similar to: Help

Hello, I was wondering if anyone knew of an R function/R code to plot bivariate (3 dimensional) partial dependence plots in random forests (randomForest package). It is apparently possible using the rgl package (http://esapubs.org/archive/ecol/E088/173/appendix-C.htm) or there may be a more direct function such as the pairplot() in MART (multiple additive regression trees)? Many

Hi, I have been trying to plot density plots using the example on: http://addictedtor.free.fr/graphiques/graphcode.php?graph=139 I used to use this function, but I cannot get any old code or even the example to work. library("geneplotter") require("RColorBrewer") x1 <- matrix(rnorm(1e4), ncol=2) x2 <- matrix(rnorm(1e4, mean=3, sd=1.5), ncol=2) x <-

Hello, I have posted this mail a few days ago but I did it wrong, I hope is right now: I have the following doubts related with lm.ridge, from MASS package. To show the problem using the Longley example, I have the following doubts: First: I think coefficients from lm(Employed~.,data=longley) should be equal coefficients from lm.ridge(Employed~.,data=longley, lambda=0) why it does not happen?

Hi there I'm trying to construct a model of mortality risk in 2D space that requires numerical integration of a hazard function, for which I'm using the integrate function. I'm occasionally encountering parameter combinations that cause integrate to terminate with error "Error in integrate... the integral is probably divergent", which I'm not sure how to interpret. The

I want to apply Nyblom-Hansen test with the strucchange package, but I don't know how is the correct way and what is the difference between the following two approaches (leeding to different results): data("longley") # 1. Approach: sctest(Employed ~ Year + GNP.deflator + GNP + Armed.Forces, data = longley, type = "Nyblom-Hansen") #results in: # Score-based CUSUM

Dear r-devel list members, On a couple of occasions I've encountered the issue illustrated by the following examples: --------- snip ----------- > mod.1 <- lm(Employed ~ GNP.deflator + GNP + Unemployed + + Armed.Forces + Population + Year, data=longley) > mod.2 <- update(mod.1, . ~ . - Year + Year) > all.equal(mod.1, mod.2) [1] TRUE > > f <-

Dear all, I would like to suggest changing the pairs.formula command such that a command like pairs(GNP ~ . - Year - GNP.deflator, longley) would behave in a similar fashion as lm(GNP ~ . - Year - GNP.deflator, longley) i.e., make a pairwise scatterplot of GNP and all other variables in the (longley) dataframe except for Year and GNP.deflator. The above command, with the

Thanks to Andy and Thomas, Reading help(hclust) more carefully would have done it but sometimes you do not see the wood for the trees... So hc$merge does exactly what I want. I have never been aware of the command str to get the structure of an R-object. It seems pretty useful to me. Thanks, Arne > -----Original Message----- > From: Liaw, Andy [mailto:andy_liaw at merck.com] >

Dear R-users, I tried to receive the merge details of a clustering by using the summary function of hclust. For illustration I use the Longley data as done by Prof Ripley (Wed 11 Apr 2001) d <- dist(longley.y) d <- d/max(d) hc <- hclust(d, "ave") But instead of getting a matrix for $merge I get: >summary(hc) Length Class Mode merge 30 -none- numeric

Douglas Bates wrote: >From 0.62 onward you should not have to create a symbolic link in >/usr/local/bin. It should be that you can run > cd $RSOURCE > ./configure --prefix=/usr/local > make install >and you will end up with the R script installed in /usr/local/bin and >all the files needed to run R in /usr/local/lib/R. > >Can you tell what the prefix is set to after

Dear Terry, Even the behaviour of lm() and glm() isn't entirely consistent. In both cases, singularity results in NA coefficients by default, and these are reported in the model summary and coefficient vector, but not in the coefficient covariance matrix: ---------------- > mod.lm <- lm(Employed ~ GNP + Population + I(GNP + Population), + data=longley) >

Dear all I'm familiarising myself with Ridge Regressions in R and the following is bugging me: How does one get p-values for the coefficients obtained from MASS::lm.ridge() output (for a given lambda)? Consider the example below (adapted from PRA [1]): > require(MASS) > data(longley) > gr <- lm.ridge(Employed ~ .,longley,lambda = seq(0,0.1,0.001)) > plot(gr) > select(gr)

Dear list members, This is probably a naive question (from a new user), but I'm having problems getting column names to be recognised in loop that is supposed to apply a set of calculations in turn to each column of a dataframe. The dataframe has 33 columns of output from a simulation. The columns each have a text name and I want to apply the same set of calculations to each column in turn.

>>>>> Fox, John <jfox at mcmaster.ca> >>>>> on Thu, 14 Sep 2017 13:46:44 +0000 writes: > Dear Martin, I made three points which likely got lost > because of the way I presented them: > (1) Singularity is an unusual situation and should be made > more prominent. It typically reflects a problem with the > data or the

>>>>> Martin Maechler <maechler at stat.math.ethz.ch> >>>>> on Thu, 14 Sep 2017 10:13:02 +0200 writes: >>>>> Fox, John <jfox at mcmaster.ca> >>>>> on Wed, 13 Sep 2017 22:45:07 +0000 writes: >> Dear Terry, >> Even the behaviour of lm() and glm() isn't entirely consistent. In both cases,

I am trying to set up R-0.62.1 on an alphastation running digital unix 3.2D. I had R-0.61.1 working fine, but have had a number of problems getting 0.62.1 going.... The next problem is that plotting to any device causes R to hang. Looking at processor usage it would appear that it gets stuck in an infinite loop rather than pausing for input. I have managed to track the problem down to plot.new().

I was wondering if someone could help me with a problem getting locfit running. I am running Digital Unix 3.2D on an alphastation, with R 0.61.1 I downloaded the CRAN distribution of locfit. I had no problems installing it with R INSTALL (I installed it into my library rather than the default base library; the output is attached at the end of this email message). I then tried to test it with the

Hi all, I just cannot think of how to do it: I want to take the first variable (column) of a data frame and regress it against all other variables. bla <- function (dat) { reg <- lm(whateverthefirstofthevariablenamesis ~., data=dat) return(reg) } What kind of function do I have to take instead of the whateverthefirstofthevariablenamesis, eval(), substitute(), get(), ... to

> 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"

Hello dear R help group, I have the following setup to analyse: We have about 150 subjects, and for each subject we performed a pair of tests (under different conditions) 18 times. The 18 different conditions of the test are complementary, in such a way so that if we where to average over the tests (for each subject), we would get no correlation between the tests (between subjects). What we wish