similar to: generate fourth vector based on known correlations

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

Hello all, I believe this can be done using bootstrap, but I am wondering if there is some other way that might be used to tackle this. #Let's say I have two pairs of samples: set.seed(100) s1 <- rnorm(100) s2 <- s1 + rnorm(100) x1 <- s1[1:99] y1 <- s2[1:99] x2 <- x1 y2 <- s2[2:100] #And both yield the following two correlations: cor(x1,y1) # 0.7568969 (cor1) cor(x2,y2)

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,

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! 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 - - -- -

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 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

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

[HVM] Pick up SMBIOS entry point in the ROMBIOS. Signed-off-by: Andrew D. Ball <aball@us.ibm.com> diff -r 1d817bfc5ed9 tools/firmware/rombios/rombios.c --- a/tools/firmware/rombios/rombios.c Fri Aug 11 14:22:54 2006 +0100 +++ b/tools/firmware/rombios/rombios.c Fri Aug 11 14:55:22 2006 -0400 @@ -9443,6 +9443,43 @@ rom_scan_increment: mov ds, ax ret +#ifdef HVMASSIST + +; Copy the

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 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",

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),

Suppose I have a list where I want to extract only the elements that occur in every component. For instance in the list foo I want to know that the numbers 2 and 3 occur in every component. The solution I have seems unnecessarily clunky. TIA, Andy foo <- list(x = 1:10, y=2:11, z=1:3) bar <-unlist(foo) bartab <- table(bar) as.numeric(names(bartab)[bartab==length(foo)])

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)

hello everyone, I'm wondering whether llvm can work together with simplescalar. Would anyone please give me some clarification? Thanks a lot!

hello everyone, I'm wondering whether llvm can work together with simplescalar. Would anyone please give me some clarification? Thanks a lot!

Dear Bert, thank you for your response. here it is the piece of R code : given 3 data frames below --- N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"),

#I am trying to understand how R fits models for contrasts in a #simple one-way anova. This is an example, I am not stupid enough to want #to simultaneously apply all of these contrasts to real data. With a few #exceptions, the tests that I would compute by hand (or by other software) #will give the same t or F statistics. It is the contrast estimates that R produces #that I can't seem to

> On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > > Hi Bogdan, > Kinda messy, but: > > N <- data.frame(N=c("n1","n2","n3","n4")) > M <- data.frame(M=c("m1","m2","m3","m4","m5")) > C <-