similar to: 'partial' in sort() inefficient?

Hi, This is essentially a reposting of http://tolstoy.newcastle.edu.au/R/devel/05/11/3305.html which had no responses, and the behaviour reported there persists in r-devel as of yesterday. (1) sort() with non-null partial > x = rnorm(100000) > keep = as.integer(ppoints(10000) * 100000) > system.time(sort(x)) [1] 0.05 0.00 0.04 0.00 0.00 > system.time(sort(x, partial = keep)) [1]

Sadly you failed to set your email program to send plain text and the data is corrupted at my end. I also think you need to reduce the size of the data set... the intent here is to increase your understanding, not debug your particular analysis. I will say that I am having a very challenging time understanding what you are trying to accomplish though. What are the equations that you think need

When I look at ppoints I see: ppoints<-function (x) { n <- length(x) if (n == 1) n <- x (1:n - 0.5)/n } However Venables & Ripley (2nd ed, p 165) say ppoints() should return (i-1/2)/n for n>=11; (i-3/8)/(n+1/4) for n<=10. The version below should work as described: ppoints<-function (x) { n <- length(x) if (n <= 10) (1:n - 0.375)/(n + 0.25) else (1:n - 0.5)/n

Dear r-bugs, Whilst playing with ppoints I discovered that when one uses it directly, occasional NA's in a vector also become data fractions: ppoints(c(1,2,NA,4)) Would it be a good idea to add a warning message as in: ppoints <- function (n, a = ifelse(n <= 10, 3/8, 1/2)) { if(any(is.na(n))) warning("'n' contains NA's") if(length(n) > 1) n <-

hi, I wrote a set of R functions for estimating what is the probability function that best fits a set of data. I wrote them based in this response: /http://tolstoy.newcastle.edu.au/R/help/03b/1714.html/ I extracted the relevant segment of the link above: //> PPCC <- function(shape, scale, x) { # only for weibull / + x <- sort(x) + pp <- ppoints(x) + cor( qweibull(pp, shape=shape,

OK, Let's try this again! Here is the reproducible script; it is long because I had to copy the panel dataset here. My question is related to systemfit; I don't know how to get the result for the entire panel. #Reproducible script Empdata<- read.csv("/Users/ngwinuiazenui/Documents/UPLOADemp.csv") View(Empdata) install.packages("systemfit")

Dear R community. I am using the identify() function to identify outliers in my dataset. This is the code I am using: #################################################################### # Function to allow identifying points in the QQ plot (by mouseclicking) qqInteractive <- function(..., IDENTIFY = TRUE) { qqplot(...) -> X abline(a=0,b=1) if(IDENTIFY) return(identify(X))

I think qqline does not do exactly what it is advertised to do ("`qqline' adds a line to a normal quantile-quantile plot which passes through the first and third quartiles."). Consider the graph: tmp <- qnorm(ppoints(10)) qqnorm(tmp) qqline(tmp) The line (which I expected go through all the points), has a slightly shallower slope than does the points plotted by qqnorm. I think

This command cdmoutcome<- glm(log(value)~factor(year) > +log(gdppcpppconst)+log(gdppcpppconstAII) > +log(co2eemisspc)+log(co2eemisspcAII) > +log(dist) > +fdiboth > +odapartnertohost > +corrupt > +log(infraindex) > +litrate > +africa >

Dear all, A very basic terrain calculated as a matrix from Spatial Points Patterns: #interpolate using the akima package library(akima) terrain=interp(ppoints$x,ppoints$y,ppoints$marks,xo=x0,yo=y0, linear=F) > class(terrain) [1] "list" > class(terrain$x) #these are the x-coord i.e: [1...1000] [1] "numeric" > class(terrain$y)#these are the y-coord i.e: [1...1000] [1]

>>>>> On Tue, 31 Aug 1999 14:57, Werner Stahel <stahel@stat.math.ethz.ch> said: WSt> Here is a suggestion. It seems that qqplots, comparing a sample WSt> to a distribution other than the normal, are not explicitly WSt> available in S or R. I found (in S-plus / Trellis it is, see below) WSt> qqplot(y, rt(300, df = 5)) WSt> as an

Hi, I am trying to write a function with the following codes and I would like it to return the values for "alpha beta para parab " seperately. Then I would like to use this funstion for "variable" with factor "a" and "b". But the result turns out to be a matrix with element like "Numeric,2" ... I guess they are just the values for

Do you know of a reference that discusses alternative choices for plotting positions for a normal probability plot? The documentation for qqnorm says it calls ppoints, which returns qnorm((1:m-a)/(m+1-2*a)) with "a" = ifelse(n<=10, 3/8, 1/2)? The help pages for qqnorm and ppoints just refer to Becker, Chambers and Wilks (1988) The New S Language (Wadsworth & Brooks/Cole),

Hello! I want to plot a P-P plot. So I've implemented this function: ppplot <- function(x,dist,...) { pdf <- get(paste("p",dist,sep=""),mode="function"); x <- sort(x); plot( pdf(x,...), ecdf(x)(x)); } I have two questions: 1. Is it right to draw as reference line the following: xx <- pdf(x,...); yy <- ecdf(x)(x); l <- lm(

Hi all. I am seeking for a statistical suggestion. My data set comprises 382 measures each having 169 variables. Each measure is the outcome of a nuclear magnetic resonance experiment, so each of the 169 points has the same unit. As I want to do some multivariate calibration using these data, I checked whether some multivariate outliers existed. I calculated Mahalanobis distances and did a

Hello, The following script allows for Weibull plots using R. Its output is similar to the output of the wblplot function (or weibplot function) in MATLAB. As opposed to the previously mentioned function it does not require proprietary software. Instead, it is based on R. My code also allows for a graphical visualization of weibull fitted data. In particular, data can be represented by a

Hi list members, I saw a couple lavaan posts here so I think I?m sending this to the correct list. I am trying to run a multigroup analysis with lavaan in order to compare behavioural correlations across two populations. I?m following the method suggested in the paper by Dingemanse et al. (2010) in Behavioural Ecology. In one of the groups, lavaan returns negative variance for one path and I?m

Hi, I want to create upper and lower 95% confidence intervals for a p-p plot of an empirical distribution with a theoretical gamma distribution. This is my code: x<-rgamma(100,shape=2, rate=1) # empirical data fitdistr(x,"gamma") # fit a gamma distribution dist<-pgamma(x,shape=1.9884256 ,rate=0.8765314 ) # fitted distribution, using the loglikelihood estimated parameters

Dear List, As a way to learn R, I am trying out some of the examples shown in the Reference Cards. I use the following to read a column of numbers from Excel: x <- read.delim("clipboard") My questions are: 1. Why is it that the first number is omitted from the selected data range? How do I tell R to pick up the first number as part of the entire selection? 2. The next thing I

Dear List, invoking qqnorm-plots in Version 63.3 produces funny things: using the option `type="s"´ on qqnorm should give a nice *line* of observed quantiles. Now, the line is walking along in order to the points index instead from lowest to highest, wich makes funny slopes. try x <- table(rnorm(1000) # or similar and qqnorm(x,type="s") # in 0.63.2 and 63.3 Well, the