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...s 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 me know. probs <- c(.1, .3, .2, .4, .7, .9, .3, .4, .5, .6) p10 <- 0.6 p00 <- 0.4 vec1 <- rep(0, 10) for (i in 2:10) { vec1[i] <- ifelse(vec1[i-1] == 0, ifelse(probs[i]<p10, 0, 1), ifelse(probs[i]<p00, 0, 1)) } Thanks GG [[alternative HTML version...

Dear friends, Now, when i use the argument return(x=x,y=y,prob=prob) , R displays the waring message: Warning message: The return value for multiple variables wasn't used in: return(x = x, y = gy, prob = prob) I used the methods of "help.search("return")" and "?return" to get some help, but didn't find info on it. Any...

I got a vector of probabilities like, probs<-c(0.001,0.5,0.02,1,.....) Is there any nice and easy builtin function to get the number of occurences within some specified probabality range. Like with 2% it would be occur[1] = sum(probs[probs>0&probs<0.02]) occur[2] = sum(probs[probs>0.02&probs<0.04...

Hi, The code below is giving me this error message: Error in while (err > eps) { : missing value where TRUE/FALSE needed In addition: Warning messages: 1: In dnbinom(x, size, prob, log) : NaNs produced 2: In dnbinom(x, size, prob, log) : NaNs produced I know from the help files that for dnbinom "Invalid size or prob will result in return value NaN, with a warning", but I am not able to find a workaround for this in my code to avoid it. I appreciate the help. Thank...

Hi all, I should have given a better explanation of my problem. Here it is. I extracted from my code the bit that gives the error. Place this in a file called test.c #include <math.h> #include <R.h> #include <Rmath.h> #include <float.h> #include <R_ext/Print.h> int main(){ double prob[3] = {0.0, 0.0, 0.0}; double prob_t...

Hello all, Consider the following problem: There is a matrix of probabilities: > set.seed(1) > probs <- array(abs(rnorm(25, sd = 0.33)), dim = c(5,5), dimnames = list(1:5, letters[1:5])) > probs a b c d e 1 0.21 0.27 0.50 0.0148 0.303 2 0.06 0.16 0.13 0.0053 0.258 3 0.28 0.24 0.21 0.3115 0...

...ot;ctest") catSignifTest <- function( catFile ) { ############################################################### ## Get the data sets from text file catData <- read.table( catFile ) ncols <- length(catData) nrows <- length(catData[,1]) ncol1 <- ncols - 1 probeNbr <- catData[1,] Achip <- catData[,ncols] for ( row in 2:nrows ) { prob <- Achip[ row ] / Achip[ 1 ] if ( prob <= 0 ) prob <- 0.0000001 if ( prob >= 1 ) prob <- 0.9999999 chip <- catData[row,] for ( col in 1:ncol1 ) { succ <- chip[col]...

Hello, I want to estimate the parameters of a binomial distributed rv using MLE. Other distributions will follow. The equation system to solve is not very complex, but I've never done such work in R and don't have any idea how to start... The system is: (1) n*P = X (2) [sum {from j=0 to J-1} Y{j} /(n-j)] = -n * ln (1-X / n) where * only X is given (empirical mean)

Dear R-helpers, I need to compute probabilties of multinomial observations, eg by doing the following: y=sample(1:3,15,1) prob=matrix(runif(45),15) prob=prob/rowSums(prob) diag(prob[,y]) However, my question is whether this is the most efficient way to do this. In the call prob[,y] a whole matrix is computed which seems a bit of a was...

Hi, I noticed this problem on my home desktop running FC4 and again on my laptop running FC5. Both have previously compiled and passed make check-all on 2.3.1 svn revisions from 10 days ago or so. On both these machines, make check-all is consistently failing (4 out of 4 attempts on the FC 4 desktop and 3 out of 3 on the...

...num[8]+1):bjyearnum[9]] bjerr9<-bjerrdata$tyerr[(bjyearnum[9]+1):bjyearnum[10]] bjerr10<-bjerrdata$tyerr[(bjyearnum[10]+1):bjyearnum[11]] bjerr11<-bjerrdata$tyerr[(bjyearnum[11]+1):bjyearnum[12]] bjerr12<-bjerrdata$tyerr[(bjyearnum[12]+1):length(bjerryear)] bjavee<-data.frame(5,12) cprob=c(0.05,0.25,0.5,0.75,0.95) bjavee[1:5,1]<-quantile(bjerr1,prob=cprob) bjavee[1:5,2]<-quantile(bjerr2,prob=cprob) bjavee[1:5,3]<-quantile(bjerr3,prob=cprob) bjavee[1:5,4]<-quantile(bjerr4,prob=cprob) bjavee[1:5,5]<-quantile(bjerr5,prob=cprob) bjavee[1:5,6]<-quantile(bjerr6,prob=cpr...

