similar to: Inconsistency among mean, median, max, var

I have found a problem in R version 1.1.1 when using apply with the median function. The problem can be illustrated with the following data matrix: X1 X2 X3 1 2 3 4 5 6 7 8 NA Enter this data matrix as X and then try apply(X,2,median,na.rm=T) The problem here is that the median function returns a named scalar if the number of observations is odd, but returns an

Here are some data frames: df3.2 <- data.frame(1:3, 7:9) df4.2 <- data.frame(1:4, 7:10) df3.3 <- data.frame(1:3, 7:9, 10:12) df4.3 <- data.frame(1:4, 7:10, 10:13) df3.4 <- data.frame(1:3, 7:9, 10:12, 15:17) df4.4 <- data.frame(1:4, 7:10, 10:13, 15:18) Now here are some commands and their answers: > median(df3.2) [1] 2 8 > median(df4.2) [1] 2.5 8.5 > median(df3.3)

This inconsistency recently came to my attention: > df <- data.frame(A = 1:10, B = rnorm(10)) > min(df) [1] -1.768958 > max(df) [1] 10 > mean(df) [1] NA Warning message: In mean.default(df) : argument is not numeric or logical: returning NA I recall the times where `mean(df)` would give `colMeans(df)` and this behaviour was deemed inconsistent. It seems though that the change has

Hello, I'm having the following questionable behavior: > summary(m) Min. 1st Qu. Median Mean 3rd Qu. Max. 1 13000 26280 25890 38550 50910 > max(m) [1] 50912 > typeof(m) [1] "integer" > class(m) [1] "integer" ...it seems to me like max() and summary(m)[6] ought to return the same number. Am I doing something wrong? I'm

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")

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, I try to transform a data frame like A B C 1 1 2,5 2 2 NA 3 1 1,0 4 1 56 5 2 23 6 1 NA 7 2 46 to the following dataframe, calculating the variance and median of the C-column, group by B, so the result will be: B C(median) D (var) 1 19,83

Hello, all, I have a dataframe of three rows and umpteen columns. I want to show the maximum, minimum, and median with a vertical line and a central dot (I'd use a boxplot, but with only three data points, that's overkill; I can't just use points, because of overlap and some of the other data plotted on the graph). This works: > boxplot(data_frame,

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)

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

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 All, I need to calculate the median for even number of data points.However instead of calculating the arithmetic mean of the two middle values,I need to calculate their geometric mean. Though I can code this in R, possibly in a few lines, but wondering if there is already some built in function. Can somebody give a hint? Thanks in advance [[alternative HTML version deleted]]

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,

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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 List, I have worked through the examples given in the help on boxplot(). If I am reading it right, only the median is produced (as a default), which is represented by the horizontal line inside the box. What argument do I need to specify if I want to show the mean as well? Thanks, platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status

Dear all, I'm trying to solve the following problem, and hoping to get some advise here from the group I have a dataframe in which the same sample was measured more than once on the same day. I would reorganize the dataframe to get a single value (mean /median) for one day Patient Day Sample Test A 1 A 23 A 1 A 36 A 5 B 44 A 5 B 23 B 2 C 10 B 2 C 5 mean Patient Day Sample Test.mean A

Dear Team - I would like to get your help on adding the values of mean and median of RTF as mentioned in the below snapshot.? Please guide me out.?Thanks for your help in advance.

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: 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.