search for: 0.85

Hi,I want to generate a dataset, which have more than one clusters (say 2) and the elements in each cluster have high correlation (say 0.85) andelements among different clusters have low (say 0.1) or zero correlation.The correlation structure of final dataset should have a block-diagonal structure, that likes 0.85 0.85 .. 0.85 0 0 0 0 ...0 0 0...0 0 0

I have 4 years of data and for each year, I have initialize the value so now for fitting the model, I want to remove the initial value and get the model based on remaining data set. Could anyone can help on this? I want to get linear model based on fourth column and 13th column but need to remove the initial value for each year and each treatment ( the second column I have 1:36) . Thank you,

Folks, The contour() function wants x and y to be in increasing order. I have a situation where I have a grid in x and y, and associated z values, which looks like this: x y z [1,] 0.00 20 1.000 [2,] 0.00 30 1.000 [3,] 0.00 40 1.000 [4,] 0.00 50 1.000 [5,] 0.00 60 1.000 [6,] 0.00 70 1.000 [7,] 0.00 80 0.000 [8,] 0.00 90

Folks, A few days ago, I had asked a question on this mailing list about making a contour plot where a function z(x,y) is evaluated on a grid of (x,y) points, and the data structure at hand is a simple table of (x,y,z) points. As usual, R has wonderful resources (and subtle complexity) in doing this, and the gurus of the list showed me the way. Here's a complete working example. One might

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

Hi all, HAVP 0.85 has been released and also added support for ClamAV 0.90 library. But there is still no rpm available for HAVP 0.85. Dag has HAVP 0.82. (havp-0.82-1.el4.test.i386.rpm<http://dag.wieers.com/rpm/packages/havp/havp-0.82-1.el4.test.i386.rpm>) . Does it support ClamAV 0.90 library ? Can I make a RPM out of current version of Havp Version 0.85 ? If so, How can do it? --

Of course (and unsurprisingly) Duncan is correct. I see that behavior in R 3.1.0, as well as the modern ones Duncan mentioned. What I said is true, as far as it goes, but the symbol being resolved is FUN, so when looking for a function it doesn't find the function version of round. Did you perhaps have a function named FUN in your global environment? If so you are being bitten by what I

Hi, I am trying to add text to the bottom of a lattice bwplot with multiple panels. I would like to add a label below each boxplot, but the labels do not come from the data. I've tried the following, code: f1 <- c(rep(c(rep("a", 3), rep("b", 3), rep("c", 3)), 2)) f2 <- c(rep("A", 9), rep("B", 9)) dv <- c(0.9, 0.8, 0.85, 0.6, 0.65,

Habría que buscar la vuelta, yo no lo se, pero posiblemente lo siguiente da una pista. Nota: al mismo código le sume una línea al final datos<-c(2,3,4,5,6,7,8) quantile(datos) quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95)) as.matrix(quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95))) as.data.frame(quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95))) # ¿ y si solo solicita

Hi All- This seems like such a pathetic problem to be posting about, but I have no idea why this testcase does not work. I have tried this using R 2.4.1, 2.4.0, 2.3.0, and 2.0.0 on several different computers (Mac OS 10.4.8, Windows XP, Linux). Below the signature, you will find my test case R code. My point in this folly is to take a dataframe of 300,000 rows, create a filter based on two

Dear all, I?m having some problems getting optim with method="Nelder-Mead" to work properly. It seems like there is no way of controlling the step size, and the step size seems to depend on the *difference* between the initial values, which makes no sense. Example: f=function(xy, mu1, mu2) { print(xy) dnorm(xy[1]-mu1)*dnorm(xy[2]-mu2) } f1=function(xy) -f(xy, 0,

Dear all ?Hmisc::rcorr states that it takes as main argument "a numeric matrix". But is it normal that it fails in such an ugly way on a data frame? (See below.) If the function didn't attempt any conversion to a matrix, I would have expected it to state that in the error message that it didn't accept 'data.frame' objects in its input. Also, I vaguely remember having used

Hi list, I'd like to make a factor with seven 1s and three 2s using the factor() function. That is, 1 1 1 1 1 1 1 2 2 2 I will then bind this factor to the matrix below using cbind.data.frame(). 0.56 0.48 0.22 0.59 0.32 0.64 0.26 0.60 0.25 0.38 0.24 0.45 0.56 0.67 0.78 0.97 0.87 0.79 0.82 0.85 I am new to R and have been using various manuals and have made many attempts without

Hello all, I'm new in R, and I have a data-frame like this (dput information below): Specie Fooditem Occurrence Volume 1 Schizodon vegetal 1 0.05 2 Schizodon sediment 1 0.60 3 Schizodon vegetal 1 0.15 4 Schizodon alga 1 0.05 5 Schizodon sediment 1 0.90 6 Schizodon

Estimado Mauricio Monsalvo Le paso una idea, no es un código muy lindo que digamos, pero al correrlo seguramente se dará cuenta de mi sugerencia. datos<-c(2,3,4,5,6,7,8) quantile(datos) quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95)) as.matrix(quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95))) as.data.frame(quantile(datos,probs = c(0.25, 0.75, 0.85, 0.90, 0.95))) # ¿ y si solo

Dear list I'm testing a predictor and I produced nice performance plots with ROCR package utilizing the 3 standard command pred <- prediction(predictions, labels) perf <- performance(pred, measure = "tpr", x.measure = "fpr") plot(perf, col=rainbow(10)) The pred object and the perfo object are S4 with the following slots An object of class "performance"

Hi all, When teaching this year's class, I was quite amazed that one of my examples didn't work any longer. I wanted to illustrate the importance of match.fun() with following code: myfun <- function(x, FUN, ...){ FUN(x, ...) } round <- 2 myfun(0.85, FUN = round, digits=1) I expected to see an error, but this code doesn't generate one. It seems as if in the current R

Hi all, Quick question: What function can I use to draw a line in the margin of a plot? segments() and lines() both stop at the margin. In case the answer depends on exactly what I'm trying to do, see below. I'm using R v. 2.8.1 on Windows XP. Cheers, Alan I'm trying to make a horizontal barplot with a column of numbers on the right side. I'd like to put a line between the

> as.character(seq(-.25,.95,.1)) [1] "-0.25" "-0.15" "-0.05" "0.05" "0.15" "0.25" "0.35" "0.45" "0.55" "0.65" "0.75" "0.85" "0.95" > as.character(seq(-.35,.95,.1)) [1] "-0.35" "-0.25"

Dear all, Lets say I have the following data frame: > df1 <- data.frame(Show=c('Star Trek', 'Babylon 5', 'Dr Who'), Size=c(0.7, 0.0, 0.701), Date=as.Date(c('2007-08-03', '2007-08-03', '2007-08-03'), format='%Y-%m-%d')) > df2 <- data.frame(Show=c('Star Trek', 'Dr Who', 'Torchwood'), Size=c(0.8, 0.85,