similar to: sample with equal probabilities

Hello If I have x=c(3,2,6,1) and n=length(x), are the following codes equivalent?? sample(x,1,replace=TRUE)    and       sample(x,1,replace=TRUE,prob=rep(1/n , n) ) Regards [[alternative HTML version deleted]]

Hello I have a loop to sample 20 samples and I want to put them in one list, how I can make this??   Regards Sulafah [[alternative HTML version deleted]]

Hello I have a loop to draw few samples and I want to but this samples in one list, how I can make this? when I use it , the list show just the last sample. I want to know the correct way to put all samples in one list.   Regards Sulafah [[alternative HTML version deleted]]

Hello If i want to resampl from the tails of normal distribution , are these commans equivelant??   upper tail:qnorm(runif(n,pnorm(b),1))  if b is an upper tail boundary   or   upper tail:qnorm((1-p)+p(runif(n))  if p is the probability of each interval (the observatins are divided to intervals)   Regards [[alternative HTML version deleted]]

Hello, Frank was nice enough to point me to the val.prob function of the Design library. It creates a beautiful graph that really helps me visualize how well my model is predicting probabilities. By default, there are two lines on the graph 1) fitted logistic calibration curve 2) nonparametric fit using lowess Right now, the nonparametric line doesn't look very good. The

Hello I used SIGN.test function in R 2.12.2 to apply one sample sign test and it is worked well ,but I want to put statistic in a variable and I could not get this value, I used : x=rnorm(100) t=SIGN.test(x,md=0,alt="less") t$rval[1]$statistic   the last command work in the old version of R but it does not work in R 2.12.2. what can I do to get value of statistic??   Regards Sulafah

Hi all, I have a vector xm say: xm = c(1,2,3,4,5,5,5,6,6) I want to return a vector with the corresponding probabilities based on the amount of times the numbers occurred. For example, I should get the following vector for xm: prob.xm = c(1/9, 1/9, 1/9, 1/9, 3/9, 3/9, 3/9, 2/9, 2/9) Any help greatly appreciated. -- Thanks, Jim. [[alternative HTML version deleted]]

i am trying to use the sample command and have one question about it: i am getting the error Error in sample(length(x), size, replace, prob) : insufficient positive probabilities when i use something like: sample (1:4, prob=c(0,0,1,0))[1] i was expecting that to return a 3 every time while this is not exactly what i am using it for, i need the capabilities to deal with zeros (as the

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),

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_tot = 0.; prob[0] = 0.3*dnorm(2, 0,

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 ,0.93 ,0.05 )) Error in

// R 2.3.1 Can someone please explain why this error returns? > y=numeric(100) > x=matrix(runif(16),4,4) > for(i in 2:100) + { + y[i]=which(rmultinom(1, size = 1, prob = x[y[i-1], ])==1) + } Error in rmultinom(n, size, prob) : too few positive probabilities thx much ej

  Dear R helpers   Suppose I have two sets of ranges (interest rates) as   Range 1 : (7 – 7.50, 7.50 – 8.50, 8.50 – 10.00) with respective probabilities 0.42, 0.22 and 0.36.     Range II : (11-12, 12-14, 14-21) with respective probabilities 0.14, 0.56 and 0.30 respectively.     My problem is to form the combinations of these ranges in a decreasing order of joint probabilities. It is assumed that

hello, I want to simulate 200 times the mean of a joint probability (y1) and 200 times the mean of another joint distribution (y2), that is I'm expecting to get 200 means of y1 and 200 means of y2. y1 and y2 are probabilities that I calculate from the marginal prob. (z1 and z2 respectively) multiple by the conditional prob. (x1 and x2 respectively), which I generaterd from the binomial

This is in followup to a thread on R-help with subject "Sampling". I claim that R does the wrong thing by default when sampling with unequal probabilities without replacement - the selection probabilities are not proportional to 'prob', for any draw after the first: I suggest that R do what S-PLUS now does (though you're free to choose a better implementation). What S-PLUS

Hello How to compare  two statistical histograms? How i can know if these histograms are equivalent or not??   Regards [[alternative HTML version deleted]]

Hi Folks, Given a series of n independent Bernoulli trials with outcomes Yi (i=1...n) and Prob[Yi = 1] = Pi, I want P = Prob[sum(Yi) = r] (r = 0,1,...,n) I can certainly find a way to do it: Let p be the vector c(P1,P2,...,Pn). The cases r=0 and r=n are trivial (and also are exceptions for the following routine). For a given value of r in (1:(n-1)), library(combinat) Set <- (1:n)

I want to construct a logit model, plot the probability curve with the confidence intervals, and then I want to print out a data frame with the predictor, response value, predicted value, the low ci predicted value, and the high ci predicted value. So it should look something like: value low_ci prob hi_ci 5 0.10 0.12 0.13 6 0.11 0.13 0.16 7 0.13 0.15

> library(ROCR) > library(e1071) svmres.prob <- svm(traindx, traindy, probability=TRUE) svmpred.prob <- predict(svmres.prob, testdx, probability = TRUE, decision.values = TRUE, type="prob") > print(length(attr(svmpred.prob, "probabilities"))) [1] 0 > print(attr(svmpred.prob, "probabilities")) NULL > print(attributes(svmpred.prob)$decision.values)

Dear R Community, I am pleased to announce the beta-release of the prob package. The source code is now on CRAN, and binaries should be generated there before long. In the meantime, you can get it with install.packages("prob", repos = "http://r-forge.r-project.org") The prob package gives a framework for doing elementary probability on finite sample spaces in R. The