Re: [R] For logical i and length(i) > length(x), x[i] <- value makes length(x) == length(i)
Splus 8.2.0 (c. 2010) and Splus6.0 (c. 2001) act like R in this respect. x <- c(10.0, 11.0, 12.0) i <- c(FALSE,TRUE,FALSE,FALSE,TRUE) x[i] #[1] 11 NA x[i] <- 1:2 x #[1] 10 1 12 NA 2 I no longer have access to a running version of Splus 3.4, but it does not look like a feature we would have changed. Bill Dunlap TIBCO Software wdunlap tibco.com On Sat, Sep 12, 2015 at 9:17 AM, Suharto Anggono Suharto Anggono via R-help wrote: > I'll make my points clearer. > > - The case of logical index vector longer than the original vector is not > usual case for me. As I said, I encountered it in the code of function 'rank' > in R. > > - Before, I have read "S-PLUS help" (version 3.4, it seems), > http://www.uni-muenster.de/ZIV.BennoSueselbeck/s-html/helpfiles/Subscript.html. > The following is the relevant part. There are missing pieces there that I > fill by guessing. > > Vector subscripts are generated with when i and x are both vectors. () > The result of the expression is to extract or replace elements of x > corresponding to a vector of positive indices computed according to the value > of i. > > If i is logical the indices are produced by starting at 1 and selecting > the numbers for which the corresponding element is T. If is shorter than it > is extended by cyclic repetition. It can be longer than as well, with no > change in the computation of indices. > > > > For replacements, x[i] <- value the rule is that the length of will be set to > the largest value in the indices, if that is bigger than the current length > of x. > > > >From it, I infer that, if i is logical and length(i) >= length(x), x[i] <- > >value has the same effect to x[which(i)] <- value, where which(i) takes > >indices where i is T. > > - In R, if i is logical and length(i) > length(x), length(x) after x[i] <- > value may be different from after x[which(i)] <- value. > > - So, I wonder if R inherits the behavior from S or not. > > - The behavior is not clearly documented in R. I just find "R Language > Definition", "3.4.1 Indexing by vectors", that can be interpreted to imply > the behavior. > > - However, for a particular case, function 'rank' in R relies on the behavior. > > However, it seems that relying on the behavior is not on purpose. Previously, > at least until R 3.1.3, the code of function 'rank' has the following before > yy <- NA . > yy <- integer(length(x)) > storage.mode(yy) <- storage.mode(y) > > It seems that yy[] <- NA is what is intended. > > - However, for me, the behavior is plausible. The assumption is that indices > from 1 to length(i) exist. > > -- > > I think this behavior is consistent with typical indexing behaviour in R... I > would ask you what result you thought you should get? I, for one, can think > of all sorts of uses for numeric indexes that have different lengths than the > vector, but am stumped to think of any use for what you are proposing. > --- > Jeff NewmillerThe . . Go Live... > DCN:<[hidden email]>Basics: ##.#. ##.#. Live Go... > Live: OO#.. Dead: OO#.. Playing > Research Engineer (Solar/BatteriesO.O#. #.O#. with > /Software/Embedded Controllers) .OO#. .OO#. rocks...1k > --- > Sent from my phone. Please excuse my brevity. > > On September 5, 2015 10:02:05 AM PDT, Suharto Anggono Suharto Anggono via > R-help <[hidden email]> wrote: > >>I came across this behavior when I followed the code of function 'rank' >>in R. >> >>It seems that subassignment of a vector by a logical index vector that >>is longer than the original vector always results in expanding the >>original vector to the length of the index vector. >> >>The resulting length may be different from the result of subassignment >>by the equivalent numeric vector. For subassignment of a vector by a >>numeric index vector, the original vector is expanded to the maximum >>index, if it is larger than the length of the original vector. >> >>This is an example. >> >>> x <- NA >>> x[c(FALSE,TRUE,FALSE)] <- 1 >>> x >>[1] NA 1 NA >> >>Compare to this. >> >>> x <- NA >>> x[which(c(FALSE,TRUE,FALSE))] <- 1 >>> x >>[1] NA 1 >> >>Does S exhibit the same behavior? >> >>Currently, if there is NA and na.last = "keep", function 'rank' in R >>relies on this behavior to give correct result length. >> >>In "R Language Definition", "3.4.1 Indexing by vectors" says: "Logical. >>The indexing i should generally have the same length as x. If it >>is longer, then x is conceptually extended with NAs. " The >>statement can be taught to support the observed behavior. >> >>> sessionInfo() >>R version 3.2.2 (2015-08-14) >>Platform: i386-w64-mingw32/i386 (32-bit) >>Running
[R] R CRAN check works great on local Mac OS, but failed on Windows (win-builder web). Why?
