similar to: missing value where TRUE/FALSE needed

Hi All, I am counting the number of occurrences of the terms listed in one vector in another vector. My code runs: for( i in 1:length(vector3)){ vector3[i] = sum(1*is.element(vector2, vector1[i])) } where vector1 = vector containing the terms whose occurrences I want to count vector2 = made up of a number of repetitions of all the elements of vector1 vector3 = a vector of NAs that is

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

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

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

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 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 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

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

Hi, I got two data point vectors. Now I want to make a ks.test(). I you print both vectors you will see, that they fit pretty fine. Here is a picture: http://www.jochen-bauer.net/downloads/kstest-r-help-list-plot.png As you can see there is one histogram and moreover there is the gumbel density function plotted. Now I took to bin-mids and the bin-height for vector1 and computed the

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

Dear R users, I have a problem computing time differences using R. I have a date that are given using the following format: 20080620.00, where the 4 first digits represent the year, the next 2 ones the month and the last 2 ones the day. I would need to compute time differences between two vectors of this given format. I tried around trying to change this format into any type of time serie

por favor como se calculan las probabilidades para conformar el segundo vector SI.Distr, en este ejemplo puse la existente en SARS2003 #Análisis paso a paso rm(list = ls()) #existe información de la incidencia de casos de una enfermedad incidencia <-c (1,0,0,0,0,0,0,0,2,1,1,0,1,2,0,2,0,2,2,2,3, 1, 2, 0,2, 0, 1, 1, 2, 1, 1, 2, 3, 0, 1, 2, 0, 2, 1, 3, 1, 6, 3, 3,7, 4, 4, 3, 6,

Hello everyone. I use double for loops to fill in matrices, but there are surely better (and computationally faster) ways to perform that task. Could someone show me, given the following example of a double for loop, how this could be done? It is much easier to learn by examples. Val <- matrix(0, nrow=n+1, ncol=n+1) for( i in 0:n){ for(j in 0:i){ Val[j+1, i+1] <- u^j*d^(i-j)

Dear all, I have a surprising problem with the representation of frequencies in a histogram. Consider, for example, the R code: b<-rnorm(2000,3.5,0.3) hist(b,freq=F) When I plotted the histogram, I expected that values in the y-axis (the probability) varied between 0 and 1. Instead, they varied within the range 0-1.3. Have you got any suggestion for obtaining a correct graph with