similar to: My very first loop!! I failed. May I have some start-up aid?

Displaying 20 results from an estimated 200 matches similar to: "My very first loop!! I failed. May I have some start-up aid?"

2017 Aug 19
0
My very first loop!! I failed. May I have some start-up aid?
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
2017 Aug 19
0
My very first loop!! I failed. May I have some start-up aid?
[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"),
2012 Mar 07
5
Sampling problems
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
2011 Aug 12
1
Which Durbin-Watson is correct? (weights involved) - using durbinWatsonTest and dwtest (packages car and lmtest)
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
2009 Dec 21
3
Question About Repeat Random Sampling from a Data Frame
Good Morning: I've read many, many posts on the r-help system and I feel compelled to quickly admit that I am relatively new to R, I do have several reference books around me, but I cannot count myself among the fortunate who seem to strong programming intuition. I have a data set consisting of 1637 observations of five variables: tensile strength, yield strength, elongation, hardness and a
2012 Jun 14
1
Question about sampling
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
2011 Apr 08
3
random sampling with levels and with replacement
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 <-
2009 Aug 20
2
boxplot with log="y" and values starting at 0
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,
2009 Nov 13
2
help sample from large dataset - misleading error?
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
2012 Nov 27
5
loop with date
Hello, I tried to construct my very first loop today and completly failed :-( Maybe someone can help me? I have a dataframe somewhat like this one: myframe <- data.frame (Timestamp=c("24.09.2012 09:00", "24.09.2012 10:00", "24.09.2012 11:00", "25.09.2012 09:00", "25.09.2012 10:00", "25.09.2012
2003 Jun 18
2
Private: Problem with tapply/lapply and sample (PR#3286)
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
2009 May 21
3
Problems with sample variance
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
2010 Aug 17
1
using color palette in persp
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:
2007 Feb 27
3
looping
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 <-
2005 Feb 28
1
draw random samples from empirical distribution
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.
2012 Dec 07
0
apply a function at: dateX, dateX+1, dateX+2, ....
Dear knowing people, Dennis Murphy helped me a lot with my first loop last week. Thanks again - I could have made more than 10 "Thank-You cakes" in the time it saved me! But now I want to complicate the thing. My ideas didn't work. Let's see if anyone is smarter ;-) The following packages are needed: library(adehabitatHR) library(rgdal) library(plyr) # My dataframe looks
2007 Oct 10
3
simulated data using empirical distribution
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.
2001 Dec 13
2
sample (PR#1212)
Full_Name: Antonio Possolo Version: 1.3.1 OS: Linux (RH 7.1), Windows 2000 Submission from: (NULL) (192.35.44.3) A FEATURE THAT EASILY GENERATES BUGS sample(pi, size=1) produces 1, 2, or 3. sample(c(pi, pi), size=1) produces 3.141593 always. Although this conforms with the behavior explained in the help page for "sample", the behavior for the case where x (in sample(x, ...)) has
2012 Sep 26
2
average environmental data if AnimalID and Time is duplicated
Hello, I tried for about three hours now to solve this problem but I can't figure it out. I am sure someone knows how do it. At least I hope so. I have a data frame somewhat like this: myframe <- data.frame (ID=c("Ernie", "Ernie", "Bert", "Bert"), Timestamp=c("24.09.2012 09:00", "24.09.2012 09:00", "24.09.2012 10:00",
2012 Dec 17
2
calculate a "rolling mean" including standard deviation
Hello everyone, I have a data frame somewhat like this one: myframe <- data.frame (Timestamp=c( "24.09.2012 06:00", "24.09.2012 07:00", "24.09.2012 08:00", "24.09.2012 09:00", "24.09.2012 10:00", "24.09.2012 11:00", "24.09.2012 12:00",