similar to: vectorisation suggestion

Hello everyone, I have two vectors having only one element different: vector1 vector2 vector1 TWC TWC TWC VFC TWX NA VIA/B VFC VFC

Dear Statisticians--- This is not even an R question, so please forgive me. I have so much ignorance in this matter that I do not know where to begin. I hope someone can point me to documentation and/or a sample. I want to compute a covariance as quickly as non-humanly possible on an Intel core processor (up to SSE4) under linux. Alas, I have no idea how to engage CPU vectorization. Do I need

Another question on functions - I have something that looks like plotter<-function(i){ temp.i<-rwb[rwb$vector1 <=(i*.10),] with(temp.i, plot(vector2, vector3, main=(i*.10),)) mod<-lm(vector3~vector3-1,data=temp.i) r2<-summary(mod)$adj.r.squared rsqrd[i]<-r2 legend("bottomright", legend=signif(r2), col="black") abline(mod) rsqrd<<-rsqrd } I'd

Windows binaries at CRAN (bin/windows/Windows-NT/base) have been updated to R-1.1.0. See below for a list of Windows-specific changes. We thanks all the people who checked over pre-test versions. guido masarotto (for the R-core team) Windows-specific changes to R ============================= There is now a GUI preferences editor on the Edit menu in Rgui. A data entry editor is now

Windows binaries at CRAN (bin/windows/Windows-NT/base) have been updated to R-1.1.0. See below for a list of Windows-specific changes. We thanks all the people who checked over pre-test versions. guido masarotto (for the R-core team) Windows-specific changes to R ============================= There is now a GUI preferences editor on the Edit menu in Rgui. A data entry editor is now

Hello dear R people, for my MSc thesis I need to program some functions, and some of them simply do not work. In the following example, I made sure both vectors have the same length (10), but R gives me the following error: Error in if (vector1[i] == vector2[j]) { : missing value where TRUE/FALSE needed I googled for possible solutions, but I did not find a good explanation for this...

Hi all, I am having a bit of trouble using the levels() function. I have a factor with many elements, and when I use the function levels() to extract the list of unique elements, some of the elements returned are not actually in the factor. For example I would have this: > vector <- dataset$Benchmark > class(vector) [1] "factor" > length(vector) [1] 35615 > vector2

Dear all, I was encountering with a typical Matching problem and was wondering whether R can help me to solve it directly. Let say, I have 2 vectors of equal length: vector1 <- LETTERS[1:6] vector2 <- letters[1:6] Now I need to match these 2 vectors with all possible ways like: (A,B,C,D,E) & (a,b,c,d,e) is 1 match. Another match can be (A,B,C,D,E) & (b,a,c,d,e), however there

hi is there a faster way to do this? i <- 1 for(x in vector1) for(y in vector2) { m[[i]] <- (x - y) i <- i + 1 } regards soren -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in

Looking for help with the following problem. Given a sample of zeros and ones, for example: > VECTOR1<-rep(c(1,0),c(15,10)) > VECTOR1 [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 How would I create a new sample (VECTOR2) also containing zeros and ones, in which the phi-coefficient between the two sample vectors was drawn from a population with a known

Hello, My name is Ick Hoon Jin and I am Ph. D. student in Texas A & M Univ.. When I run the C embedded in R in the Linux system, I confront the following error after 6,000 iteration. By googling I found this error is from the problem in C. *** caught segfault *** address (nil), cause 'memory not mapped' My C code is following:

Hi, I have many files containing one column of data. I like to use the scan function to parse the data. Next I like to bind to a large vector. I try this like: count<-1 files <- list.files() # all files in the working directory for(i in files) { tmp <- scan(i) assign(files[count], tmp) count<-count+1 } This part works! Now I like to plot the data in a boxplot.

Hola Jose, si CONCATENAR significa APILAR, es decir, concantenar verticalmente, por decirlo de algun modo, podrias hacerlo con rbind(): nuevovector <- rbind(vector1,vector2) Si ademas quieres que cada valor de los vectores originales sea identificado en el nuevovector, puedes usar: nuevovector <- stack(vector1,vector2) en este ultimo caso se agrega una columna adicional tipo factor, con

Dear all, meanwhile i found out how to handle the coordinate thing and plot the lines like i like. the remaining problem is, i need something like names.arg does in barplot for my plot. my plot connects several dots with several lines. and now i like characters as the names of the intersects of the x-axis. vector1=c(a,b) vector2=c(c,d) # this vector makes sure the yellow line starts at

Dear all, I hope you can forgive me my stupid questions, but I am a very new R user (; So, this is my question: I have two matrices, those are: matrix1 <- matrix(cbind(vector1, vector2), 1,2, dimnames = list(c("values"), c("T value", "p value"))) matrix2 <- matrix(dcbind,2,6,dimnames =

HI All, I have written and succesfully uploaded a new package. The licence it is under is 'GPL' --no version. My assumption is, since all the code is written in R the licence R used for R would affect the code (hence my "GPL" stands for "whatever version of the GPL R is under") I am happy with the licencing I used, but I'd like to ask if there is any transitive

Hi All, I'd like to ask for a few clarifications. I am doing some calculations over some biggish datasets. One has ~ 23000 rows, and 6 columns, the other has ~620000 rows and 6 columns. I am using these datasets to perform a simulation of of haplotype coalescence over a pedigree (the datestes themselves are pedigree information). I created a new dataset (same number of rows as the pedigree

Dear All, just a quick example: > x = 1:25 > x[12] = NA > x [1] 1 2 3 4 5 6 7 8 9 10 11 NA 13 14 15 16 17 18 19 20 21 22 23 24 25 > y = x[x<10] > y [1] 1 2 3 4 5 6 7 8 9 NA Is there any way of NOT getting NA for y = x[x<10]? Similarly > y = x[x<15] > y [1] 1 2 3 4 5 6 7 8 9 10 11 NA 13 14 How do I get rid of the NA (not post

Hi All, I am trying to match two character arrays (email lists) using either pmatch(), match() or charmatch() functions. However the function is "missing" some matches due to differences in the cases of some letters between the two arrays. Is there any way to disable case sensitivity or is there an entirely better way to match two character arrays that have identical entries but written

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