similar to: newbie:looking for an efficient way to compute distance vector

I have a program which needs to compute squared Euclidean distance between two vectors million of times, which the Rprof shows is the bottleneck. I wondered if there is any faster way than my own simple function distance2 = function(x1, x2) { temp = x1-x2 sum(temp*temp) } I have searched the R-help archives and can not find anything except when the arguments are matrices. Thanks for any

Hello, Im working on a 50933 point count bird abundance dataset. I've succeeded in calculating a distance matrix for this entire set, but I don't have sufficient memory to convert this to a matrix, as below... abun.dist <- dist(abun.mat[1:50993,1:235) test <- rowMeans(as.matrix(abun.dist)) Error in matrix(0, size, size) : too many elements specified ive been able to run a hclust()

I am pretty new to R. So this may be an easy question for most of you. ? I would like to calculate the squared distances of a large set (let's say 20000) of vectors (let's say dimension of 5) to a fixed vector. ? Say I have a data frame MY_VECTORS with 20000 rows and 5 columns, and one 5x1 vector y. I would like to efficiently calculate the squared distances?between each of the 20000

Dear all, I would like to compute power of a square non symmetric matrix. This is a part of a simulation study. Matrices are quite large (e.g., 900 by 900), and contains many 0 (more than 99 %). I have try the function mtx.exp of the Biodem package: library(Biodem) m <- matrix(0, 900, 900) i <- sample(1:900, 3000, replace = T) j <- sample(1:900, 3000, replace = T) for(x in 1:3000)

Dear All, Suppose I have a program as below: Outside is a loop for simulation (with random generated data), inside there are several sapply()'s (10~100) over the data and something else, but these sapply's have to be sequential. And each sapply do not involve very intensive calculation (a few seconds only). So the outside loop takes minutes to finish one iteration. I guess the better way

Hi All, I'm working on analyzing a large data set, lets asume that dim(Data)=c(1000,8700). I want to calculate the canberra distance between the columns of this matrix, and using a toy example ('test' is a matrix filled with random numbers 0-1): > system.time(d<-as.matrix(dist(t(test), method = "canberra", diag = FALSE, upper = FALSE, p = 2))) user system

Hi, I just started to use R for my Phd. I have to write a Formula by witch, given two matrices of vectors (electrostatic potential map) I shall calculate the similarity between them. Could you please tell me if I'am doing it right. The complecte formula is attached to an image file the script #!/usr/bin/env Rscript options(max.print=2000000) x<-read.table("a")

Hi All, I would like to compute the simple finite-difference approximation to the gradient of a scalar function of a large number of variables (on the order of 1000). Although a one-time computation using the following function grad() is fast and simple enough, the overhead for repeated evaluation of gradient in iterative schemes is quite significant. I was wondering whether there are

Hi all, I'm to inexperienced to come up with the matrix solution elusively appearing on the horizon for the following problem and would appreciate if you could give me a nudge ... I have two vectors a, and b and need to find the closest match for each value of a in b. How to do that efficiently? Thanks, Joh

On the surface this seems pretty simple, but I flummoxed. I have two sets of numbers they bounce around zero, positive one and negative one. They have a relationship between them, where one diverges away from the other. I want create a second set of numbers that tracks that divergence. #Lets make some data like mine, kinda Firstset <- runif(100, min = -1 , max =1) Secondset <- runif(100,

Hi all, Does anybody have a magic trick handy to isolate directly consecutive integers from something like this: c(1,2,3,4,7,8,9,10,12,13) The result should be, that groups 1-4, 7-10 and 12-13 are consecutive integers ... Thanks for any hints, Joh

I have the following: @lot = Lot.find(params[:id]) part_nums = Part.all(:conditions => ["id <> ?", @lot.part.id]) I guess I should mention that Lot :belongs_to => :part I was looking at the log following the execution of these two statements and I saw something like this: Lot Load (0.4ms) SELECT * FROM "lots" WHERE ("lots"."id" = 13) Part

Hello. I'd like to compute the associate vector of distances between leaves in a binary non-rooted tree. The definition of a distance between two leaves in a binary non-rooted tree is the number of edges in the path joining the two leaves. I've tried the ape package but I'm unable to find this vector. For example, using rtree(5,rooted=F) I've obtained the following tree: $edge

Dear All, I'm dealing with a case, where 'manhattan' distance of each of 100 vectors is calculated from 10000 other vectors. For achieving this, following 4 scenarios are tested: 1) scenario 1: > x<-read.table("query.vec") > v<-read.table("query.vec2") > d<-matrix(nrow=nrow(v),ncol=nrow(x)) > for (i in 1:nrow(v)){ + d[i,]<-

Hi, I misunderstand the definition of Canberra distance in R. On Internet and in function description pages of dist() from stats and Dist() from amap, Canberra distance between vectors x and y, d(x,y), is : d(x,y) = sum(abs(x-y)/(x+y)) But in use, through simple examples, we find that the formula is : d(x,y) = (NZ + 1)/NZ * sum(abs(x-y)/(x+y)) with NZ = nb of pairs of coordinates that are

Hi I have just (finally) started to poke around in R and wanted to analyse a stream fish dataset with 28 sites and 18 species. When trying to follow the Vegan manual to run nmds from distance measures calculated by the vegdist function it turns out that I have two sites (streams) with the exactly the same four species (I have used pres-abs data in this case). When I try to run isoMDS I get an

Dear UseRs, MY PROBLEM IS A SMALL PIECE OF A REAL BIG AND A COMPLICATED PROBLEM. IF I DELIBERATE IN A VERY SIMPLE WAY THEN ALL I WANT IS TO PUT ALL THE POSSIBLE COMBINATIONS OF 75 DISTANCE MATRICES (BY TAKING 4 MATRICES, MORE COMMONLY 75C4), in the following equation. t<-as.matrix((MAT1)^2+(MAT2)^2+(MAT3)^2+(MAT4)^2+,upper=T,diag=T)) Then "1215450" values of "t"(one for

Hi all, given this example #start a<-c(0,70,50,100,70,650,1300,6900,1780,4930,1120,700,190,940, 760,100,300,36270,5610,249680,1760,4040,164890,17230,75140,1870,22380,5890,2430) length(a) b<-c(0,0,10,30,50,440,1000,140,70,90,60,60,20,90,180,30,90, 3220,490,20790,290,740,5350,940,3910,0,640,850,260) length(b) out<-ks.test(log10(a+1),log10(b+1)) # max distance D

Dear All, I would like to ask clarifications on the gower distnce matrix calculated by the function gdistin the library mvpart. Here is a dummy example: > library(mvpart) Loading required package: survival Loading required package: splines mvpart package loaded: extends rpart to include multivariate and distance-based partitioning > x=matrix(1:6, byrow=T, ncol=2) > x [,1]

Hi Elisa, I hope this is what you wanted. dat1<-read.csv("peaks.csv",sep=",") #Subset dat2<-dat1[1:5,] res1<-do.call(cbind,lapply(seq_len(nrow(dat2)),function(i) do.call(rbind,lapply(split(rbind(dat2[i,],dat2[-i,]),1:nrow(rbind(dat2[i,],dat2[-i,]))), function(x) {x1<-rbind(dat2[i,],x);