similar to: [patch] Fix typo in 'rank' documentation

Hi, I ran into a problem where I actually need rank(, ties.method="last"). It would be great to have this feature in base and it's also simple to get (see below). Thanks & cheers, Marius rank2 <- function (x, na.last = TRUE, ties.method = c("average", "first", "last", # new "last" "random", "max",

Full_Name: Doug Grove Version: R.1.9.1 OS: Linux Submission from: (NULL) (140.107.156.61) I just found that rank() has a 'decreasing' argument that is not documented in its help page. I checked my version of 2.0.0 (original release hence unpatched) and it is not documented there. For curiousity I also went back to version 1.8.1 and checked the function (not the documentation)and at

Hi All, I want to give ranks to elements in a column so I used: total_list$field1.rank <- rank(total_list$field1,ties.method="min") But this gives me the rank in increasing order. How do I get the ranks in decreasing order? I know decreasing = FALSE is not a legal argument here. Thanks. Jiong The email message (and any attachments) is for the sole use of the intended recipient(s)

This will start off sounding very easy, but I think it will be very complicated. Let's say that I have a matrix, which shows the number of apples that each person in a group has. OriginalMatrix<-matrix(c(2,3,5,4,6),nrow=5,ncol=1,byrow=T,dimnames=list(c("Bob","Frank","Joe","Jim","David"),c("Apples"))) Apples Bob 2

------------------ >>>>> Henric Winell <[hidden email]> >>>>> on Wed, 21 Oct 2015 13:43:02 +0200 writes: > Den 2015-10-21 kl. 07:24, skrev Suharto Anggono Suharto Anggono via R-devel: >> Marius Hofert-4------------------------------ >>> Den 2015-10-09 kl. 12:14, skrev Martin Maechler: >>> I think so: the code above

Hello, I'd like to rank rows of a data frame similar to what rank() does for vectors. However, ties should be broken by columns that I specify. If it is not possible to break a ties (because the row data is essentially the same), I'd like to have the same flexibility that rank() offers. Is there an elegant solution to this simple problem in R? Basically, what I need is a mixture of

Dear R-developers, I just realized that rank() behaves inconsistent if combining one of na.last in {TRUE|FALSE} with a ties.method in {"average"|"random"|"max"|"min"}. The documentation suggests that e.g. with na.last=TRUE NAs are treated like the last (=highest) value, which obviously is not the case: > rank(c(1,2,2,NA,NA), na.last = TRUE, ties.method

SAS proc rank has ties options of high and low that would allow producing ranks of the type found in the sports pages, e.g., rank (c(1,1,2,2,2,2,3)) == 1 1 3 3 3 3 7 Could R support these ties.methods?

I don't know the difference between rank and order.For example: > x=c(10,30,30,20,10,20) > x[rank(x,ties.method="first")] [1] 10 10 20 30 30 20 > x[order(x)] [1] 10 10 20 20 30 30 the result is quite different, x[rank(x,ties.method="first")] [1] 10 10 20 30 30 20 It is not sorted,why? -- View this message in context:

Hi. I am using R 2.3.1 on WIndows XP, and I am having trouble with the rank function in the presence of numerical NA data. I want the NA's all to get the same rank, but they don't. Here is an example from my session: >ct_align_rets_f2$liq[6851:6859] [1] 115396 NA 362595 NA 242986 340805 NA 692905 251533

Hello R gurus, I want to calculate the Spearman rho between two ranked lists. I am getting results with cor.test that differ in comparison to my own spearman function: > my.spearman function(l1, l2) { if(length(l1) != length(l2)) stop("lists must have same length") r1 <- rank(l1) r2 <- rank(l2) dsq <- sapply(r1-r2,function(x) x^2) 1 - ((6 * sum(dsq))

Suppose I have two columns, x,y. I can use order(x,y) to calculate a permutation that puts them into increasing order of x, with ties broken by y. I'd like instead to calculate the rank of each pair under the same ordering, but the rank() function doesn't take multiple values as input. Is there a simple way to get what I want? E.g. > x <- c(1,2,3,4,1,2,3,4) > y <-

Hi, Is there a function like rank but that solves the ties by randomly assigning a value (doesn't average ranks of ties). This is what I actually need: I want to make NA all elements of each column in an array that are ranked in a position larger that rankmax for each column. # Say I've got an array b: b<-cbind(c(1:5,5:1),c(1,12,14,2,5,4:8)) #> b # [,1] [,2] #[1,] 1 1

Dear List, after updating the exactRanksumTests package I receive a warning that the package is not developed any further and that one should consider the coin package. I don't find the signed rank test in the coin package, only the Wilcoxon Mann Whitney U-Test. I only found a signed rank test in the stats package (wilcox.test) which is able to calculate the exact pvalues but unfortunately

Hi, Is there any simple solution to get ranks in descending order? Example, a <- c(10, 98, 98, 98, 99, 100) r <- rank(a, ties.method="average") produces 1 3 3 3 5 6 I would want this instead: 6 5 3 3 3 1 Note that reversing r doesn't work but in small examples. Thanks, -Jose -- jquesada at andrew.cmu.edu Research associate http://lsa.colorado.edu/~quesadaj Dept. of

Dear All, Is there any simple way to way to produce "rank", for a given list, but in a descending order? E.G: x = list(a=c(1,5,2,4)); rank(x$a); produces 1,4,2,3 However I am looking for a way to generate (4,1,3,2). It would be particularly nice if the proposed solution has all the niceties of rank function (like NA handling and ties.method functionality) TIA Manoj

Hi All, I had 12766 elements in a column, 12566 are values and 200 are "NA"s. I used the following line to get the ranks: total_list$MB.rank <- rank(-total_list$MB,ties.method="min",na.last=NA) but I got an error message: Error in `$<-.data.frame`(`*tmp*`, "BCRP_PW_F.rank", value = c(3949, 6182, : replacement has 12199 rows, data has 12766 What

I have a question concerning the Wilcoxon signed-rank test, and specifically, which R subroutine I should use for my particular dataset. There are three different commands in R (that I'm aware of) that calculate the Wilcoxon signed-rank test; wilcox.test, wilcox.exact, and wilcoxsign_test. When I run the three commands on the same dataset, I get different p-values. I'm hoping that

Hi, I'm wondering if anyone can point me to a function that will allow me to do a ranking that treats ties differently than rank() provides for? I'd like a method that will assign to the elements of each tie group the largest rank. An example: For the vector 'v', I'd like the method to return 'rv' v: 1 2 3 3 3 4 5 5 6 7 rv: 1 2 5 5 5 6 8 8 9 10 Thanks,

Hi everyone! I need to perform a Wilcoxon rank sum test, but I have some ties and the groups have different size also. When I deal with ties I use the wilcox.exact function, how can I solve the different size problem using this function? thanks net -- View this message in context: http://r.789695.n4.nabble.com/Wilcoxon-rank-sum-in-unbalanced-design-tp3447400p3447400.html Sent from the R help