Dea'R' helpers I have following data - prob = c(0.1, 0.2, 0.3, 0.4, 0.5) frequency = c(100, 75, 45, 30, 25) no_trials = c(10, 8, 6, 4, 2) freq1 = rbinom(frequency[1], no_trials[1], prob[1]) freq2 = rbinom(frequency[2], no_trials[2], prob[2]) freq3 = rbinom(frequency[3], no_trials[3], prob[3]) freq4 = rbinom(frequency[4], no_trials[4], prob[...

Dear R-users >prob <- c(0.5,0.4,0.3,0.1,0.0) >cal <- prob * log(prob,base=2) >cal [1] -0.5000000 -0.5287712 -0.5210897 -0.3321928 NaN Is there any way to change NaN to zero ? I did come up with this by applying Ripley's relpy to my previous question cal <-prob*log(pmax(prob,0.00000001),ba...

I am trying to reset the default arguments in the rnbinom function with the following example code: params <- c("size" = 1, "mu" = 1) formals(rnbinom)[names(params)] <- params rnbinom(n = 10) It returns the following: Error in rnbinom(n = 10) : argument "prob" is missing, with no default If I set the defaults with this code: params <- c("size" = 1, "prob" = .5) formals(rnbinom)[names(params)] <- params rnbinom(n = 10) The function works correctly. The documentation specifies that you can set mu or prob with size. I unde...

Dear r users, I have 4 variables x1,x2,x3,x4 and each one has two levels, for example Y and N. For x1: prob(Y)=0.6, prob(N)=0.4; For x2: prob(Y)=0.5, prob(N)=0.5; For x3: prob(Y)=0.8, prob(N)=0.2; For x4: prob(Y)=0.9, prob(N)=0.1; Therefore, the sample space for (x1, x2, x3, x4)={YYYY, YYYN, YYNY,......} (16 possible combination) and the corresponding probabilities are {(0.6)(0.5)(0.8)(0.9), (0.6)(0.5...

...the time being, and I''m confused. This may have more to do with statistics than R itself, but since I''m getting results from R I find counterintuitive (well, yeah, my statistical intuition has not been properly sharpened), I feel like asking. The point first: If I do > rgeom(1,prob=1) I get: [1] NaN Warning message: NAs produced in: rgeom(n, prob) And if I do: > rgeom(1,prob=0) I get: [1] NaN Warning message: NAs produced in: rgeom(n, prob) I was expecting to get 0 and Inf respectively.... Should I expect that? Going back to my textbooks (Dudewicz and Mishra 1988, pr...

Hi, In Splus7 this statement xlrmN1 <- sample(c(0,1,2),400 ,prob=c(0.02 ,0.93 ,0.05 )) worked fine, but in R the interpreter reports that the length of the vector to chose c(0,1,2) is shorter than the size of many times I want to be selected from the vector c(0,1,2). Any good reason? See below the error. > xlrmN1 <- sample(c(0,1,2),400 ,prob=c(0.02 ,0....

Habría que buscar la vuelta, yo no lo se, pero posiblemente lo siguiente da una pista. Nota: al mismo código le sume una línea al final datos<-c(2,3,4,5,6,7,8) quantile(datos) quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95)) as.matrix(quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95))) as.data.frame(quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95))) # ¿ y si solo solicita uno por vez ? as.matrix(quantile(datos,probs = c(0.25))) as.matrix(quantile(datos,probs = c(0.75))) cb...

I do: x<-rnorm(1000) quantile(x,c(.025,.975)) 2% 98% -1.844753 1.931762 Since I want to find a 95% confidence interval, I take the .025 and .975 quantiles. HOWEVER R says I have the 2% (not 2.5%) and 98% (not 97.5%) points. Is it just rounding the printed 2% and 98%, or is it REALLY finding .02 and .98 points instead of .025 and .975? Thanks for any help. Bill Simpson

I do: x<-rnorm(1000) quantile(x,c(.025,.975)) 2% 98% -1.844753 1.931762 Since I want to find a 95% confidence interval, I take the .025 and .975 quantiles. HOWEVER R says I have the 2% (not 2.5%) and 98% (not 97.5%) points. Is it just rounding the printed 2% and 98%, or is it REALLY finding .02 and .98 points instead of .025 and .975? Thanks for any help. Bill Simpson