similar to: Loop with random sampling and write.table

Displaying 20 results from an estimated 10000 matches similar to: "Loop with random sampling and write.table"

2012 Mar 29
1
Random sample from a data frame where ID column values don't match the values in an ID column in a second data frame
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
2008 Nov 12
1
sampling and testing
Hello everyone, I have a dataset in the following format: col1 col2 # # # # # # # # # # # # What I want to do is: loop a random sample 10 times, and for each time it is sampled I want to run a correlation between both columns. What I have so far is this: >feb <- read.csv("corr.csv") # where the dataset is for February >attach(feb) >for(i in
2006 Sep 27
3
Converting text to numbers
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
2011 Mar 26
2
simple if question
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
2012 Jul 31
2
phantom NA/NaN/Inf in foreign function call (or something altogether different?)
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
2012 Dec 04
2
computing marginal values based on multiple columns?
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
2011 Sep 26
1
How to Store the executed values in a dataframe & rle function
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
2011 Oct 08
2
Permutation or Bootstrap to obtain p-value for one sample
Hi, I am having trouble understanding how to approach a simulation: I have a sample of n=250 from a population of N=2,000 individuals, and I would like to use either permutation test or bootstrap to test whether this particular sample is significantly different from the values of any other random samples of the same population. I thought I needed to take random samples (but I am not sure how
2010 Sep 07
1
average columns of data frame corresponding to replicates
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
2012 Jan 25
4
help to slip a file name using "strsplit" function
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",
2011 Aug 14
2
conditional filter resulting in 2 new dataframes
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
2009 Feb 02
1
A question regarding bootstrap
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
2010 May 02
1
Re :argument is not numeric or logical
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 <-
2007 May 10
1
how to pass "arguments" to a function within a function?
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.)
2013 Nov 21
2
Running R embedded in an mpiexec spawned process - Fatal error: you must specify '--save', '--no-save' or '--vanilla'
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
2010 Nov 04
2
Converting Strings to Variable names
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
2009 Sep 16
2
T-test to check equality, unable to interpret the results.
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,
2006 Jun 29
1
kmeans clustering
Hello R list members, I'm a bio informatics student from the Leiden university (netherlands). We were asked to make a program with different clustering methods. The problem we are experiencing is the following. we have a matrix with data like the following research1 research2 research3 enz sample1 0.5 0.2 0.4 sample2 0.4
2008 May 21
2
Problem in converting natural numbers to bits and others
Hi, I just started using R for about one week and I have few problems. i)I have a problem in finding right function to convert a table of natural numbers to bitwise. For a simple example; I have the below table:- Column Col1 Col2 Col3 Sample1 5 7 10 Sample2 0 2 1 Sample3 4 0 0 Supposedly i wanted to convert to :- Column Col1 Col2 Col3
2009 Jun 03
2
Create a time interval from a single time variable
I am trying to set up a data set for a survival analysis with time-varying covariates. The data is already in a long format, but does not have a variable to signify the stopping point for the interval. The variable DaysEnrolled is the variable I would like to use to form this interval. This is what I have now: ID Age DaysEnrolled HAZ WAZ WHZ Food onARV