15 Name1 25 10 Name2 15 18 Name2 35 24 Name2 27 45 Name3 17 21 Name3 30 22 Name3 15 40 Name3 32 55 With...

..."SAM") > V [1] "Fred" "Mary" "SAM" > class(V) [1] "character" I would like to change it to a list: > L=as.list(V) > L [[1]] [1] "Fred" [[2]] [1] "Mary" [[3]] [1] "SAM" but I need to name the components as name1, name2, name3, … so it should look like this: $name1 [1] "Fred" $name2 [1] "Mary" $name3 [1] "SAM" Any help will be much appreciated Joseph [[alternative HTML version deleted]]

I have a series of columns that need to be evaluated in various tables. I need to apply a function in the following manner somers2(name1,name2). name1 is a vector of x inputs for the function which correspond to a vector of y inputs in name2. y<-rep(c(3,4,5,8),6) z<-rep(c(23,24,25,26,27,28),4) name1<-sprintf("Pred_pres_%s_indpdt[,%s,,]",x,y) name2<-sprintf("population[,%s]",z) rank<-function(i,j...

Hi, I want to construct a list as Lst <- list(name_1=object_1, ..., name_m=object_m) If name_1 is a variable with value "NAME1", how can I ask R to use "NAME1" instead of 'name_1' as the name of the list element? Thanks! Yupu [[alternative HTML version deleted]]

Here is a small reproducible example: data <- structure(list(V1 = structure(1:3, .Label = c("Name1", "Name2", "Name3"), class = "factor"), V2 = structure(c(1L, 3L, 2L), .Label = c("nam1", "name-1", "name_12"), class = "factor"), V3 = structure(1:3, .Label = c("nam2", "nam_34", "name-2"), clas...

...question that apparently received no response from the oblivion of collective silence, and besides I'm also curious about the answer > From: Griffith Feeney (gfeeney at hawaii.edu) > Date: Fri 28 Jan 2000 - 07:48:45 EST wrote (to R-help) > Constructing lists with > > list(name1=name1, name2=name2, ...) > > is tedious when there are many objects and names are long. Is there > an R > function that takes a character vector of object names as an > argument and > returns a list with each objected tagged by its name? > The idiom lapply(ls(pat = "...

...ported. It would be helpful if you were to use dput to give us the sample data since you say you have already imported it. > The imported data frame looks something like this (the actual file is > huge, so this is example data) > > DF: > IDKey X1 Y1 X2 Y2 X3 Y3 X4 Y4 > Name1 21 15 25 10 > Name2 15 18 35 24 27 45 > Name3 17 21 30 22 15 40 32 55 That data is missing in X3 etc, but would be NA in an actual data frame, so I don't know if my workaround was the same as your workaround. Dput would have clarified the starting point. > I would...