search for: cor2

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, c...

...,18,14,13,12,12) ? # Function to calculate the difference in Spearman coefficients pearson_diff <- function(data, indices) { ? # Sample the data ? d <- data[indices, ] ? ?# Calculate the Spearman correlation coefficients for every sample ? cor1 <- cor(x1, y1, method="spearman") ? cor2 <- cor(x2, y2, method="spearman") ? # Return the difference ? return(cor1 - cor2) } ? # Create a data.frame with the data data <- data.frame(x1, y1, x2, y2) ? # Use the boot function to apply the bootstrap set.seed(123) # For reproducibility bootstrap_results <- boot(data = data,...

...nd(x1,x2,x3,x4) ) cor1 # create 1st version of z z <- rnorm(100) # combine in a matrix m1 <- cbind( x1, x2, x3, x4, z ) # center and scale m2 <- scale(m1) # find cholesky decomp c1 <- chol(var(m2)) # force to be independent m3 <- m2 %*% solve(c1) # create new correlation matrix: cor2 <- cbind( rbind( cor1, z=c(.5,.3,.1,.05) ), z=c(.5,.3,.1,.05,1) ) # create new matrix m4 <- m3 %*% chol(cor2) # uncenter and unscale m5 <- sweep( m4, 2, attr(m2, 'scaled:scale'), '*') m5 <- sweep( m5, 2, attr(m2, 'scaled:center'), '+') ##Check they are...

...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 "rho=0.74", or whatever. But if I use 'substitute' or 'evaluate' commands in order to get a real Greek letter, I lose the ability to paste it with a data value. Any ideas? Thanks, Jo Jo Hardin Assistant Professor Department of...

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 - - -- -...

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

...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) # -0.2055501 (cor2) Now for my questions: 1) is cor1 larger then cor2? (CI for the diff ?) 2) With what P value? 3) What if the values of s1 are not independent ? I found an older thread discussing such issues: http://tolstoy.newcastle.edu.au/R/e2/help/06/09/1035.html But wasn't sure how much this might be re...

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

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

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 , integer() , size = 2 , n = 360 * 720 , signe...

...quot;) } } group <- as.factor(eval(obj$call[[3]])) fac.levels <- levels(group) x <- eval(obj$call[[2]]) x1 <- subset(x, group==fac.levels[1]) x2 <- subset(x, group==fac.levels[2]) s1 <- cov(x1) s2 <- cov(x2) cor1 <- cor(x1) cor2 <- cor(x2) n1 <- nrow(x1) n2 <- nrow(x2) n.vars <- ncol(x) V <- (1/(n1+n2-2))*(((n1-1)*s1)+((n2-1)*s2)) Vcor <- (1/(n1+n2-2))*(((n1-1)*cor1)+((n2-1)*cor2)) mu1 <- obj$means[1,] mu2 <- obj$means[2,] d2 <- mahalanobis(mu1, mu2, V) d <-...

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

...44862745098039,0.458156862745098,0.647843137254902) ImageWidth = size(data,2); dim(A)[2] # i.e. number of col MaxOffset = 99; # defined variable, =2R ImageWidthToProcess = ImageWidth-MaxOffset; # =2R defined variable for Offset = 1:MaxOffset ; loop to calculate the autocorrelation. basicly, they do cor2, which I think is Spearman correlation, between every column and store it in the vector position??? OffsetPlaquette = data(:,1+Offset:ImageWidthToProcess+Offset); AutoCData(Offset) = corr2(data(:,1:ImageWidthToProcess), OffsetPlaquette) end # IN R would be... but it is not correct! What wo...

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 for values >= 0.85 using, corm <- corm[abs(corm[,3]) >= 0.86, ] library(network); library(sna) net <- network(corm, directed = F) cd <- component.dist(net) delete...

...696 0.8676857 ST210 0.8104837 0.9335584 1.0000000 0.8304132 0.9141465 ST211 0.8899451 0.8392696 0.8304132 1.0000000 0.8064669 ST212 0.7486417 0.8676857 0.9141465 0.8064669 1.0000000 ", header=TRUE) It works perfectly. If I have a correlation matrix with some NAs (but not only NAs) like this: cor2 <- read.table(text=" ST208 ST209 ST210 ST211 ST212 ST208 1.0000000 NA 0.9666491 0.9573701 0.9233598 ST209 NA 1.0000000 0.9744054 0.9577192 0.9346706 ST210 0.9666491 0.9744054 1.0000000 0.9460145 0.9582683 ST211 0.9573701 0.9577192 0.9460145 1.0000000 NA ST212 0.9233598 0.934...

...route2_pond, acc2, NA hydro->ind_plando, NA, 1 hydro->long_sup15, eau2, NA hydro->long_inf15, eau3, NA topog->pente, NA, 1 topog->est, top2, NA topog->sud, top3, NA topog->nord, top4, NA topog->ouest, top5, NA dist_protect-> urbanisation, cor1,NA dist_protect-> adj_99, cor2, NA dist_protect-> etat_hexa, cor3, NA topog-> urbanisation, cor4, NA topog-> adj_99, cor5, NA topog-> etat_hexa, cor6, NA topog-> access_hexa, cor7, NA topog<->hydro, cor8, NA topog<->pedo, cor9, NA pedo-> urbanisation, cor10, NA pedo-> adj_99, cor11, NA pedo->...

Dear R users graphing with plot(x) seams to work for a small length(x), when length(x) is too large it seams to clutter the display, a solution would be to display subsets of x at a time, yet a better way which I hope R supports is to place a sliding bar on the display window to control length(x) and thus the resolution, which will involve auto scaling the y axis as well as automatically

Hi useRs, I had posted about Adobe FLEX talking to R for rich visualisation. Reply from Jeffery Horner contained links to the revolution-computing.com webpage which had information pertaining to the Bay Users R group Meetup on Web Dashboards with R. I have a very specific project that I need to implement. I wish to use the graphics capabilities provided by Adobe FLEX to visualise outputs from R.

Good morning to all, I am encountering a blocking issue when using the function 'breackpoints' from package 'strucchange'. *Context:* I use a data frame, 248 observations of 5 variables, no NA. I compute a linear model, as y~x1+...+x4 x4 is a dummy variable (0 or 1). I want to check this model for structural changes. *Process & issues:* *First, I used function Fstats.* It

...t; Hi Harsh, > > If your project is so specific on the regression model, my suggestion > is to use tcl/tk or Gtk2 interface instead of FLEX, because your > requirement on interaction is simple (e.g. you don't need to > drag-and-drop points). > > I think the example 'run.cor2.examp()' in 'TeachingDemos' package can > give you enough inspiration. > > Or if you want to do the job (to change the slope) automatically, you > may want to see the example in the 'animation' package: > > ## > install.packages("animation") > l...