similar to: histogram

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}))

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)

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

Dear all, I think this is an easy task, but I don't know how to do it. Specifically, I have 69 columns with 300.000 rows. In each cell there is a code like "2E3", "4RR", etc. I now have a list that replaces this with values, e.g., old new 2E3 5 4RR 3 etc. The list ist about 1600 rows long, so also to extensive for normal solutions Do you how to do that? It would

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, You may try this: dat1<- read.table(text=" CustID TripDate Store Bread Butter Milk Eggs 1 2-Jan-12 a 2 0 2 1 1 6-Jan-12 c 0 3 3 0 1 9-Jan-12 a 3 3 0 0 1 31-Mar-13 a 3 0 0 0 2 31-Aug-12 a 0 3 3 0 2 24-Sep-12 a 3 3 0 0 2 25-Sep-12 b 3 0 0 0 ",sep="",header=TRUE,stringsAsFactors=FALSE) dat2<- dat1[,-c(1:3)] res<- lapply(seq_len(ncol(dat2)),function(i)

Hello! I have a column in my data frame that I have to split: I have to distill the numbers from the text. Below is my example and my solution. x<-data.frame(x=c("aaa1_bbb1_ccc3","aaa2_bbb3_ccc2","aaa3_bbb2_ccc1")) x library(stringr) out<-as.data.frame(str_split_fixed(x$x,"aaa",2)) out2<-as.data.frame(str_split_fixed(out$V2,"_bbb",2))

I would like to know whether there is a faster way to do the below operation (updating vec1). My objective is to update the elements of a vector (vec1), where a particular element i is dependent on the previous one. I need to do this on vectors that are 1 million or longer and need to repeat that process several hundred times. The for loop works but is slow. If there is a faster way, please let

Hello R-users, I need to create a vector inserting an 1 after each value of another vector. For example: vec1<-c(2,3,4) I need to create a vector with the values 2,1,3,1,4 Does anyone know how create this vector without loops (vec1 could have 1000 elements) Thank you, Juan -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

This can be vectorized. Try ix <- seq_along(vec2) S_diff2 <- sapply(seq_len(N1-(N2-1)*ratio_sampling), \(j) sum((vec1[(ix-1)*ratio_sampling+j] - vec2[ix])**2)) On Sun, Jun 16, 2024 at 11:27?AM Laurent Rhelp <laurentRHelp at free.fr> wrote: > > Dear RHelp-list, > > I try to use the package comprehenr to replace a for loop by a list > comprehension. > > I

Hello R specialists, I have a base vector called vBase with 102 elements. I have another vector vec1 which elements are only part of vBase but is shorter. I transform vec1 so I get a vector with the same size as vBase and with each common element on the same indexed row. If a value is missing in vec1 then I put a Na like this: vec1 vBase Amsterdam Amsterdam Na

Thanks. Yes. Your approach can identify: Glaxy ace S 5830 and S 5830 Glaxy ace But you can not identify using same program: Iphone 4S 16 G Iphone 4S 16G How should I solve both in same time. Kind regards,Tammy [[alternative HTML version deleted]]

Is there any built-in way to lexicographically compare two vectors of the same length in R? The textbook algorithm could be coded as follows: lex.cmp <- function (vec1,vec2) { for (j in 1:length(vec1)) { if (vec1[j] < vec2[j]) { return(-1) } if (vec1[j] > vec2[j]) { return(1) } } return(0) } Thanks, Gabriel

Hi, In the example you showed: m1<- matrix(0,length(vec),max(vec)) 1*!upper.tri(m1) #or ?m1[!upper.tri(m1)] <-? rep(rep(1,length(vec)),vec) #But, in a case like below, perhaps: vec1<- c(3,4,5) ?m2<- matrix(0,length(vec1),max(vec1)) ?indx <- cbind(rep(seq_along(vec1),vec1),unlist(tapply(vec1,list(vec1),FUN=seq),use.names=FALSE)) m2[indx]<- 1 ?m2 #???? [,1] [,2] [,3] [,4] [,5]

Simple one, have read and googled, still no luck! I want to create several empty vectors all of the same length. I would like multiple empty vectors (vec1, vec2, vec3) and want to create them all in one line. I've tried vec1,vec2,vec3 <- vector(length=5) and c(vec1,vec2,vec3) <- vector(length=5) and several other attempts but nothing seems to work ... suggestions? Thanks Jim

Hallo all Please help me. I am lost and do not know what is the problem. I have a factor called kvartaly. > attributes(kvartaly) $levels [1] "1Q.04" "2Q.04" "3Q.04" "4Q.04" "1Q.05" "2Q.05" "3Q.05" "4Q.05" $class [1] "factor" > mode(kvartaly) [1] "numeric" > str(kvartaly) Factor w/ 8

Hello again, Let say I have following user defined function: fn <- function(x, y) { Vec1 <- letters[1:6] Vec2 <- 1:5 return(list(ifelse(x > 0, Vec1, NA), ifelse(y > 0, Vec2, NA))) } Now I have following calculation: > fn(-3, -3) [[1]] [1] NA [[2]] [1] NA > fn(3, -3) [[1]] [1] "a" [[2]] [1] NA Here I can not understand why in the second case, I get

Dear all, Without a loop, I would like transform 3 numeric vectors empty of 0/1 of same length Vec1 : transform 1 to A and 0 to "" Vec2 : transform 1 to B and 0 to "" Vec3 : transform 1 to C and 0 to "" to obtain only 1 vector Vec who is the paste of the 3 vectors (Ex : ABC, BC, AC, AB,...) Any idea ? Thank you for your help -- Michel ARNAUD

Dear RHelp-list, ?? I try to use the package comprehenr to replace a for loop by a list comprehension. ?I wrote the code but I certainly miss something because it is very slower compared to the for loops. May you please explain to me why the list comprehension is slower in my case. Here is my example. I do the calculation of the square difference between the values of two vectors vec1 and

Dear R-helpers: I have a question related to using do.call to call cbind and get. #the following works vec1 <- c(1,2) vec2 <- c(3,4) ColNameVec <- c('vec1','vec2') mat <- do.call("cbind",lapply(ColNameVec,get)) mat #put code above into a function then it does not work #before doing so, first remove vec1 and vec2 from global environment rm(vec1,vec2) test