similar to: reading table

Hi, i have a problem in assigning 2 conditions to grep() , my data look like this: DA 24 N7 Rad= 3.4 20 Sac= 0.93 Acc= 4.76 DA 24 N7 Rad= 3.4 14 Sac= 0.65 Acc= 3.33 DA 24 N7 Rad= 3.4 3 Sac= 0.14 Acc= 0.71 DA 24 N7 Rad= 3.4 11 Sac= 0.51 Acc= 2.62 DG 23 N7 Rad= 3.4 8 Sac= 0.37 Acc= 1.91 DG 23 N7 Rad= 3.4 5 Sac= 0.23 Acc= 1.19 DG 23 N7 Rad= 3.4 0 Sac= 0.00 Acc= 0.00 DG 23 N7 Rad= 3.4 3 Sac=

To anyone who can help: I have two brief questions concerning sapply. Following below is the code for my example. The two problems are described at the end of the code: site <- rep(2:6, each = 12) tillage <- rep(c(1,-1), each = 6, times = 5) carbon <- c(18.23, 16.06, 17.81, 16.07, 17.26, 17.08, 14.92, 15.88, 12.11, 14.23, 16.99, 13.57, 20.34, 20.3,

I've only begun investigating R as a substitute for SPSS. I have a need to identify for each CASE the closest (or most similar) 5 other CASES (not including itself as it is automatically the closest). I have a fairly large matrix (50000 cases by 50 vars). In SPSS, I can use Correlate > Distances to generate a matrix of similarity, but only on a small sample. The entire matrix can not

hi Geoffrey, Since you are saying it happens on all types of volumes, lets do the following: 1) Create a dist-repl volume 2) Set the options etc you need. 3) enable gluster volume profile using "gluster volume profile <volname> start" 4) run the work load 5) give output of "gluster volume profile <volname> info" Repeat the steps above on new and old

On Sat, 2011-10-29 at 15:16 -0500, Hal Finkel wrote: > On Sat, 2011-10-29 at 14:02 -0500, Hal Finkel wrote: > > On Sat, 2011-10-29 at 12:30 -0500, Hal Finkel wrote: > > > Ralf, et al., > > > > > > Attached is the latest version of my autovectorization patch. llvmdev > > > has been CC'd (as had been suggested to me); this e-mail contains > >

Hi, I have a question about limma. I have data from spotted arrays. I have class A and Class B on the same slide. In Limma if I put class A as the reference (ref), are the list of genes I get in the output for class A or class B? #design <- modelMatrix(targets, ref="classA") design Josephine Brennan [[alternative HTML version deleted]]

Hallo R-experts, I produced a R graphic with the help of the following commands: plot(0:4,0:4, type = "n", xlab = "X-Werte", ylab = "Y-Werte") points(rnorm(100), rnorm(100), col = "black") locator(2, type= "l") locator(2, type = "l", col = "red") I could display this graphic by using windows() before entering those commands.

On Sat, 2011-10-29 at 14:02 -0500, Hal Finkel wrote: > On Sat, 2011-10-29 at 12:30 -0500, Hal Finkel wrote: > > Ralf, et al., > > > > Attached is the latest version of my autovectorization patch. llvmdev > > has been CC'd (as had been suggested to me); this e-mail contains > > additional benchmark results. > > > > First, these are preliminary

R Help - I'm trying to run a correlation matrix with a covariate of "age" and will at some point will also want to covary other variables concurrently. I'm using the "psych" package and have tried other methods such as writing a loop to extract semi-partial correlations, but it does not seem to be working. How can I accomplish this? library(psych) > set.cor(y =

Hi, i'd like to plot some data that I have with the value on the x axis and freq on the y axis. So, I need to calculate the freq a value is seen within my data vector for example, say i have a vector of data data=c(1,1,1,4,5,5,6) I want values<-c(1,4,5,6) freq<-c(3,1,2,1) in order to enable me to plot this. Sorry, i'm new to R. What is standard

I've figured out how to use optim (barely) to estimate 2 parameter = weibull distributions. I can't get over how easy this is. What I need to = do is use a weight in the observations..... For example,=20 the tree diameters and weights are are=20 4.70 , 100 6.00, 98 7.10, 75.0 8.10, 86.3 8.60, 80.456 8.90, 20.5 9.50, 16.6 11.40, 12.657 11.80, 12.47 14.50,

The aggregate function for some reason will now work for me. The error I'm getting is: "Error in sort.list(y) : 'x' must be atomic for 'sort.list' Have you called 'sort' on a list?" agPriceList=aggregate(PriceList$Size, list(PriceList$bandNum),sum) *Price list dataframe:* dput(PriceList) structure(list(Price = c(0, 8.18, 8.27, 10.42, 10.5, 10.6, 11.13,

I apologize that I've asked a similar question before, but being new to R I don't think I did a very good job of formating the question. I've included a text file since the date set is somewhat large. What I have is a huge string of numbers in a text file. The numbers are all separated by comma's and the groups are separated by a semicolon. What I would like to do is read each

Ralf, et al., Attached is the latest version of my autovectorization patch. llvmdev has been CC'd (as had been suggested to me); this e-mail contains additional benchmark results. First, these are preliminary results because I did not do the things necessary to make them real (explicitly quiet the machine, bind the processes to one cpu, etc.). But they should be good enough for discussion.

On Sat, 2011-10-29 at 12:30 -0500, Hal Finkel wrote: > Ralf, et al., > > Attached is the latest version of my autovectorization patch. llvmdev > has been CC'd (as had been suggested to me); this e-mail contains > additional benchmark results. > > First, these are preliminary results because I did not do the things > necessary to make them real (explicitly quiet the

I tried toy problems and there doesn't seem to be a basic problem between lm() and ts objects: X = data.frame(x=c(1,2,7,9), y=c(7,2,3,1)) lm(y ~ x, X) X <- lapply(X, function(x) ts(x, frequency=12, start=c(1994,7))) lm(y ~ x, X) and this works fine - whether you do an lm() before or after making ts objects, it's okay. But I have a situation where things aren't okay.

I am trying to get stock metadata from Yahoo finance (or maybe there is a better source?) here is what I did so far: yahoo.url <- "http://finance.yahoo.com/d/quotes.csv?f=j1jka2&s="; stocks <- c("IBM","NOIZ","MSFT","LNN","C","BODY","F"); # just some samples socket <-

Hi! I want to choose part of the data. But it has a question. > attach(PIKO) > PIKO[1:5,] code species dbh x y tag status branch 92 10142 PIKO 38.9 6.05 12.81 10165 alive 0 109 10213 PIKO 41.0 6.71 26.21 10202 alive 0 135 10222 PIKO 48.3 18.98 21.28 10214 alive 0 146 10223 PIKO 47.0 19.45 22.83 10224 alive 0 147 10223 PIKO 20.0 19.74

I am using R for the first time in one of my classes. My students have alerted me to a problem for which we have not found an answer. We find that some means returned by the model.tables function are not correct when missing data is present in analysis of variance problems. We have duplicated the problem using R 1.2.0, 1.2.1, and 1.2.2 under Windows 98 and several distributions of Linux (Redhat

How to optimize the for-loop to be reasonably fast for sample.size=100000000 ? You may want to change sample.size=1000 to have an idea what I am achieving. set.seed(143) A <- matrix(sample(0:1, sample.size, TRUE), ncol=10, dimnames=list(NULL, LETTERS[1:10])) B <- list() for(i in 1:10) { B[[i]] <- apply(combn(LETTERS[1:10], i), 2, function(x) { sum(apply(data.frame(A[,x]), 1,