similar to: Correlation of ranks of labels?

I'm sure this is a simple task, but how to do it has escaped me. I have imported data from two separate files (each file contains the results from an information retrieval algorithm) organized into a list. They are organized by File,Query, and Rank (in that order): [[1]] Doc Query Rank 5 1 1 9 1 2 7 1 3 5 2 1 7 2 2 9 2 3 [[2]]

Dear R groups: I have the data as follows, I want to plot the "Rank1 ~ obs30*Cases" and "Rank2 ~ obs30*Cases" on the same plot as one 3-D scatter plot, how to do that? Any help is highly appreciated. ID obs30 Cases RANK1 RANK2 1 0.03175 63 82 81 2 0.00000 34 1 34 3 0.00000 36 2 41 4 0.00000 54 3 26 5 0.00000 22 4 42 6 0.00746 134 39 32 7 0.00000 2 5 53 8 0.01190 168 46 31

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",

Full_Name: Dirk Koschuetzki Version: 2.1.1 OS: source code Submission from: (NULL) (194.94.136.34) Hello, >From the source code (R-2.1.1, file: .../R-2.1.1/src/library/stats/R/) ****************************** cor.test.default <- function(x, y, alternative = c("two.sided", "less", "greater"), method = c("pearson", "kendall",

hi, i have 2 lists of ranks for which i'd like to compute kendall tau. there are ties in the ranks which (to the best of my knowledge) means i cant use tau a but rather b or c. how does R handle that? are ties automatically detected (using corr.test()) and is tau b/c computed instead of tau a? also kendall does not work when values in list 1 do not occur in list 2 (and vice versa) - how does

Hi R-users I have a data set. There are 10 products and the numbers of people who ranked the products. The format of the data set is productID rank1 rank2 rank3 rank4 rank5 rank6 rank7 rank8 rank9 rank10 ------------------------------------------------------------------------------------------------------- 1 10 2 3 3 6 4 2 5

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, I'm looking for a function to measure the dispersion of a set of values ranging from 0 to 1. This function should be 0 if all the values are evenly spaced within the interval and it should be > 0 if values are clustered. The more clustered the values are, the higher should the function be. An example: [0; 0.2; 0.4; 0.6; 0.8; 1] - function should be ~ 0 [0; 0.1; 0.1; 0.15; 1] -

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 all, I would like to ask a question related to Kendall package. I ran Kendall (x,y) and saw the results. But I am not sure which tau values R reported. I have ties in my data set, so I want tau-b. Can anybody tell how Kendall package is calculating tau values? I have looked at the package PDF, but I could not find any useful information. As long as I see from the following link, there

Colleagues, I ran some nonparametric regressions in R (run in RedHat Linux), then a colleague repeated the analyses in SAS. When we obtained different results, I tested S-Plus (same Linux box). And, got yet different results. I replicated the results with a small dataset: DATA: 37.5 23 37.5 13 25 16 25 12 100 15 12.5 19 50 20 100 13 100 10 100 10 100 16 50 10 87.5

Hi all, I am trying to write a script that will compute Kendall's tau for a 75 time series (using the Kendall package) and will then write the tau and p values from the Kendall test to a text file table that can be read into Excel. I am having no problem calculating Kendall's tau and the associated p value for each time series, but I am having trouble figuring out how aggregate the

> library(Kendall) > Kendall(1:3,1:3) WARNING: Error exit, tauk2. IFAULT = 12 <<<<<< tau = 1, 2-sided pvalue =1 I believe Kendall tau is well-defined for this case and the reported value is correct; isn't it a bug to give a warning? (And if, e.g., the pvalue is not well-defined in this case, wouldn't it be better to return NA or NaN or something?) Also,

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

Dear R developers, Currently many (all?) test functions in R describe the alternative hypothesis, but not the the null hypothesis being tested. For example, cor.test: > require(boot) > data(mtcars) > with(mtcars, cor.test(mpg, wt, met="kendall")) Kendall's rank correlation tau data: mpg and wt z = -5.7981, p-value = 0.000000006706 alternative hypothesis: true tau is not

Hi, I recently ran into the problem that I needed a Siegel-Tukey test for equal variability based on ranks. Maybe there is a package that has it implemented, but I could not find it. So I programmed an R function to do it. The Siegel-Tukey test requires to recode the ranks so that they express variability rather than ascending order. This is essentially what the code further below does. After the