I have my R package checked on local Mac OS via R CMD check --as-cran. Everything works great: no errors, no warnings, just 1 note about first submission, all test files passed! However, when I submit my .tar.gz package file to the http://win-builder.r-project.org to check on Windows. Some errors occurred when running the examples. The main issue is related to the object created by Matrix. All errors indicated *invalid class "dgCMatrix" object: Dimnames[1] is not a character vector. * Detailed check log is attached below. I checked the examples on local machine line-by-line, but cannot identify the problem. Is the error caused by that maybe the Matrix package was written differently for Mac OS and Windows platforms? /* checking examples ... ** running examples for arch 'i386' ... ERROR Running examples in 'grpregOverlap-Ex.R' failed The error most likely occurred in: > base::assign(".ptime", proc.time(), pos = "CheckExEnv") > ### Name: cv.grpregOverlap > ### Title: Cross-validation for choosing regularization parameter lambda > ### Aliases: cv.grpregOverlap > > ### ** Examples > > ## linear regression, a simulation demo. > set.seed(123) > group <- list(gr1 = c(1, 2, 3), + gr2 = c(1, 4), + gr3 = c(2, 4, 5), + gr4 = c(3, 5), + gr5 = c(6)) > beta.latent.T <- c(5, 5, 5, 0, 0, 0, 0, 0, 5, 5, 0) # true latent > coefficients. > # beta.T <- c(5, 5, 10, 0, 5, 0), true variables: 1, 2, 3, 5; true groups: > 1, 4. > X <- matrix(rnorm(n = 6*100), ncol = 6) > X.latent <- expandX(X, group) Error in validObject(.Object) : invalid class "dgCMatrix" object: Dimnames[1] is not a character vector Calls: expandX ... initialize -> callNextMethod -> .nextMethod -> validObject Execution halted/ *I also attached source code for the two related functions.* /expandX <- function(X, group) { incidence.mat <- incidenceMatrix(X, group) # group membership incidence matrix over.mat <- Matrix(incidence.mat %*% t(incidence.mat), sparse = TRUE, dimnames = dimnames(incidence.mat)) # overlap matrix grp.vec <- rep(1:nrow(over.mat), times = diag(over.mat)) # group index vector # expand X to X.latent X.latent <- NULL names <- NULL ## the following code will automatically remove variables not included in 'group' for(i in 1:nrow(incidence.mat)) { idx <- incidence.mat[i,]==1 X.latent <- cbind(X.latent, X[, idx, drop=FALSE]) names <- c(names, colnames(incidence.mat)[idx]) } colnames(X.latent) <- paste('grp', grp.vec, '_', names, sep = "") X.latent } incidenceMatrix <- function(X, group) { n <- nrow(X) p <- ncol(X) if (class(group) != 'list') { stop("Argument 'group' must be a list of integer indices or character names of variables!") } J <- length(group) grp.mat <- Matrix(0, nrow = J, ncol = p, dimnames=list(rep(NA, J), rep(NA, p))) if(is.null(colnames(X))) { colnames(X) <- paste("V", 1:ncol(X), sep="") } if (is.null(names(group))) { names(group) <- paste("grp", 1:J, sep="") } if (class(group[[1]]) == 'numeric') { for (i in 1:J) { ind <- group[[i]] grp.mat[i, ind] <- 1 colnames(grp.mat)[ind] <- colnames(X)[ind] } } else { ## character, names of variables for (i in 1:J) { grp.i <- as.character(group[[i]]) ind <- colnames(X) %in% grp.i grp.mat[i, ] <- 1*ind colnames(grp.mat)[ind] <- colnames(X)[ind] } } rownames(grp.mat) <- as.character(names(group)) # check grp.mat if (all(grp.mat == 0)) { stop("The names of variables in X don't match with names in group!") } grp.mat } / -- View this message in context: http://r.789695.n4.nabble.com/R-CRAN-check-works-great-on-local-Mac-OS-but-failed-on-Windows-win-builder-web-Why-tp4712179.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] For logical i and length(i) > length(x), x[i] <- value makes length(x) == length(i)
