similar to: documentation for rank() (PR#7298)

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, 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",

Marius Hofert-4------------------------------ > Den 2015-10-09 kl. 12:14, skrev Martin Maechler: > I think so: the code above doesn't seem to do the right thing. Consider > the following example: > > > x <- c(1, 1, 2, 3) > > rank2(x, ties.method = "last") > [1] 1 2 4 3 > > That doesn't look right to me -- I had expected > > >

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

I noticed a typo in the documentation for the 'rank' function. Specifically, it describes ties.method="first" and contrasts with... ties.method="first", when it should be ties.method="last". Thanks, Jon -------------- next part -------------- Index: src/library/base/man/rank.Rd =================================================================== ---

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

Hi, >I would like to rank a time-series of data, extract the top ten data items from this series, determine the >corresponding row numbers for each value in the sample, and take a mean of these *row numbers* (not the data). >I would like to do this in R, rather than pre-process the data on the UNIX command line if possible, as I need to >calculate other statistics for the series.

------------------ >>>>> 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

Hi I want to use order() to get the order of a vector. But I would need a different behavior when ties occur: similar to the parameter ties.method = "random" in the rank() function, I would need to randomise the ties. Is this possible? Example: x <- rep(1:10, 2) order(x) [1] 1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20 order(x) [1] 1 11 2 12 3 13 4 14 5 15

Hi, I have a data.frame which is ordered by score, and has a factor column: Browse[1]> wc[c("report","score")] report score 9 ADEA 0.96 8 ADEA 0.90 11 Asylum_FED9 0.86 3 ADEA 0.75 14 Asylum_FED9 0.60 5 ADEA 0.56 13 Asylum_FED9 0.51 16 Asylum_FED9 0.51 2 ADEA 0.42 7 ADEA 0.31 17

Hi. I'm going to introduce the R-package for a group of medical doctors later this week and is a little confused about there use of a test named "willcoxon-pratt" for testing if the clinical and biochemical markers has decreased significantly after the use of some drugs for a group of patients. Looking into the R-functions I would in R recommand using a matched-pairs Wilcoxon

Hi, I have noticed different rank values calculated by qr() depending on LAPACK parameter. When it is FALSE (default) a true rank is estimated and returned. Unfortunately, when LAPACK is set to TRUE, the min(nrow(A), ncol(A)) is returned which is only occasionally a true rank. Would not it be more consistent to replace the rank in the latter case by something based on the following pseudo code ?

Hi, This report deals with p-values coming from chisq.test using the simulate.p=TRUE option. The issue is numerical accuracy and was brought up in previous bug reports 3486 and 3896. The bug was considered fixed but apparently was only mostly fixed. Just the typical problem of two values that are mathematically equal not ending up numerically equivalent. Consider this series of three 2x2

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

Le 22/01/2018 ? 17:40, Keith O'Hara a ?crit?: > This behavior is noted in the qr documentation, no? > > rank - the rank of x as computed by the decomposition(*): always full rank in the LAPACK case. For a me a "full rank matrix" is a matrix the rank of which is indeed min(nrow(A), ncol(A)) but here the meaning of "always is full rank" is somewhat confusing. Does it

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?

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 <-

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

Hello, I am looking for a way to subset a data frame by choosing the top ten maximum values from that dataframe. As well this occurs within some factor levels. ## I've used plyr here but I'm not married to this approach require(plyr) ## I've created a data.frame with two groups and then a id variable (y) df <- data.frame(x=rnorm(400, mean=20), y=1:400,