search for: mysamples

Hello! I have a data frame mysample (sorry for a long way of creating it below - but I need it in this form, and it works). I regress Y onto X1 through X11 - first without weights, then with weights: regtest1<-lm(Y~., data=mysample[-13])) regtest2<-lm(Y~., data=mysample[-13]),weights=mysample$weight) summary(regtest1) summary(regtest2) Then I calculate Durbin-Watson for both regressions

Dear all, I have a data similar to this: myframe<- data.frame (ID=c("Ernie", "Ernie","Ernie","Ernie"), Timestamp=c("24.09.2012 08:00", "24.09.2012 09:00", "24.09.2012 10:00", "25.09.2012 10:00"), Longitude=c("8.481","8.482","8.483","8.481"),

Hi, I need to sample randomly my dataset for 1000 times. The sample need to be the 80%. I know how to do that, my problem is that not only I need the 80%, but I also need the corresponding 20% each time. Is there any way to do that? Alternatively, I was thinking to something like setdiff () function to compare my 80% sample to the original dataset and obtain the corresponding 20%, unfortunately

...lution to give different numbers is using 1 instead of 8 in registerDoSNOW(makeCluster(8, type = "SOCK")) but that seems to defeat the point. library(foreach) library(doSNOW) registerDoSNOW(makeCluster(8, type = "SOCK")) getDoParWorkers() getDoParName() getDoParVersion() mySamples <- foreach (jj = 1:4, .combine=cbind) %dopar% { return(sample(1:10,10,replace=TRUE)) } mySamples ########## r 2.14.1 ########## > library(foreach) > library(doSNOW) > registerDoSNOW(makeCluster(8, type = "SOCK")) > getDoParWorkers() [1] 8 > getDoParName() [1] &q...

Thank you for providing the example code... for the request of running it multiple times it would have helped if you could have confirmed that the example ran through without errors... there were a lot of mistakes in it. Look into using the reprex package to check your example next time. I don't do this kind of analysis... I really don't know what to expect from the functions. The

[answers inline] On 18 August 2017 at 20:08, Dagmar <Ramgad82 at gmx.net> wrote: > > myframe<- data.frame (ID=c("Ernie", "Ernie","Ernie","Ernie"), > Timestamp=c("24.09.2012 08:00", "24.09.2012 09:00", "24.09.2012 10:00", > "25.09.2012 10:00"),

...ach variable within each subset for my characteristic of interest. I am familiar with sampling and saving single columns of data to do this sort of thing, but I am not sure how to accomplish this with a multiple-variable data set. For example, I am currently iterating this using a clunky process: mysamples<-list() for (i in 1:10){ mysamples[[i]] <- dataset[ sample(1:1637,100,prob=c(rep(163.7/1637,513),rep(245.5/1637,197),rep(1227.8/1637,927)),replace = TRUE), ] } But this leaves me with the additional task of defining each mysample[i] iteration and converting it to a form on which I can apply...

Dear list I wish to extract from a population genotypized for 10 SNP a subsample of the same population of size n with similar allele frequencies. Essentially i have a matrix of 200 rows (df) like this Name,Condition,rs1385699_X,rs6625163_X,rs962458_X,Rs4658627_1, sample01,Case,1,1,1,-1 sample02,Control,1,1,1,1 sample06,Control,1,-1,1,0 sample10,Case,1,1,1,0 sample11,Control,1,1,1,1

...ample # i have commented out the write.table statement below testData <- read.table(textConnection("rs9999853,cv0084,A,A rs999986,cv0084,C,B rs9999883,cv0084,E,F rs9999853,cv0085,G,H rs999986,cv0085,I,J rs9999883,cv0085,K,L"),header=FALSE,sep=",") closeAllConnections() mysamples <- unique(testData$V2) for (one_ind in mysamples) { one_sample <- testData[testData$V2==one_ind,] mywide <- reshape(one_sample, timevar = "V1", idvar = "V2",direction = "wide") # write.table(mywide,file ="newdata.txt",append=TRUE,row.names=F...

Greetings: I am looking for some help (probably really basic) with looping. What I want to do is repeatedly sample observations (about 100 per sample) from a large dataset (100,000 observations). I would like the samples labelled sample.1, sample.2, and so on (or some other suitably simple naming scheme). To do this manually I would >smp.1 <- sample(100000, 100) >sample.1 <-

Dear all, i have a dataset of about 400 records , with a variable that has  two levels 40 bad and 360 good among other variables,how do i come up  with10 random samples that have the composition of as the main sample  but maintaining the 40 bad 360 good with replacement, i recently discovered that my random samples generated dont maintain the ratio. My code is as  : mysample <-

Hi, I'm working with a data.frame containing values between 0 and 22000. Most of the values are actually between 0 and 50 and the high ones are outliers. I want to generate a boxplot and since the outliers are extremely high, I need to scale the y scale logarithmically. Otherwise one wouldn't really see the boxes of the boxplot. boxplot(dat, log="y", ylim=c(0,

Hi All, I want to take a simple random sample from a large dataset, gly, but I'm getting an error message. Any help? dim(gly) [1] 112371 37 > s1 <- sample(gly,100) Error in `[.data.frame`(x, .Internal(sample(length(x), size, replace, : cannot take a sample larger than the population when 'replace = FALSE' Thanks, Rachel [[alternative HTML version

Hello! I am using persp to get a plot of a 32x48 matrix. data <- as.matrix(read.table(file="mySample.dat")) persp(data) And I'm having trouble with trying to give a z-dependent palette to it like rainbow (or any built in one would do). How to combine that with the "col" argument? nice day Szilvi -- View this message in context:

Full_Name: Peter Gedeck Version: R1.6.2 and R1.7.0 OS: Windows XP Submission from: (NULL) (194.191.169.72) Hello, I marked the bug report Private, as I don't want my email address on the web server. The problem that I found is best explained using an example. index <- 1:6 cluster <- c(1,1,1,2,2,3) tapply(index,cluster,sample) gives $"1" [1] 2 1 3 $"2" [1] 4 5

Dear R users, I am a beginner to R. I generated 1000 samples with 15 data in each sample I tried finding the variance for each sample I used the code: m=1000;n=15 > r<-rnorm(15000) > for(i in 1:m){ x=data[,i] v=var(x)} what I got was just the variance for the last sample i.e. the 1000th sample but what I want is 1000 variance. Does anyone know what I did wrong? Thanks Chloe Smith

Dear All, I am trying to build a program which will take repeated samples (w/o replacement) from a population of values. The interesting catch is that I would like the sample values to be removed from the population, after each sample is taken. For example: pop<-c(1,5,14,7,9,12,18,19,65,54) sample(pop, 2) = lets say, (5,54) ## This is where I would like values (5, 54) to be removed from

Dear useRs, I have an empirical distribution (not normal etc) and I want to draw random samples from it. One solution I can think of is to compute let's say 100 quantiles, then use runif() to draw a random number Q between 1 and 100, and finally run runif() again to pull a random value from the quantile Q. Is there perhaps a better/more elegant way of doing this? Thank you, b.

Hello all: I'm sure this is a trivial request, but I'm still a beginner at this, and haven't been able to find it. I need to create simulated data based on some empirical distributions of a single variable. I've found R functions to help me simulate data based on analytical distributions, or to make simulations based on correlation matrices, but nothing so simple as what I need.

Dear experts, I am a beginner of R. I'm looking for experts to guide me how to do programming in R in order to randomly replace 5 observations in X explanatory variable with outliers drawn from U(15,20) in sample size n=100. The replacement subject to y < 15. The ultimate goal of my study is to compare the std of y with and without the presence of outliers based on average of 1000