I'll make my points clearer. - The case of logical index vector longer than the original vector is not usual case for me. As I said, I encountered it in the code of function 'rank' in R. - Before, I have read "S-PLUS help" (version 3.4, it seems), http://www.uni-muenster.de/ZIV.BennoSueselbeck/s-html/helpfiles/Subscript.html. The following is the relevant part. There are missing pieces there that I fill by guessing. Vector subscripts are generated with when i and x are both vectors. () The result of the expression is to extract or replace elements of x corresponding to a vector of positive indices computed according to the value of i. If i is logical the indices are produced by starting at 1 and selecting the numbers for which the corresponding element is T. If is shorter than it is extended by cyclic repetition. It can be longer than as well, with no change in the computation of indices. For replacements, x[i] <- value the rule is that the length of will be set to the largest value in the indices, if that is bigger than the current length of x. >From it, I infer that, if i is logical and length(i) >= length(x), x[i] <- >value has the same effect to x[which(i)] <- value, where which(i) takes >indices where i is T. - In R, if i is logical and length(i) > length(x), length(x) after x[i] <- value may be different from after x[which(i)] <- value. - So, I wonder if R inherits the behavior from S or not. - The behavior is not clearly documented in R. I just find "R Language Definition", "3.4.1 Indexing by vectors", that can be interpreted to imply the behavior. - However, for a particular case, function 'rank' in R relies on the behavior. However, it seems that relying on the behavior is not on purpose. Previously, at least until R 3.1.3, the code of function 'rank' has the following before yy <- NA . yy <- integer(length(x)) storage.mode(yy) <- storage.mode(y) It seems that yy[] <- NA is what is intended. - However, for me, the behavior is plausible. The assumption is that indices from 1 to length(i) exist. -- I think this behavior is consistent with typical indexing behaviour in R... I would ask you what result you thought you should get? I, for one, can think of all sorts of uses for numeric indexes that have different lengths than the vector, but am stumped to think of any use for what you are proposing. --- Jeff NewmillerThe . . Go Live... DCN:<[hidden email]>Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. On September 5, 2015 10:02:05 AM PDT, Suharto Anggono Suharto Anggono via R-help <[hidden email]> wrote: >I came across this behavior when I followed the code of function 'rank' >in R. > >It seems that subassignment of a vector by a logical index vector that >is longer than the original vector always results in expanding the >original vector to the length of the index vector. > >The resulting length may be different from the result of subassignment >by the equivalent numeric vector. For subassignment of a vector by a >numeric index vector, the original vector is expanded to the maximum >index, if it is larger than the length of the original vector. > >This is an example. > >> x <- NA >> x[c(FALSE,TRUE,FALSE)] <- 1 >> x >[1] NA 1 NA > >Compare to this. > >> x <- NA >> x[which(c(FALSE,TRUE,FALSE))] <- 1 >> x >[1] NA 1 > >Does S exhibit the same behavior? > >Currently, if there is NA and na.last = "keep", function 'rank' in R >relies on this behavior to give correct result length. > >In "R Language Definition", "3.4.1 Indexing by vectors" says: "Logical. >The indexing i should generally have the same length as x. If it >is longer, then x is conceptually extended with NAs. " The >statement can be taught to support the observed behavior. > >> sessionInfo() >R version 3.2.2 (2015-08-14) >Platform: i386-w64-mingw32/i386 (32-bit) >Running under: Windows XP (build 2600) Service Pack 2 > >locale: >[1] LC_COLLATE=English_United States.1252 >[2] LC_CTYPE=English_United States.1252 >[3] LC_MONETARY=English_United States.1252 >[4] LC_NUMERIC=C >[5] LC_TIME=English_United States.1252 > >attached base packages: >[1] stats graphics grDevices utils datasets methods base > >__ >[hidden email] mailing list -- To UNSUBSCRIBE and more, see >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide >http://www.R-project.o
Re: [R] Adding a second Y axis on a dotplot
