similar to: Comparing the correlations coefficient of two (very) dependent samples

I believe the problem is here: cor1 <- cor(x1, y1, method="spearman") cor2 <- cor(x2, y2, method="spearman") The x's and y's are not looked for in data (i.e. NSE) but in the environment where the function was defined, which is standard evaluation. Change the above to: cor1 <- with(d, cor(x1, y1, method="spearman")) cor2 <- with(d, cor(x2, y2,

Given four known and fixed vectors, x1,x2,x3,x4, I am trying to generate a fifth vector,z, with specified known and fixed partial correlations. How can I do this? In the past I have used the following (thanks to Greg Snow) to generate a fifth vector based on zero order correlations---however I'd like to modify it so that it can generate a fifth vector with specific partial

R-Experts, Here below my R code working without error message but I don't get the results I am expecting. Here is the result I get: [1] "All values of t are equal to 0.28611928397257 \n Cannot calculate confidence intervals" NULL If someone knows how to solve my problem, really appreciate. Best, S ######################################################### # Difference in Spearman

Hi! All, I have 2 correlation matrices of 4000x4000 both with same row names and column names say cor1 and cor2. I have extracted some information from 1st matrix cor1 which is something like this: rowname colname cor1_value a b 0.8 b a 0.8 c f 0.62 d k 0.59 - - -- -

I am trying to generate a fourth vector,z, given three known and fixed vectors, x1,x2,x3 with corresponding known and fixed correlations with themeselves and with z. That is, all correlations are known and prespecified. How can I do this? Thank you, ben

Hi, what I would need are 2 vector pairs (x,y) and (x1,y1). x and x1 must have the same sd. y and y1 should also exhibit the same sd's but different ones as x and x1. Plotting x,y and x1,y1 should produce a plot with 2 vectors having a different slope. Plotting both vector pairs in one plot with fixed axes should reveal the different slope. many thanks syrvn -- View this message in

Hi useRs, I'm trying for the first time to use a sem. The model finally runs, but gives a warning saying : "In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names = vars, : Could not compute QR decomposition of Hessian. Optimization probably did not converge. " I found in R-help some posts on this warning, but my attemps to modify the code didn't change

Hello, I have a data set with 15 variables, and use "pairs" to plot the scatterplot of this data set. Then I want to plot some circles on the small pictures with high correlation(e.g. > 0.9). First, I use "cor" to obtain the corresponding correlation matrix (x) for this scatterplot. Second, use "seq(along = x)[x > 0.9]" to find the positions of the small

Hi I have a couple of functions that work on the object created by another R command and then print out or summarise the results of this work. The main function is defined as: hotelling.t <- function(obj) { #internal commands } I then have print.hotelling.t() that takes the list returned by hotelling.t and prints it with some extra significance calculations, formatting, etc. I want to

Hello, I have a customized pairs () fonction as follows that displays correctely my data. ------------------------------------------------------------------------ panel.cor1 <- function (x, y, digits=2, prefix="") { usr <- par("usr"); on.exit(par(usr)) par(usr = c(0, 1, 0, 1)) r <- cor(x, y,use="pairwise.complete.obs",

Hi Now a more theoretical question. I have two correlation matrices - one of a set of variables under a particular condition, the other of the same set of variables under a different condition. Is there a statistical test I can use to see if these correlation matrices are "different"? Thanks Mick

I have a raster file(1440*720 rows) contains values of 1 ,2 , and 3. when I plot the file , I got a map of three colors but I do not know which is which. How can I put those colors as as I want : 1=red 2=blue 3=green code: pvm <- file("C:\\User_sm-das.bin","rb") cor1<- readBin(pvm, numeric(), size=4, n=1440*720, signed=TRUE) r <-raster(t(matrix((data=cor1),

Has the loess.smooth() function changed? It used to work, but now it causes R to abort with a segmentation fault. I stole the function points.lines() from V&R 1st ed. pp. 67--68, but now it only works if I remove the line with loess.smooth. Here's the function I'm using: points.lines <- function(x, y, ...) { cor1 <-round(cor(x, y, use="pairwise"), digits=2)

Hi, I am trying to convert a 2D correlation matrix to 3 columns for graphical representation: rdata = replicate(100, rnorm(15)) #construct a 2D matrix c1 = cor(rdata) #outputs a correlation matrix Now I want to convert the 2D c1 to (row#, col#, correlation) 1 1 cor1 1 2 cor2 1 3 cor3 ... 2 1 cor.. Is there a way to do this? The main reason I am doing this is to find a correlation based graph

Dear users, I have for the moment a function which looks for the best correlation for each file I have in my correlation matrix. I'm working on a list.files. Here's the function: get.max.cor <- function(station, mat){ mat[row(mat) == col(mat)] <- -Inf which( mat[station, ] == max(mat[station, ],na.rm=TRUE) ) } If I have a correlation matrix like this

Hi, all. I know that this is probably something that others have asked, but I can't find a reference in either the FAQ or the help pages. I'm trying to find a way to put Greek letters as a label of the plot *with* a value from the data. Previously I've used pasted and the word "rho". * paste("rho=", cor2[i]) will produce a label of

Hello, I am wondering whether there is a way to test whether two non-independent correlation coefficients are significantly different, in R? I have an experimentally measure variable Y, and two different variables X1, and X2, which are predictions of Y that were predicted using two different computational models. I would like to see whether the correlation of Y and X1, and Y and X2 is

dir1 <- list.files("C:\\Users\\aalyaari\\Desktop\\cor", "*.bin", full.names = TRUE) dir2 <- list.files("C:\\Users\\aalyaari\\Desktop\\cor2", "*.bin", full.names = TRUE) results <- list() for (.files in dir1){ # read in the 365 files as a vector of numbers for dir1 file1 <- do.call(rbind,(lapply(.files, readBin ,

I have tried these steps and the encryption on the console still does not seem to validate? Any other thoughts? Thanks, Tom -----Original Message----- From: Jerome Alet [mailto:alet@unice.fr] Sent: Tuesday, November 20, 2001 3:50 PM To: icecast@xiph.org Subject: Re: [icecast] Encrypted password problem with icecast 1.3.11 On Mon, Nov 19, 2001 at 12:03:53PM -0500, Thomas Steiger wrote: > I

Dear list members, I need to translate 3 lines of matlab code to R (a loop, to be specific), and I don't know what would be the results in matlab or how to do it in R-- I don't realise if they are doing to the col, vector or what. if the results are a vector or a value or a matrix :-( Anyone with matlab, can run it and give me the result? Any ideias what am I doing wrong? The code is