similar to: mean (PR#10864)

Dear,?? The script'formula("x")' loops infinitely. More specifically, it throws thefollowing error:? ? Error: evaluationnested too deeply: infinite recursion / options(expressions=)?? ? As a side effect,this makes an IDE like RStudio to crash. On the other hand,the script 'formula("y")'?works as expected : if ?y? does notexist in the global environment,

Full_Name: Version: 2.9.0 OS: windows, linux Submission from: (NULL) (128.231.21.125) In NEWS, it says "median.default() was altered in 2.8.1 to use sum() rather than mean(), although it was still documented to use mean(). This caused problems for POSIXt objects, for which mean() but not sum() makes sense, so the change has been reverted." But it's not reverted yet.

Folks: I know this may be overreaching, but are we missing what's important? WHY do the zeros occur? Are they values less then a known or unknown LOD? -- and/or is there positive mass on zero? In either case, using logs to calculate a geometric mean may not make sense. Paraphrasing Greg Snow, what is the scientific question? What is the model? Cheers, Bert On Mon, Jan 17, 2011 at 9:13 AM,

Dear R-helpers, I am dealing with an Oracle database (using package RODBC). I use R in order to transform some Oracle tables into .txt files (using function sqlFetch from package RODBC and then function write.table). However, I cannot do it without creating an R object, which is rather restrictive for very big Oracle tables. Indeed, any R Object is stored into RAM, which can be of limited

By considering this reproducible example #start code library(lattice) dotplot(variety ~ yield | site, data = barley, layout = c(1,6), index.cond= function(x,y){median(x)}, panel = function(x,y,...) { panel.dotplot(x,y,...) median.values <- median(x) panel.abline(v=median.values, col.line="red")

Full_Name: Richard Huggins Version: 1.7.1 OS: windows 2000 Submission from: (NULL) (131.172.4.44) > x<-rnorm(100,2,1) > mean(x) [1] 1.73299 > summary(fivenum(x)) Min. 1st Qu. Median Mean 3rd Qu. Max. -0.3655 1.1070 1.7430 1.7320 2.3840 3.7910 > summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max. -0.3655 1.1070 1.7430 1.7330 2.3830 3.7910 >

Okay, so I have a homework projecr for R, and we had to input the following link as some sort of data: nb10 <- read.table("http://www.adjoint-functors.net/su/web/314/R/NB10"). Afterwards, we have to use fivenum(nb10) to find max, min, quantiles, and sd, but I'm okay with this. The next question is where I'm stuck. The question is as follows; Compare the median (use the

Hello, following functions doesnt work correct with my data: median(), geo.mean(). My datafiles contain more than 10.000 lines and six columns from a flow-cytometer-measurment. I need the arithmetic and geometric mean and median. For the calculation of the geometric mean i wrote following function: fix(geo.mean) function(x) { n<-length(x)

Given this example mean.values<-colMeans(VADeaths) mean.values<-apply(VADeaths, 2, mean) median.values<-apply(VADeaths, 2, median) dotchart(VADeaths, gdata=mean.values) dotchart(VADeaths, gdata=median.values) is it possible to ?combine? a single dotchart showing both the mean and the median for each single group (with different plotting symbols)? ?is it that possible with the use of

Hi together, below is a small example which produces outcome I do not understand, namely that the median function works fine on a data.frame without negative numbers, but doesn't work on a data.frame with one negative number. I'm sure there is a reasonable explanation for that or better, that I'm doing something wrong and someone could guide me how to solve it. I tried googling it,

Hi! Suppose I have a dataset as follows pd = c(10,7,10,11,7,11,7,6,8,3,12,7,7,10,10) I wish to calculate the mean, standard deviation, median, skewness and kurtosis i.e. regular standard statistical measures. average = mean(pd) stdev    = sd(pd) median = median(pd) skew    = skewness(pd) kurt     =  kurtosis(pd) Q. No (1) How do I get these at a stretch using some R package? I came across

I think median() behaves as designed: As long as the argument can be ordered, the "middle observation" makes sense, except when the middle falls between two categories, and you can't define and average of the two candidates for a median. The "sick man" would seem to be var(). Notice that it is also inconsistent with cov(): >

The data sets I am working with all have a weight variable--e.g., each row doesn't mean 1 observation. With that in mind, nearly all of the graphs and summary statistics are incorrect for my data, because they don't take into account the weight. **** For example "median" is incorrect, as the quantiles aren't calculated with weights: sum( weights[X < median(X)] )

R gurus, On Aug 20, 2002, I asked in R-help about calculating a running 5-day median on a large matrix. Thanks to Martin Maechler <maechler@stat.math.ethz.ch> and Ray Brownrigg <Ray.Brownrigg@mcs.vuw.ac.nz> for responding. I ended up writing C code (and an R interface) to do it, which is about 1000x faster than the naive method! (72s became .09s on a 223 x 520 matrix). I added a

Full_Name: Kevin Wright Version: 1.4 OS: Windows 95 Submission from: (NULL) (170.54.59.160) Note: This was the 1.4 build for Windows that Brian Ripley made available. The first example in the help for reshape doesn't work for me. When I cut and paste, this is what happens: > data(Indometh,package="nls") > summary(Indometh) Subject time conc

R-2.1.1 on windows XP I just noticed something unpleasant when using bwplot (from lattice). In order to satisfy a wish from a client, I needed to produce sets of boxplots conditioned by another factor. My client didn't like the look of the boxplots (by default, they have a star to mark the median, instead of the commonly used line). I told him "no problem" dumped panel.bwplot,

I found a strange inconsistency: If m is a matrix and d is a data frame then - mean(m), median(m), max(m) and max(d) all return a single value but - mean(d) returns the column means - median(d) fails - both var(m) and var(d) return the variance covariance matrix You pretty much have to experiment to figure this out since much of this behavior is not readily obvious from the help files.

Hi! I have a data set of 86 values that are non-normally distributed (counts). The median value is 10. I want to get an estimate of the 95% confidence interval for this median value. I tried to use a one-sample Wiolcoxin test: wilcox.test(Comps,mu=10,conf.int=TRUE) and got the following output: Wilcoxon signed rank test with continuity correction data: Comps V = 2111, p-value = 0.05846

Full_Name: Sanford Weisberg Version: 1.9.1 OS: Win XP Submission from: (NULL) (160.94.148.2) > wm <- read.table(url("http://www.stat.umn.edu/~sandy/wmdata0.txt"), header=TRUE) > mean(wm$Spd1) [1] 7.7773 > summary(wm$Spd1) Min. 1st Qu. Median Mean 3rd Qu. Max. 0.222 4.780 7.550 7.780 10.200 21.600 The mean of this variable DOES NOT ROUND to the value

what does this mean in R-1.8.1 release notes? o median() no longer `works' for odd-length factor variables.