I forgot to put a line about latticeExtra's doubleYScale library(latticeExtra) ? doubleYScale Duncan -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Duncan Mackay Sent: Sunday, 13 September 2015 00:56 To: R Subject: Re: [R] Adding a second Y axis on a dotplot Hi see https://stat.ethz.ch/pipermail/r-help/2007-June/134524.html to get you started. Its toolate or too early here Regards Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England Armidale NSW 2351 Email: home: mac...@northnet.com.au -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of li li Sent: Saturday, 12 September 2015 03:30 To: r-help Subject: [R] Adding a second Y axis on a dotplot Hi all, I plotted a dotplot based on the data below and code below. I would like to add another yaxis on the right with a different col, different tickmarks and a different label. Can anyone give some help?Thanks very much!! Hanna > tmp1 result lot trt trtsymb trtcol 1 98 lot1 trt1 1 blue 2 99 lot2 trt1 1 blue 3 98 lot3 trt1 1 blue 4 100 lot4 trt1 1 blue 5 100 lot5 trt1 1 blue 6 101 lot6 trt1 1 blue 7 101 lot7 trt1 1 blue 8 99 lot8 trt1 1 blue 9 100 lot9 trt1 1 blue 10 94 lot1 trt2 16red 11105 lot2 trt2 16red 12 87 lot3 trt2 16red 13119 lot4 trt2 16red 14 96 lot5 trt2 16red 15113 lot6 trt2 16red 16106 lot7 trt2 16red 17 71 lot8 trt2 16red 18 95 lot9 trt2 16red library(lattice) dotplot(result ~ lot, tmp1, cex=1.1, ylab = "values", xlab="lot", jitter.y = F, aspect=1.0, pch=tmp1$trtsymb, col=tmp1$trtcol, scales=list(rot=30), main="", key = list(text = list(labels = c("trt1", "trt2"),cex=c(0.9,0.9)), points = list(pch =c(1,12), col =c("blue", "red")), space = "right")) __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] maximum-likelihood-estimation with mle()
peter dalgaard gmail.com> writes: > > You are being over-optimistic with your starting values, and/or > with constrains on the parameter space. > Your fit is diverging in sigma for some reason known > only to nonlinear-optimizer gurus... > > For me, it works either to put in an explicit > constraint or to reparametrize with log(sigma). [snip] Another few strategies: set.seed(101) x <- 1:10 y <- 3*x - 1 + rnorm(length(x), mean=0, sd=0.5) library("stats4") nLL <- function(a, b, sigma) { -sum(dnorm(y, mean=a*x+b, sd=sigma, log=TRUE)) } fit <- mle(nLL, start=list(a=0, b=0, sigma=1), nobs=length(y)) Nelder-Mead is generally more robust than BFGS, although slower: fit2 <- mle(nLL, start=list(a=0, b=0, sigma=1), method="Nelder-Mead",nobs=length(y)) bbmle can be used to simplify things slightly (this repeats Peter Dalgaard's transformation of sigma): library("bbmle") fit3 <- mle2(y~dnorm(mean=mu,sd=exp(logsigma)), parameters=list(mu~x), data=data.frame(x,y), start=list(mu=0, logsigma=0)) You can also profile out the sigma: ## dnorm with sd profiled out dnorm2 <- function(x,mean,log=FALSE) { ssq <- sum((x-mean)^2) dnorm(x,mean,sd=sqrt(ssq/length(x)),log=log) } fit4 <- mle2(y~dnorm2(mean=mu), parameters=list(mu~x), data=data.frame(x,y), start=list(mu=0)) __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] removing outlier
On Sep 12, 2015, at 2:32 AM, Juli wrote: > Hi Jim, > > thank you for your help. :) > > My point is, that there are outlier and I don´t really know how to deal with > that. > > I need the dataframe for a regression and read often that only a few outlier > can change your results very much. In addition, regression diacnostics > didn´t indcate me the best results. > Yes, and I know its not the core of statistics to work in a way you get > results you would like to have ;). > > So what is your suggestion? > > And if I remove the outliers, my problem ist, that as you said, they differ > in length. I need the data frame for a regression, so can I remove the whole > column or is there a call to exclude the data? Most regression methods have a 'subset' parameter which would allow you to distort the data to your desired specification. But why not think about examining a different statistical model or using robust methods? That way you can keep all your data. (Sounds like you don't really have a lot.) -- David. > > JULI > > > > -- > View this message in context: > http://r.789695.n4.nabble.com/removing-outlier-tp4712137p4712170.html > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Installation R package Rssa from directory
Many thanks, Jeff! I install package with install.packages('d:\\Rssa\\Rssa_0.13.zip', repos=NULL) -- View this message in context: http://r.789695.n4.nabble.com/Installation-R-package-Rssa-from-directory-tp4712160p4712174.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] removing outlier
