similar to: Fast Kendall's Tau

I've noticed that the implementation of Kendall's Tau in R is O(N^2). The following reference describes how it can be done in O(N log N): A Computer Method for Calculating Kendall's Tau with Ungrouped Data William R. Knight Journal of the American Statistical Association, Vol. 61, No. 314, Part 1 (Jun., 1966), pp. 436-439 http://www.jstor.org/pss/2282833 I'm interested in

Hello, I can't find a package which calculates Kendall's tau-c. There is the package Kendall, but it only calcuates Kendall's tau-b. Here is the example from ttp://www2.chass.ncsu.edu/garson/pa765/assocordinal.htm. cityriots <- data.frame(citysize=c(1,1,2,2,3,3), riotsize=c(1,2,1,2,1,2), weight=c(4,2,2,3,0,4)) cityriots <- data.frame(lapply(cityriots,function(x)

Hi, I need to calculate Kendall's tau for large data vectors (length > 100'000). Is somebody aware of a faster algorithm or package function than "cor(, method="kendall")"? There are ties in the data to be considered (Kendall's tau-b). Any suggestions? Regards Ferdinand

I discovered that the Kendall's tau calculation in R uses all pairwise comparisons which is O(n^2) and takes a long time for large vectors. I implemented a O(n*log(n)) algorithm based on merge-sort. Is this of interest to be included in core R? The code (fortran and R wrapper) is available in my package clinfun v0.9.7 (not exported in NAMESPACE). Thanks, Venkat -- Venkatraman E. Seshan,

I have 3 variables relating to the successful introductions of species to 95 different areas: introduction frequency; number of successes pre 1906; number of successes post 1906 The data are not normal, nor homo-skedatic, so I am using non-parametric statistics. I have calculated Kendall's tau between both introduction & successes pre 1906 (tau=0.3903) and introduction & successes

Full_Name: Dan Field Version: 1.6.2 OS: N/A Submission from: (NULL) (209.115.168.187) In kendall.c (library is ctest), the limits for the first loop in routine kendall_tau run from 0 through n-1, and the inner loop runs from 0 through i-1. This causes the each pair at index i to be compared with itself; my understanding is that there should only be n*(n-1)/2 pairs under consideration for

Full_Name: Marek Ancukiewicz Version: 2.10.1 OS: Linux Submission from: (NULL) (74.0.49.2) Both cor() and cor.test() incorrectly handle ordered variables with method="kendall", cor() incorrectly handles ordered variables for method="spearman" (method="person" always works correctly, while method="spearman" works for cor.test, but not for cor()). In

A search of the archives did not reveal an answer: For basic tests of association, where one has no a priori knowledge of the form of the relation or of the distributions of the variables, rank correlation seems like a good start. Why is cor.test() with Kendall and Spearman options relegated to the ctest package, rather than in the base package? Does this suggest that the developers consider

Dear I estimate a sample selection model using the Clayton copula and Burr and Gaussian marginal. I need to derive ther Kendall'sw tau from the concordance coefficient by integration. I came across a way to do that in R long time ago but cannot find it again. Can somewone tell me what to read and what to use? Thank you. Steven Yen

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

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 try to calculate Kendall's W coefficient and I have a bizarre error. little.app.mat<-matrix(c(1,3,4,2,6,5,2,4,3,1,5,6,3,2,5,1,5,4),nrow=3,byrow=TRUE) print(kendall.w(little.app.mat[-1,])) >>> Kendall's W for ordinal data >>> W = 0.7753623Error in if (is.na(x$p.table)) { : argument is of length zero

How do you prove usefulness of a feature? Do you have an example of a feature that has been added after proving to be useful in the package space first? Thank you, Morgan On Fri, 11 Oct 2019 13:53 Michael Lawrence, <lawrence.michael at gene.com> wrote: > Thanks for this interesting suggestion, Morgan. While there is no strict > criteria for base R inclusion, one criterion relevant

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

Dear Prof. Therneau I was impressed to discover that the 'survConcordance' now handles Surv() objects in counting format (example below to clarify what I mean). This is not documented in the help page for the function. I am very curious to see how a c-index is estimated in this case, using just the linear predictors. It was my impression that with left truncation the ordering of

hello r users, i'm trying to use the apply method on a function with several inputs, but cannot figure out how to send multiple arguments to the function (not multiple runs of the same function, but one run of the function including two variables - each used within the function). a <- c(1:10,999,999,999) b <- c(11:20,999,999,999) tfun <- function(x,y){ if( (x = 1 & y !=999)

The survexp function can fail when called from another function. The "why" of this has me baffled, however. Here is a simple test case, using a very stripped down version of survexp: survexp.test <- function(formula, data, weights, subset, na.action, rmap, times, cohort=TRUE, conditional=FALSE, ratetable=survexp.us, scale=1, npoints, se.fit,

Could someone point me in the right direction for the following issue: A function is defined as follows: tfun <- function(dat) { fmla <- as.formula("y~x+z") dat2 <- dat mdl <- lm(fmla,dat2) mdl <- step(mdl) } Then the following code dat <- data.frame(x=1:10,z=1:10,y=(1:10)^2+10*(1:10)) tfun(dat) generates the output Start: AIC= 43.67

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 everyone! I have a problem with defining 2 fk referencing the same table. I have a Program table and a Team table. The Program should have an away team and a home team fk. From my understanding, "the fk column should be named after the class of the target table, converted to lowercase, with _id appended". But, in my case, I have 2 fk referencing the same table. How can I do this?