similar to: sampling and testing

Greetings R Users! I am posting to inquire about the proximity matrix in the randomForest R-package. I am having difficulty pushing very large data through the algorithm and it appears to hang on the building of the prox matrix. I have read on Dr. Breiman's website that in the original code a choice can be made between using an N x N matrix OR to increase the ability to compute large

Hi all, I'm working to apply a function that will generate a matrix of results only when a specific criteria is met. I want my final results to be a matrix with both the values that meet the criteria (the results of the function), and those that to do in the same positions in the matrix (the original numbers). Here's a sample of what I would like to do: t.mean.1.c <- c(-15, -20,

Hi, I'm installing rgdal but I keep having failures because I have not been able to find a good source of information for the correct configuration settings when installing GDAL. My error from the R install.packages("rgdal") is below. Can someone point me to a good source to tell me how to set after ./configure when installing GDAL? I'd like to be able to work with raster

Hello, I'm trying to develop a box plot of time series data to look at the range in the data values over the entire period of record. My data initially starts out as a list of hourly data, and then I've been using this code to make this data into the final ts array. # Read in the station list stn.list <- read.csv("/home/kbennett/fews/stnlist3", as.is=T, header=F) # Read in

Hi, I have Forecast Class and Observed Class in a data matrix as below. > Sample1 FCT OBS 1 1 5 2 2 4 3 3- 3+ 4 3 3 5 3+ 3- 6 4 2 7 5 1 I want to find the difference between Observed and Forecast Classes. How can I get this done? I tried to following to convert the 1 through 5 classes, to 1 through 7 for both OBS and FCT column. > Sample1$OBS2 <- Sample1$OBS

Dear experts, Please forgive the puzzled title and the length of this message - I thought it would be best to be as complete as possible and to show the avenues I have explored. I'm trying to fit a linear model to data with a binary dependent variable (i.e. Target.ACC: accuracy of response) using lrm, and thought I would start from the most complex model (of which "sample1.lrm1" is

Hello all, I have what feels like a simple problem, but I can't find an simple answer. Consider this data frame: > x <- data.frame(sample1=c(35,176,182,193,124), sample2=c(198,176,190,23,15), sample3=c(12,154,21,191,156), class=c('a','a','c','b','c')) > x sample1 sample2 sample3 class 1 35 198 12 a 2 176 176

Hi group, This is how my test file looks like: Chr start end sample1 sample2 chr2 9896633 9896683 0 0 chr2 9896639 9896690 0 0 chr2 14314039 14314098 0 -0.35 chr2 14404467 14404502 0 -0.35 chr2 14421718 14421777 -0.43 -0.35 chr2 16031710 16031769 -0.43 -0.35 chr2 16036178 16036237 -0.43 -0.35 chr2 16048665 16048724 -0.43 -0.35 chr2 37491676 37491735 0 0 chr2 37702947 37703009 0 0

Hi Group, I have a data frame below. Within this data frame there are samples (columns) that are measured more than once. Samples are indicated by "idx". So "id1" is present in columns 1, 3, and 5. Not every id is repeated. I would like to create a new data frame so that the repeated ids are averaged. For example, in the new data frame, columns 1, 3, and 5 of the original

Hi! I need to perform this simple sampling function several hundred times: x1=as.character(rnorm(1000, 100, 15)) x2=as.character(rnorm(1000, 150, 10)) y1=as.data.frame(x1,x2) sample1=as.data.frame(sample(y1$x1, 12, replace = FALSE, prob = NULL)) sample1 write.table(sample1, "sample1.txt", sep=" ",row.names=F,quote=F) My knowledge of loops is quite low. How can I produce 100

Dear Researchers, I have several files as this example: Myfile_MyArea1_sample1.txt i wish to split in "Myfile", "MyArea1", "sample1", and "txt", becasue i need to use "sample1" label. I try to use "strsplit" but I am able just to split as "Myfile_MyArea1_sample1" and "txt" OR "Myfile", "MyArea1",

This is what I am starting with: initial<- matrix(c(1,5,4,8,4,4,8,6,4,2,7,5,4,5,3,2,4,6), nrow=6, ncol=3,dimnames=list(c("1900","1901","1902","1903","1904","1905"), c("sample1","sample2","sample3"))) And I need to apply a filter (in this case, any value <5) to give me one dataframe with only the

Dear List Members, I have two small samples (n=20), the distributions are highly skewed. Does it make any sense to do a boostrap test to check for difference in means? And if so, could this be done like this: x <- numeric(10000) for(i in 1:10000) { x[i] <- mean(sample(sample1,replace=TRUE)) - mean(sample(sample2,replace=TRUE)) } (mean(sample1)-mean(sample2))/sd(x) Regards, Erika

Hi all, I have data size of : > dim(sample) [1] 35943 17 The first column is "stdate" - is date ( 01/11/2009 00:00:00,02/11/2009 00:00:00,02/11/2009 00:00:00 etc... ) Login is 13th column - is numbers (12,0,1 erc...) The below operation return the following error. > sample1 <- read.csv(file="sample1.csv",sep=",",header=TRUE) > avglog <-

Hello, Let's say I've drawn a random sample (sample1.df) from a large data frame (main.df), and I want to create a second random sample (sample2.df) where the values in its ID column *are not* in the equivalent ID column in the first sample (sample1.df). How would I go about doing this? In other words: The values in sample2.df$ID *are not found* in sample1.df$ID, and both samples are

I have searched the r-help files but have not been able to find an answer to this question. I apologize if this questions has been asked previously. (Please excuse the ludicrousness of this example, as I have simplified my task for the purposes of this help inquiry. Please trust me that something like this will in fact be useful what I am trying to accomplish. I am using R 2.4.1 in Windows XP.)

I'd like someone familiar with the R options initialization to comment on a difference of behavior within/without mpiexec I have a (.NET) application with embedded R that is proven to run in a single process: ./Sample1.exe on a Debian Linux with R 3.0.2 Running the same code with mpiexec, it fails at the R engine initialization: mpiexec -n 1 ./Sample1.exe Fatal error: you must

Hi everyone, I have just got different samples from a dataframe (independent and exclusive, there aren't common elements among them). I want to create a variable that indicate the sampling selection of the elements in the original dataframe (for example, 0 = no selected, 1= sample 1, 2=sample 2, etc.). I have tried to do it with ifelse command, but the problem is that the second line

Hi all, I am processing 24 samples data and combine them in single table called CombinedSamples using following: CombinedSamples<-rbind(Sample1,Sample2,Sample3) Now variables Sample1, Sample2 and Sample3 have many different columns. To make it more flexible for other samples I'm replacing above code with a for loop: #Sample is a string vector containing all 24 sample names for (k in

Hi, I have the precision values of a system on two different data sets. The snippets of these results are as shown: sample1: (total 194 samples) 0.6000000238 0.8000000119 0.6000000238 0.2000000030 0.6000000238 ... ... sample2: (total 188 samples) 0.80000001 0.20000000 0.80000001 0.00000000 0.80000001 0.40000001 ... ... I want to check if these results are statistically significant? Intuitively,