Hi Jim, thank you for your help. :) My point is, that there are outlier and I don´t really know how to deal with that. I need the dataframe for a regression and read often that only a few outlier can change your results very much. In addition, regression diacnostics didn´t indcate me the best results. Yes, and I know its not the core of statistics to work in a way you get results you would like to have ;). So what is your suggestion? And if I remove the outliers, my problem ist, that as you said, they differ in length. I need the data frame for a regression, so can I remove the whole column or is there a call to exclude the data? JULI -- View this message in context: http://r.789695.n4.nabble.com/removing-outlier-tp4712137p4712170.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help
hi In cluster analysis K-means , hirarical cluster ,DBSCAN and EM how do we calculate purity [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding a second Y axis on a dotplot
Hi see https://stat.ethz.ch/pipermail/r-help/2007-June/134524.html to get you started. Its toolate or too early here Regards Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England Armidale NSW 2351 Email: home: mac...@northnet.com.au -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of li li Sent: Saturday, 12 September 2015 03:30 To: r-help Subject: [R] Adding a second Y axis on a dotplot Hi all, I plotted a dotplot based on the data below and code below. I would like to add another yaxis on the right with a different col, different tickmarks and a different label. Can anyone give some help?Thanks very much!! Hanna > tmp1 result lot trt trtsymb trtcol 1 98 lot1 trt1 1 blue 2 99 lot2 trt1 1 blue 3 98 lot3 trt1 1 blue 4 100 lot4 trt1 1 blue 5 100 lot5 trt1 1 blue 6 101 lot6 trt1 1 blue 7 101 lot7 trt1 1 blue 8 99 lot8 trt1 1 blue 9 100 lot9 trt1 1 blue 10 94 lot1 trt2 16red 11105 lot2 trt2 16red 12 87 lot3 trt2 16red 13119 lot4 trt2 16red 14 96 lot5 trt2 16red 15113 lot6 trt2 16red 16106 lot7 trt2 16red 17 71 lot8 trt2 16red 18 95 lot9 trt2 16red library(lattice) dotplot(result ~ lot, tmp1, cex=1.1, ylab = "values", xlab="lot", jitter.y = F, aspect=1.0, pch=tmp1$trtsymb, col=tmp1$trtcol, scales=list(rot=30), main="", key = list(text = list(labels = c("trt1", "trt2"),cex=c(0.9,0.9)), points = list(pch =c(1,12), col =c("blue", "red")), space = "right")) __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Wavelets Forecast/Prediction
(internet) search! e.g. on "R package wavelets" There are several. See also the time series task view on CRAN. -- Bert Bert Gunter "Data is not information. Information is not knowledge. And knowledge is certainly not wisdom." -- Clifford Stoll On Fri, Sep 11, 2015 at 10:58 PM, Diego Ubuntu wrote: > Is there any way to forecast/predict a time series using wavelets ? Is > there any package with that functionality included? > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] kendall tau distance
many thanks for replying. I have the vectors running the code ConDisPairs( data.frame(c(1,2,3,4,5),c(3,4,1,2,5)) ) $pi.c [,1] [,2] [1,] 12 0 [2,] 8 1 [3,] 7 3 [4,] 5 6 [5,] 0 10 $pi.d [,1] [,2] [1,] 0 14 [2,] 3 12 [3,] 7 9 [4,] 8 5 [5,] 10 0 $C [1] 69 $D [1] 109 I could not find the result related to the result at wiki page in any way..looking for 4 ? dissimilar pairs? many thanks > From: dcarl...@tamu.edu > To: ragi...@hotmail.com; r-help@r-project.org > Subject: RE: [R] kendall tau distance > Date: Fri, 11 Sep 2015 15:37:19 + > > The Wikipedia article gives a simple formula based on the number of > discordant pairs. You can get that from the ConDisPairs() function in package > DescTools. > > - > David L Carlson > Department of Anthropology > Texas A&M University > College Station, TX 77840-4352 > > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ragia Ibrahim > Sent: Thursday, September 10, 2015 12:40 PM > To: r-help@r-project.org > Subject: [R] kendall tau distance > > Dear group > how to calculate kendall tau distance according to Kendall_tau_distance at > wikipedia > > href="https://en.wikipedia.org/wiki/Kendall_tau_distance" > target="_blank" > class="newlyinsertedlink">https://en.wikipedia.org/wiki/Kendall_tau_distance > > > thanks in advance > Ragia > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.