similar to: order() function, decreasing=TRUE unexpected behaviour

I've found the need to compute a version of order(x,y) where I want the sort order for *increasing* x and *decresing* y ... something we could imagine could be provided in the future as order(x,y, decreasing = c(FALSE, TRUE)) i.e., using a 'vectorized' decreasing argument. {No, I'm not volunteering right now!} I've found the following R-level solution and like to quiz

Dear R forum, I have a data.frame as defied below - df = data.frame(names = c("C", "A", "A", "B", "C", "B", "A", "B", "C"), dates = c("4/15/2013", "4/13/2013", "4/15/2013", "4/13/2013", "4/13/2013", "4/15/2013", "4/14/2013",

The NEWS file for R-devel (as of 2018-11-28 r75702) says ? order(...., decreasing=c(TRUE,FALSE)) could fail in some cases. Reported from StackOverflow via Karl Nordstr?m. However, either I don't understand the meaning of decreasing=c(TRUE,FALSE) or there are still problems. I thought order(x,y,decreasing=c(TRUE,FALSE) meant to return indices, i, such that x[i] was non-increasing and

Hello all, I'm running very simple regression but face a problem of non-homogeneity of variance, but with a decreasing variance with increasing mean...I do not know how to deal with that. this relationship doesn't seem to be strong, but it's my first time to see something like that, and would like to know what to do if one day it becomes stronger. I tested just for fun some

Hi all, I have a large data frame and would like to make a barplot of a categorical variable with the bars sorted in order of decreasing frequency. # Example: v1 = c(1.2, 1.4, 0.9, 1.0, 1.1, 1.0) v2 = c("aa", "cc", "bb", "bb", "cc", "bb") v3 = c(8, 10, 11, 9, 9, 10) df = data.frame(v1=v1, v2=v2, v3=v3) # How can I tell ggplot to sort

Hi, Please what am I doing wrong? I tried installing some views (eg MachineLearning, Multivariate) on R2.15.3, but it keeps on telling me that the package is not available for 2.15.3. Is it true? Thanks [[alternative HTML version deleted]]

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)

Hello, i'm using <R.h> & <Rmath.h>. What is the c equivalent to sort(x,deceasing=T) i.e. to get sort(x) i type R_qsort(x, 1, n), where do i set the decreasing=T flag ? More generally, is there a place where the functions in the R sources are documented ? Best, -- View this message in context: http://r.789695.n4.nabble.com/R-qsort-decreasing-T-tp3470432p3470432.html

Hi all I have a large dataframe with (among others) a categorical variable of 52 levels and would like to create a barplot with the bars ordered in decreasing frequency of the levels. I belive it is referred to as a pareto plot. Consider a subset where I keep only the categorical variable in question. # Example: v1 = c("aa", "cc", "bb", "bb",

Hi all, I have a frequency data that looks like this. 3 2 1 5 What I want to get is the "decreasing" cumulative of this data yielding 11 8 6 5 0 Is there any? I am aware of cumsum(), which will yield 3 5 6 11. But it is not what I want. - G.V.

Hi everybody, I try to do an exponential model with decreasing. In my data, there is no data missing but there negative values. I adapted this following code whose the origin code comes from in this forum: Regress<-read.table("C:\\Users\\Regression.csv",sep=";",dec=",",header=TRUE) Regress f <- function(x,a,b) {a * exp(b * x)} x<-Regress$EWT

Hi, I'm quite new to R so this question will sound quite fundamental. I need to create a vector of length 160. The first element should be (1+r)^159 and each element thereafter should decrease by a factor of (1+r) until the 160th element that should be 1. Is there a function similar to seq() but increasing or decreasing by factors? I need to do this in one step i.e, not using loops. Any help

## While in R v. 2.3.1 (the mid-July patch for Windows) ## on a Windows XP machine, this call to order() works fine... order(1:10,decreasing = TRUE) ## [1] 10 9 8 7 6 5 4 3 2 1 ## ...however, the argument name 'decreasing' ## must be typed in toto (note the missing 'g' ## in the following): > order(1:10,decreasin = TRUE) ## Error in order(na.last, decreasing, ...) :

function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: > any(diff(ppois(0:19,lambda=0.9))<0) [1] TRUE Actually, > ppois(19,lambda=0.9)<ppois(18,lambda=0.9) [1] TRUE Which could not be TRUE. Code is tested in both R 3.5.1 and Microsoft R Open 3.5.1. _

Le 04/12/2018 ? 11:27, I?aki Ucar a ?crit?: > On Tue, 4 Dec 2018 at 11:12, <qweytr1 at mail.ustc.edu.cn> wrote: >> function ppois is a function calculate the CDF of Poisson distribution, it should generate a non-decreasing result, but what I got is: >> >>> any(diff(ppois(0:19,lambda=0.9))<0) >> [1] TRUE >> >> Actually, >> >>>

Hello fellow R's, I'm sure there must be an easy way to do this. But after digging in the documentation and thinking about it for a while I couldn't figure it out. I need to get a decreasing recursive vector in. I mean something like this: if starting at 2, and ending at 6, the vector should be 2 3 4 5 6 3 4 5 6 4 5 6 5 6 6 An easy way would be to do this x <-

I've just realized that it could be handy to have a 'decreasing' argument in 'is.unsorted'. And I'm cheekily hoping someone else will implement it. It is easy enough to work around (with 'rev'), but would be less hassle with an argument. The case I have in mind uses 'is.unsorted' in 'stopifnot'. Pat -- Patrick Burns pburns at pburns.seanet.com

Function factor() in the current development version (2009-05-22) guarantees that levels are different character strings. However, they may represent the same decimal number. The following example is derived from a posting by Stavros Macrakis in thread "Match .3 in a sequence" in March nums <- 0.3 + 2e-16 * c(-2,-1,1,2) f <- factor(nums) levels(f) # [1]

Is it a bug that quantile() can return a lower sample quantile for a higher probability? > ##### quantile returns decreasing results with increasing probs (data at the end of the message) > quantile(x2, (0:5)/5) 0% 20% 40% 60% 80% -0.0014141174 -0.0009041968 -0.0009041968 -0.0007315023 -0.0005746115 100% 0.2905596324 >

We currently have a Toshiba Perception EX and I would like to start moving toward VOIP. Is there anyway we can run these two systems in parrallel? Better yet, is there anyway we can run fully on asterisk using the current PBX hardware? The current PBX has a mix of analog, digital and electronic cards in it. I tried to google for advice but I didn't find anything that pertained to this.