similar to: Sorting a matrix?

In the code of functions 'order' and 'sort.list' in R 3.5.0 alpha (in https://svn.r-project.org/R/branches/R-3-5-branch/src/library/base/R/sort.R), in "fastpass, take advantage of ALTREP metadata", there is "try the reverse since that's easy too...". If it succeeds, ties are reordered, violating stability of sorting. Example: x <- sort(c(1, 1, 3)) x # 1

I found this example in an Introductory R book in the chapter on Matrices and Arrays The array is > m [,1] [,2] [,3] [,4] [,5] [1,] 0 12 13 8 20 [2,] 12 0 15 28 88 [3,] 13 15 0 6 9 [4,] 8 28 6 0 33 [5,] 20 88 9 33 0 The code is #returns the minimum value of d[i,j], i !=j and the row attaining #the minimum, for square

Dear R users, I have 32 observations in data x. After sorting this, I want to compute means and variances of 3 groups divided by "nr". Actually, the number of groups is flexible. Any suggestion will be greatly appreciated. Kathryn Lord --------------------------------------------------------------------------- x=rnorm(32) y=sort(x) nr=matrix(c(12,11,10,10,10,11),2,3) > nr

Hi, I'm trying to present a table of some experimental data, and I want to order the rows by the instance names. The issue I've got is that there are a variety of conventions for the instance names (e.g. competition01, competition13, small_1, big_20, med_9). What I want to be able to sort them first in category order so: competition < small < med < big, and then perform the

sort() and order() are not quite the same, as "one knows": o order allows breaking ties by more than one argument; o sort allows a 'partial' and 'na.last' argument Still, the following timing (on a `simple' UltraSparc I) suggest that actually two different algorithms are used > N <- 10000 > typeof(x0 <- 1:N) # --- x0 : already sorted --- [1]

>>>>> Karolis Koncevi?ius >>>>> on Mon, 9 Dec 2019 23:43:36 +0200 writes: > So I tried adding Infinity support for all cases. > And it is (as could be expected) more complicated than I thought. "Of course !" Thank you, Karolis, in any case! > It is easy to add Inf support for the test. The problems start with conf.int=TRUE.

Hello everybody, I need to calculate seasonal means with temperature data for my work. I have 70 files coming from weather stations, which looks like this for example: startdate <- as.POSIXct("01/01/2006", format = "%d/%m/%Y") enddate <- as.POSIXct("05/01/2006", format = "%d/%m/%Y") date <- seq(from = startdate, to = enddate, by =

Hi Andres, Postgres uses removable code support and Orcv1. I does make me quite > worried to see a phase where there'll be no viable way of using both in > llvm. Why isn't the right answer here to at lest develop the > replacement as a set of patches / as a branch that then can be merged as > a whole / shortly after each other, rather than just starting to develop > a

Dear list, order(x,y,z) returns a permutation to order x, ties broken by y, remaining ties broken by z. (And so on.) What I'd like to do is order(X), where X is a matrix (or a list or data frame if necessary) of unspecified size, which orders X[,1], ties broken by X[,2], remaining ties broken by X[,3] and so on - without having to know and to write down how many columns X has. Any

Hi, I have a character vector as below: a<-c('10','3R','4','4R','5','5R','6','6R','7','8','9','7R','1','10R','11'

Hi, I'm trying to translate some Matlab code to R and I'm trying to implement the behavior of Matlab's sort() which when applied to a matrix will sort the columns returning the column sorted matrix as well as a matrix of permuted indices. Doing: > x <- matrix(c(3,4,2,6,3,4,8,7,5), nr=3) > x [,1] [,2] [,3] [1,] 3 6 8 [2,] 4 3 7 [3,] 2 4 5

Hi. I am using the Matching package for propensity score matching. For each treated unit, I want to find all control units whose propensity scores lie within a certain distance from the treated unit. The sample code is as follows: > library(Matching) > x <- rnorm(100000) > y <- rnorm(100000) > z <- rbinom(100000,1,0.002) > logit.reg <-

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 colleagues, I have really heavy problems in sorting the results of a table according to certain features of the levels in the table. Prerequisites: It all starts with a fairly simple data set, which stores observations of 21 observers (horizontally from 1 to 21; 21 is reference/goldstandard) who diagnosed 42 videos (vertically from 1 to 42). See dump of R-object "input" below in

Dear R-help, I was comparing SAS (I do not know what version it is) and R (version 2.6.0 (2007-10-03) on Linux) survival analyses with time-dependent covariates. The results differed significantly so I tried to understand on a short example where I went wrong. The following example shows that even when argument 'method' in R function coxph and argument 'ties' in SAS procedure

Thank you for a fast response. Nice to see this mailing list being so alive. Regarding Inf issue: I agree with your assessment that Inf should not be removed. The code gave me an impression that Inf values were intentionally removed (since is.finite() was used everywhere, except for paired case). I will try to adjust my patch according to your feedback. One more thing: it seems like you

Hello, Writing to share some things I've found about wilcox.test() that seem a a bit inconsistent. 1. Inf values are not removed if paired=TRUE # returns different results (Inf is removed): wilcox.test(c(1,2,3,4), c(0,9,8,7)) wilcox.test(c(1,2,3,4), c(0,9,8,Inf)) # returns the same result (Inf is left as value with highest rank): wilcox.test(c(1,2,3,4), c(0,9,8,7), paired=TRUE)

I am trying to determine whether two samples are identical or not. I'm aware that somebody can use the Kolmogorov-Smirnoff test to compare empirical distributions, but since my samples have ties I'm not sure if I'm getting the right p-values for the comparison. Can the Kolmogorov-Smirnoff test be adjusted for the case when ties exists and are there any functions that already

Oh, this seemed so simple (and I'm sure the answer will be, as usual, so thanks in advance for enlightening me). I need to sort each row of a matrix independent of the others. For example, > test <- matrix(c(8,7,1,2,6,5,9,4,3),nrow=3) > test [,1] [,2] [,3] [1,] 8 2 9 [2,] 7 6 4 [3,] 1 5 3 I can get each row sorted well enough. > sort(test[1,]) [1]

I define the following function: (Please don't wonder about the use of this function, this is just a simplified version of my actual function. And please don't spend your time in finding an alternate way of doing the same as the following does not exactly represent my function. I am only interested in a good explanation) > f1 =