similar to: howto get the number of columns and column names of multiply data frames

Hi, May be this helps: m1<- as.matrix(read.table(text=" y1 g24 y1 0 1 g24 1 0 ",sep="",header=TRUE)) m2<-as.matrix(read.table(text="y1 c1 c2 l17 ?y1 0 1 1 1 ?c1 1 0 1 1 ?c2 1 1 0 1 ?l17 1 1 1 0",sep="",header=TRUE)) m3<- as.matrix(read.table(text="y1 h4??? s2???? s30 ?y1 0 1 1 1 ?h4 1 0 1 1 ?s2 1 1 0 1 ?s30 1 1 1

Hi, let's see, if someone can help my with this one: I have the string as follows: > str<-("one","two","three") Now I want to concatenate the items to one string, seperateted by space or something else, >str >"one, two, three" If possible without a loop. My actual goal ist to create string like >str.names >"female = names1, male

Hi all, probably really simple to solve, but having no background in programming I haven't been able to figure this out: I have two dataframes like df1 <- data.frame(names1=c('aa','ab', 'ac', 'ad'), var1=c(1,5,7,12)) df2 <- data.frame(names2=c('aa', 'ab', 'ac', 'ad', 'ae'), var2=c(3,6,9,12,15)) Now I want merge

Hello gurus, I have some variables, and i am creating combinations for analysis in the end i need these variables to be displayed like "LEPTIN+SAA+PTH". currently i am using loop to perform this. I would appreciate any pointers to do it without the loop. > mols=c("LEPTIN","SAA","PTH","sEGFR") > samples=mols[1:3] > samples [1]

Dear R-Helpers, this is my first post ever to a mailing list, so please feel free to point out any missunderstandings on my side regarding the conventions of this mailing list. My problem: Assuming the following character vector is given: names <- c("filia Maria", "vidua Joh Dirck Kleve (oo 02.02.1732)", "Bernardus Engelb Franciscus Linde j.u.Doktor referendarius

Hi R users: I got this code to generate a graphic for each member of a lists. list1<-list(A=data.frame(x=c(1,2),y=c(5,6)),B=data.frame(x=c(8,9),y=c(12,6))) names1<-names(list1) sapply(1:length(list1),function(i) with(list1[[i]],plot(x,y,type="l",main=paste("Graphic of",names1[i])))) Is there a more elegant solution for not to use two separate lists? I would like to

Hi R users, This seems like a simple problem but I have searched nabble for the answer and can't seem to find it. All I want to do is produce a boxplot where I have two boxes for one Individual but on the xaxis I only have one tick mark centred between the boxes so I can add the Individuals' name. I have 30 IDs and have shown the code I use below for a couple of IDs, I figure the data

Hi all, I want to execute a loop of a program: for (u in Timeframemin:Timeframe){} Imagine that Timeframemin<-10 Timefram<-10000 Is it posible to execute the loop but only proving from 10 to 10000 but jumping 10 each time, for example, execute for 10,20,30.....to Timeframe. Other question is, when a program is "heavy" and has a lot of loops to execute (how can I know where

Hi All, I am new to llvm backend. I am trying out few examples to understand backend codegen. I have ported llvm LEG @ https://github.com/frasercrmck/llvm-leg to llvm 3.9 successfully. Currently, the LEG instructions are RISC load-store type instruction. I want to generate some instructions for register indirect mode, like following: IR: @a = local_unnamed_addr global i32 0, align 4 @b =

useR's, I want to apply this function to the columns of a data frame: u[u >= range(v)[1] & u <= range(v)[2]] where u is the n column data frame under consideration and v is a data frame of values with the same number of columns as u. For example, v1 <- c(1,2,3) v2 <- c(3,4,5) v3 <- c(2,3,4) v <- as.data.frame(cbind(v1,v2,v3)) uk1 <- seq(min(v1) - .5, max(v1) + .5,

Hi everyone I have banking data set in long format with 4 columns.One of these columns is bank name which consist of 49 banks and I want the data for only 40 banks out of these 49 so can anyone help me on how to get this 40 banks data My data looks like Year Name totalliabilties assets 1990 a 90 10 1991 a 89 48 1992 a 87

Full_Name: Frank Sch?ffer Version: 2.3.1 OS: Ubuntu 6.10 (Edgy Eft) Submission from: (NULL) (86.56.0.173) frank at darwin:~$ R R : Copyright 2006, The R Foundation for Statistical Computing Version 2.3.1 (2006-06-01) ISBN 3-900051-07-0 R ist freie Software und kommt OHNE JEGLICHE GARANTIE. Sie sind eingeladen, es unter bestimmten Bedingungen weiter zu verbreiten. Tippen Sie 'license()'

Dear R family,i am trying to plot and save, simultaneously, about 1000. the name of each plot is contained in "names" file. when i run this loop, i get an error. "Error in plot.new() : Unable to open file 'C:/R/SAVEHERE/myplot_Tak.jpg' for writing". could you please correct the mistake in the loop? >names<-(names(sp)) >for(a in seq_along(names)){ >mypath

dear all, I have two dataframes dataframe1 ID a b c dataframe2 ID value a;W 100 X;c 200 Y;Z 300 I wanted to match the IDs from the two dataframes and record the values into a new column of dataframe1 at the corresponding rows. This is what I expect: dataframe1 ID value a 100 b c 200 I tried doing it like this: for (i in seq(1:nrow(dataframe1))) {

Dataframe1 contains a list of specific dates. Dataframe2 contains a large dataset, one element of which is Date. How do I create a subset of Dataframe2 that excludes the dates from Dataframe1? I know how to do it with a left outer join vs null in SQL, but I can't figure out how to do it more directly via the subcripts that already exist? Dateframe1 Date 1/1/2010 1/18/2010 Dataframe2 Date

Hello, I have the following data: gene Actual groups reps 1 1 213 1 2 2 2 123 2 3 3 3 2 1 2 4 4 12 2 3 5 1 0 1 1 6 2 34 2 2 7 3 123 1 1 8 4 12 2 2 I want to find the best way to

hello R users, I didn't find a solution for a special problem. I have two dataframes. dataframe1: X value row col ID 1 8.973498062 5512625 3460000 1 2 11.656658570 5501625 3464000 2 3 11.121777570 5495625 3473000 3 4 9.310465964 5508625 3477000 4 5 8.883483845 5515625 3496000 5 dataframe2: X value

Is it possible to set a create mask and/or a force create mode with exceptions for specified file extensions? E.g. I want shell scripts (*.sh) with the executable bit set, but other files should not be executable. AFAIK, this is currently not possible, but maybe I'm wrong. Regards, Evert

Hi, Let's say that I have a set of column names that begin with the string "Xyz". How do I extract these specific columns? I tried to do the following: dataframe1[,grep("Xyz",colnames(dataframe1))] But it does not work. What is wrong with my expression?

Hi Thanks Gabor for your suggestion. I am posting the code that worked for me. dataframe1 = data.frame(cbind(Src = c(1,1,1,2,3), Target1 = c('aaa','bbb','ccc','aaa','ddd'))); #must be data frame dataframe2 = data.frame(cbind(Src = c(2,3,4,4,4), Target2 = c('aaaa','dddd','bbbb','eeee','ffff'))); dataframe3 =