similar to: Hmisc::rcorr on a 'data.frame'?

Dear all Is there an easy way to display only one half (top-right or bottom-left) of a correlation matrix? > require(Hmisc) > rcorr(as.matrix(mtcars[ , 1:4])) mpg cyl disp hp mpg 1.00 -0.85 -0.85 -0.78 cyl -0.85 1.00 0.90 0.83 disp -0.85 0.90 1.00 0.79 hp -0.78 0.83 0.79 1.00 n= 32 P mpg cyl disp hp mpg 0 0 0 cyl 0 0 0 disp 0 0

Dear all I'm unable to find an example of extracting the rotated scores of a principal components analysis. I can do this easily for the un-rotated version. data(mtcars) .PC <- princomp(~am+carb+cyl+disp+drat+gear+hp+mpg, cor=TRUE, data=mtcars) unclass(loadings(.PC)) # component loadings summary(.PC) # proportions of variance mtcars$PC1 <- .PC$scores[,1] # extract un-rotated scores of

Dear R-help, This is an example in the {Hmisc} manual under rcorr.cens function: > set.seed(1) > x <- round(rnorm(200)) > y <- rnorm(200) > round(rcorr.cens(x, y, outx=F),4) C Index Dxy S.D. n missing uncensored Relevant Pairs Concordant Uncertain 0.4831 -0.0338 0.0462 200.0000

Dear list, I get the following message when I use rcorr in library "Hmisc" ------------------------------------------------------ > rcorr(lskPox0t30, type=c("spearman")) Error in "[<-.data.frame"(*tmp*, is.na(x), value = 1e+30) : matrix subscripts not allowed in replacement ------------------------------------------------------ I do not understand

Hi, I'm trying to calculate 95% confidence interval of C statistic of logistic regression model using rcorr.cens in rms package. I wrote a brief function for this purpose as the followings; CstatisticCI <- function(x) # x is object of rcorr.cens. { se <- x["S.D."]/sqrt(x["n"]) Low95 <- x["C Index"] - 1.96*se Upper95 <- x["C

Hi all, Sorry about posting a really novice question. I was able to run rcorr after converting the list to a matrix by your help. I'm though wondering if there is any way to find out an exact p value as the output only gave me 0 for P value as shown below. I've added options(digits=10), which doesn't seem to help at all. Any help would be appreciated. P D Prime T

Hi, I try to export the outputs of rcorr into excel. but I got error message,"cannot coerce class "rcorr" into a data.frame". Actually i just need export part of results of this analysis,e.g. p-values or stat-values. Does anyone have sort of exprience before or you can help on how to export subset of results of analysis? Many Thanks! Xin

Dear r-helpers! I want to classify my vegetation data with hierachical cluster analysis. My Dataset consist of Abundance-Values (Braun-Blanquet ordinal scale; ranked) for each plant species and relev?. I found a lot of r-packages dealing with cluster analysis, but none of them is able to calculate a distance measure for ranked data. Podani recommends the use of Goodman and Kruskals' Gamma for

Hello R-User! I have a data.frame with 82 variables (columns) and 290 rows. The variables are set to classes factor, ordered factor and numeric. I used the following code Matrix.My.data<-as.matrix(Df.My.Data[2:82]) Matrix.My.data.rcorr<-rcorr(Matrix.My.data, type="spearman") and got the following message: Fehler in storage.mode(x) <- if (.R.) "double" else

Dear all On my system html() conversion of a `latex()' object fails. Follows a dummy example: > require(Hmisc) > data(Angell) > .object <- cor(Angell[,1:2], use="complete.obs") > tmp <- latex(.object, cdec=c(2,2), title="") > class(tmp) [1] "latex" > html(tmp) /tmp/RtmprfPwzw/file7e72f7a7.tex:9: Warning: Command not found: \tabularnewline

Hello, Also, the other file, NPA.csv, is not in tabular form. Can you please reformat it? Rui Barradas Citando ruipbarradas at sapo.pt: > Hello, > > Please keep this on the list, always cc r-help. > One of the files in your attachment is empty: > > y <- read.csv(file.choose("GT.csv")) > Error in read.table(file = file, header = header, sep = sep, quote =

Hello, Please keep this on the list, always cc r-help. One of the files in your attachment is empty: y <- read.csv(file.choose("GT.csv")) Error in read.table(file = file, header = header, sep = sep, quote = quote,? : ? no lines available in input Rui Barradas ? Citando Chaitanya Ganne <Chaitanya.Ganne at jefferson.edu>: > Thank you so much for your input. > > I am

Dear R users, Does the results below make any sense? Can the the interval of the correlation coefficient be between *-1.0185* and -0.8265 at 95% confidence level? Liviu > library(boot) > data(mtcars) > with(mtcars, cor.test(mpg, wt, met="spearman")) Spearman's rank correlation rho data: mpg and wt S = 10292, p-value = 1.488e-11 alternative hypothesis: true rho is not

Dear all, Consider the model below: > x <- lm(mpg ~ cyl * disp * hp * drat, mtcars) > summary(x) Call: lm(formula = mpg ~ cyl * disp * hp * drat, data = mtcars) Residuals: Min 1Q Median 3Q Max -3.5725 -0.6603 0.0108 1.1017 2.6956 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.070e+03 3.856e+02 2.776 0.01350 * cyl

Hi! All, To find co-expressed genes from a expression matrix of dimension (9275 X 569), I used rcorr function from library(Hmisc) to calculate pearson correlation coefficient (PCC) and their corresponding p-values. From the correlation matrix (9275 X 9275) and pvalue matrix (9275 X 9275) obtained using rcorr function, I wanted to select those pairs whose PCC's are above 0.8 cut-off and then

I am trying to plot a graph but the points on the graph should be different symbols and colors. It should represent what is in the legend. I tried using the points command but this does not work. Is there another command in R that would allow me to use different symbols and colors for the points? Thank you kindly. data(mtcars) plot(mtcars$wt,mtcars$mpg,xlab= "Weight(lbs/1000)",

Dear List, I would like to calculate the Goodman-Kruskal gamma for the predicted classes obtained from an ordinal regression model using lrm in the Design package. I couldn't find a way to get gamma for predicted values in Design so have found previous positings suggesting to use : Rcorr.cens(x, S outx = TRUE) in the Hmisc package My question is, will this work for predicted vs

Dear all Is there an easier way to retrieve the name of an object? For example, > tmp <- 1:10 > as.character(quote(tmp)) [1] "tmp" > as.character(quote(mtcars$cyl)) [1] "$" "mtcars" "cyl" > as.character(quote(mtcars$cyl))[3] [1] "cyl" The last call more than anything seems a hack. Is there a better way? Thank you Liviu

Hello, When plotting a barchart with ggplot it drops the levels of the factor for which no counts are available. For example: library(ggplot) mtcars$cyl<-factor(mtcars$cyl) ggplot(mtcars[!mtcars$cyl==4,], aes(cyl))+geom_bar() levels(mtcars[!mtcars$cyl==4,]) This shows my problem. Because no counts are available for factorlevel '4', the label 4 dissapears from the plot. However, I

Dear R-helpers, I recently switched from SAS to R, in order to model the occurrence of rare events through logistic regression. Is there a package available in R to calculate the Goodman-Kruskal Gamma? After searching a bit I found a function "rcorr.cens" which should do the job, but it is not clear to me how to define the input vectors? Is "x" a vector with the fitted