similar to: splitting a string column into multiple columns faster

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="")))

Hi, (Please use ?dput() to share the example dataset. Avoid using images to show dataset. Also, please read the posting guide esp. regarding home work, assignments etc.) res <- sapply(Gene[,-1],function(x) tapply(x,list(Gene$Genotype),mean)) #or res2 <-? aggregate(.~Genotype, data=Gene,mean) #or library(plyr) ?res3 <- ddply(Gene,.(Genotype),numcolwise(mean)) identical(res2,res3)

Hi, directory<- "/home/arunksa111/data.new" #first function filelist<-function(directory,number,list1){ setwd(directory) filelist1<-dir(directory) direct<-dir(directory,pattern = paste("MSMS_",number,"PepInfo.txt",sep=""), full.names = FALSE, recursive = TRUE) list1<-lapply(direct, function(x) read.table(x,header=TRUE, sep =

Hola Jorge, muchas gracias por tu pronta respuesta, no me di cuenta que el formateo podría causar problemas, envío de nuevo el código sin formatos. La idea básica es para un set de números de columnas (desordenados) y un set de numeros de fila el loop lo que hace es ir a la fila y columna correspondiente de data, tomar el valor y luego hacer la media sobre esos.

Muchas gracias a ambos Carlos y Jorge por las respuestas. Pido disculpas en la demora de respuesta, pero estuvo complicada la semana. La pregunta era un ejercicio de ejemplo para poder entender mejor los usos, creo que me armaré una guía en markdown con ejemplos varios para ir consultando cuando me salgan dudas de como usarlos. En realidad no importaba tanto si mejorara demasiado los tiempos

Dear Eliza, Try this: Lines1<-readLines(textConnection("1911.01.01?????? 7.87 1911.01.02?????? 9.26 1911.01.03?????? 8.06 1911.01.04?????? 8.13 1911.01.05????? 12.90 1911.02.06?????? 5.45 1911.02.07?????? 3.26 1911.03.08?????? 5.70 1911.03.09?????? 9.24 1911.04.10?????? 7.60 1911.05.11????? 14.82 1911.05.12????? 14.10 1911.06.13?????? 7.87 1911.06.14?????? 9.26

Dear UseRs,Extremely sorry for a basic question. I have a matrix of 19 rows and 365 columns. what i want to do is the following...First i want to leave out column number 1 and want to calculate the row wise mean of the remaining columns, which will obviously give me 365 values in one column, and then subtracting these values from the column i left out i.e. col=1 then i want to leave out column 2

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 Group, I have a question on Stringr package I have a table like this X Y ab su - di ac pi - tu ad tu - tu I want output like this X Y ab su ab di ac pi ac tu ad tu ad tu I am wondering if this operation can be done using stringr package (only) ? [[alternative HTML version deleted]]

Perdon por no se lo suficientemente claro :( Tu codigo produce `validPairs` que tiene 7 variables y 360 observaciones. Donde > validPairs[1,] V1 V2 V3 V4 V5 V6 valid 60 e1 g1 c1 e1 g1 c2 Valid indica que un maestro tiene asignado c1 y c2 en la escuela e1 y el grado g1. Correcto? Si es asi, esto es casi lo que queira producir y creo que puedo llegar a donde quiero usando tu codigo de base.

Hi, my data frame is x<-data.frame(ID=c("abc/def","abc/def/ghi","abc","mno/pqr/st/ab")) I want to split my column ID using "/" as the place to split. How can I do that without telling the code how many sub-columns. I could use nchar(gsub("[^/]","",x$ID)) to get how many "/" are in each row of the column, but could

Hi, Try this: #mat1 is the data res<-do.call(cbind,lapply(seq_len(nrow(mat1)),function(i) {new1<-do.call(rbind,lapply(seq_len(nrow(mat1[-i,])),function(j) {x1<-rbind(mat1[i,],mat1[j,]); x2<-(abs(x1[1,1]-x1[2,1])*abs(x1[1,5]-x1[2,5]))+(abs(x1[1,2]-x1[2,2])*abs(x1[1,6]-x1[2,6]))+(abs(x1[1,3]-x1[2,3])*abs(x1[1,7]-x1[2,7]))+(abs(x1[1,4]-x1[2,4])*abs(x1[1,8]-x1[2,8]))}));new1}))

Hola, 0. La falta de 'elegancia' hace que sea mas dificil hacer cambios al codigo. Por ejemplo cambiar n.classrooms <- 4 a n.classrooms <- 20 1. Cuando tengo solo 4 puedo hacer esto: schoolGrade$A <- Teachers$Teacher.ID[1:cuttoff1] schoolGrade$B <- Teachers$Teacher.ID[1:cuttoff1] schoolGrade$C <- Teachers$Teacher.ID[(cuttoff1+1):n.teachers] schoolGrade$D <-

Gracias Carlos, Tu codigo es un gran paso en el sentido correcto pero no produce exactamente lo que estoy buscando. Mi "solucion" en stackoverflow <http://stackoverflow.com/questions/31137940/randomly-assign-teachers-to-classrooms-imposing-restrictions/31143808#31143808> produce un data frame `schoolGrade` con 240 observaciones y 7 variables. Mi objetivo es poder generar un data

Ira, obj_name<- load("arun.RData") Pred1<- get(obj_name[1]) Actual1<- get(obj_name[2]) dat2<- data.frame(S1=rep(Pred1[,1],ncol(Pred1)-1),variable=rep(colnames(Pred1)[-1],each=nrow(Pred1)),Predict=unlist(Pred1[,-1],use.names=FALSE),Actual=unlist(Actual1[,-1],use.names=FALSE),stringsAsFactors=FALSE) dat2New<- dat2[!(is.na(dat2$Predict)|is.na(dat2$Actual)),] ?dat3<-

Hi, I need to aggregate rows of a data.frame by computing the mean for rows with the same factor-level on one factor-variable; here is the sample code: x <- data.frame(rep(letters,2), rnorm(52), rnorm(52), rnorm(52)) aggregate(x, list(x[,1]), mean) Now my problem is, that the actual data-set is much bigger (120 rows and approximately 100.000 columns) ? and it takes very very long

Hi, I have a 2 x 10000 matrix of confidence intervals. The first column is the lower and the next column is the upper. I want to cont how many times a number say 12 lies in the interval. Can anyone assist? -- Thanks, Jim. [[alternative HTML version deleted]]

Hey, Is it possible that R can calculate each options under each column and return a summary table? Suppose I have a table like this: Gender Age Rate Female 0-10 Good Male 0-10 Good Female 11-20 Bad Male 11-20 Bad Male >20 N/A I want to have a summary table including the information that how many answers in each category, sth like this: X

Dear All, I am pretty new to R and thus my question may sound silly. Is there a way to automatically generate a series of separate vectors (so not arranged in a matrix), without typing and changing every time the values, and store them as separate *xlsx file, where the "*" is replaced by the name of the vector itself? What i would like to create is a total of 12 vectors,

Hi, You could try: A<- matrix(unlist(read.table(text=" 1 2 3 4 5 6 7 8 9 9 8 7 6 5 4 3 2 1 ",sep="",header=FALSE)),ncol=3,byrow=FALSE,dimnames=NULL) library(matrixStats) ?res1<-t(sapply(split(as.data.frame(A),as.numeric(gl(nrow(A),2,6))),colProds)) ?res1 #? [,1] [,2] [,3] #1??? 4?? 10?? 18 #2?? 63?? 64?? 63 #3?? 18?? 10??? 4