similar to: Measuring dispersion

Hi, Since "R" is a (very) generic name, I've been having some trouble searching the web for this topic. Due to this, I've just created a Google Custom Search Engine that includes several of the most relevant sites that have information on R. See it in action at: http://google.com/coop/cse?cx=018133866098353049407%3Aozv9awtetwy This is really a preliminary test. Feel free to

Hi, I'm having a weird result with the length() function: >a [... omited ...] [9994] NA "2003-12-03 16:37:00" "2002-06-26 18:43:00" [9997] "2005-07-04 04:00:00" "2007-02-16 22:09:00" "2007-02-24 15:49:00" [10000] NA > length(LastModified) [1] 9 > length(c(LastModified)) [1] 9 I was expecting to get

Hi, I have several files with data in this format: 20070102 20070102 20070106 20070201 ... The data is sorted and each line represents a date (YYYYMMDD). I would like to analyze this data using R. For instance, I would like to have a histogram by year, month or day. I've already made a simple Perl script that aggregates this data but I believe that R can be much more powerful and easy on

Hi, I'm trying to aggregate date values using the aggregate function. For example: aggregate(data,by=list(weekdays(LM),months(LM)),FUN=length) I would also like to aggregate by year but there seems to be no years() function. Should there be one? Is there any alternative choice? Also, a hours() function would be great. Any tip on this? Thanks in advance! S?rgio Nunes

Hi, What's the best way to average dates? I though mean.POISXct would work fine but... > a [1] "2007-04-02 19:22:00 WEST" > b [1] "2007-03-17 16:23:00 WET" > class(a) [1] "POSIXt" "POSIXct" > class(b) [1] "POSIXt" "POSIXct" > mean(a,b) [1] "2007-04-02 19:22:00 WEST" > mean(b,a) [1] "2007-03-17

Hi, I'm trying to draw a 2D plot using multiple tints of red. The (simplified) setup is the following: || year | x | y || My idea is that each year is plotted with a different tint of red. Older year (lightest) -> Later year (darkest). I've managed to plot this with different scales of grays simply by doing: palette(gray(length(years):0/length(years))) before the plot and for each

Hi, I'm pretty new to R and I've been having some problems filtering data in matrices. I have the following initial dataset: || year | name | varA || I have multiple values for "varA" for the same "year" and the same "name". Having this as the input I would like to obtain the following: || year | name | {varA mean} || Where I only have one line for each

Just for the record, here are my steps for producing a date based histogram. Data is stored in a file where each line only has a date - 2007/02/16 >d<-readLines("filename.dat") >d<-as.Date(d, format="%Y/%m/%d") >pdf(yearly.pdf) >hist(d, "years") >dev.off() Instead of "years" you can also use "days", "weeks",

Hi, It seems that hist() has a buggy behavior when breaking over "days". The bug can be reproduced in a few steps: > d=data.frame(date=c("2009-01-01", "2009-01-02", "2009-01-02")) > d$date=as.Date(d$date) > d$date [1] "2009-01-01" "2009-01-02" "2009-01-02" > h=hist(d$date, "days") > h$count [1] 3

Dear R-users, I am comparing differences in variance, skew, and kurtosis between two groups. For variance the comparison is easy: just var.test(group1, group2) I am using agostino.test() for skew, and anscombe.test() for kurtosis. However, I can't find an equivalent of the F.test or Mood.test for comparing kurtosis or skewness between two samples. Would the test just be a 1 df test on

Hello I do not know very much about statistics (and English language too :-( ), then I come in search of a clarification (explanation): I found two distinct results on KURTOSIS and I do not know which of them is the correct one. Any aid will be welcome! klebyn ################ CODE rnorm(1000) -> x library(moments) kurtosis(x) skewness(x) detach("package:moments")

Hi, I'm a newcomer to R, having previously used SPSS. One problem I have run into is computing kurtosis. A test dataset is here: http://www.whinlatter.ukfsn.org/2401.dat > library(moments) > data <- read.table("2401.dat", header=T) > attach(data) > loglen <- log10(Length) With SPSS, I get Skewness -0.320 Kurtosis -1.138 With R: > skewness(loglen) [1]

I wonder whether the functions for skewness and kurtosis in the e1071 package are based on correct formulas. The functions in the package e1071 are: # -------------------------------------------- skewness <- function (x, na.rm = FALSE) { if (na.rm) x <- x[!is.na(x)] sum((x - mean(x))^3)/(length(x) * sd(x)^3) } # -------------------------------------------- and #

Greetings Everybody: I generated a 1.2MB dta file based on the general social survey with Stata8 for linux. The file can be re-opened with Stata, but when I bring it into R, it says all the values are missing for most of the variables. This dataset is called "morgen.dta" and I dropped a copy online in case you are interested http://www.ku.edu/~pauljohn/R/morgen.dta looks like this

Hi, I have the following function: > kurtosis <-function(x) (mean((x-mean(x))^4))/(sd(x)^4) #x is a vector and data > print(mydata) V1 V2 V3 V4 V5 1 1007_s_at DDR1 2865.1 2901.3 1978.3 2 1053_at RFC2 103.6 81.6 108.0 3

Dear R users, I hope to explain the concepts of skewness and kurtosis by generating series of distributions with same skewness and different kurtosis or with same kurtosis and different skewness, but it seems that i cannot find the right functions. I have searched the mailing list, but no answers were found. Is it possible to do that in R? Which function could be used? Thanks a lot. --

hello R, This is about simulation in psychomtrics in IRT in R. I want to simulate b parameters(item difficulty) with moments of fixed values of mean, st.d, skewness and kurtosis. Is there any specific IRT package in R with those functions to control those moments? I have seen other programs that can control mean and st.d but not skewness and kurtosis. Thank you, helen L [[alternative HTML

I have a question about the D'Agostino skewness test and the Anscombe-Glynn kurtosis test. agostino.test(x, alternative = c("two.sided", "less", "greater")) anscombe.test(x, alternative = c("two.sided", "less", "greater")) The option "alternative" in those two functions seems to be the null hypothesis. In the output, the

Folks, I'm a bit puzzled by the fact that if I generate 100,000 standard normal variates using rnorm() and perform the Jarque-Bera on the resulting vector, I get p-values that vary drastically from run to run. Is this expected? Surely the p-val should be close to 1 for each test? Are 100,000 variates sufficient for this test? Or is it that rnorm() is not a robust random number generator?

Hello all, I have used describeBy to generate the following summary statistics. I simply need x and y error bars on a plot that has CQN (xaxis) and Price (yaxis). There should be four total points on the graph (one for each supplier). Using "error.crosses(desc$CQN, desc$Price)" does not work. group: a vars n mean sd median trimmed mad min max range skew CQN