similar to: Deprecated component in plot.histogram (PR#2696)

I would like to suggest a slight modification to diag(). In the case where x is a matrix with both row names and column names the same, it would be reasonable if the resulting vector also had those names. I often use diag() on variance matrices, where this modification is helpful. The modification requires replacing if (is.matrix(x) && nargs() == 1) return(c(x)[1 +

Just a minor point, but the adj argument in ?text is described wrongly (0.64.1). Perhaps the following might be added to the details. Values of 0, 0.5, and 1 specify left/top, middle and right/bottom, respectively. The default is for centred text, ie \code{adj = c(0.5, 0.5)}. Cheers, Jonathan. Jonathan Rougier Science Laboratories Department of

Hi Everyone, To complete the outer() and kronecker() functions in the base, may I suggest the following tensor() function, which allows the multiplication of arrays through sets of conformable dimensions. I am happy to write a help page if required. The code also needs a setdiff() function which prompts me to ask: what about simple set functions? I expect many of us have written our own

Hi everyone, I do a lot of work with large variance matrices, and I like to use "crossprod" for speed and to keep everything symmetric, i.e. I often compute "crossprod(Q %*% t(A))" for "A %*% Sigma %*% t(A)", where "Sigma" decomposes as "t(Q) %*% Q". I notice in the code that "crossprod", current definition > crossprod function (x,

Hi everyone, Can I suggest an addition to the help page for logical AND and OR, indicating the behaviour of NA (which is currently not mentioned at all). I think something simple like an extra paragraph in Details, eg \code{NA} is a valid logical object. Where a component of \code{x} or \code{y} is \code{NA}, the result will be \code{NA} if the outcome is ambiguous. In other words \code{NA

Might it be possible to tighten the definition of "is.qr". I noticed that after I mistakenly typed example(lm) # make lm object named lm.D9 qr.Q(lm.D9) which exhausted the heap memory and produced two warning messages. As an object of class "lm" has a "qr" component, "is.qr" failed to detect that "lm.D9" was not a "qr" object. The

Hi everyone, Can I suggest the following cosmetic change to cut.default (R-1.3.1), to adjust the labels in the case that include.lowest = TRUE: arrow:dma0jcr% diff -u cut.R my.cut.R --- cut.R Tue Nov 13 12:28:35 2001 +++ my.cut.R Tue Nov 13 12:35:16 2001 @@ -25,6 +25,11 @@ ch.br[-nb], ",", ch.br[-1], if(right)"]"

In legend() is it possibly to mix lines and boxes? I would like to do something like legend(locator(1), legend = c("al", "ben", "chris", "dave"), lty = c(1, 2, 3, NA), fill = c(NA, NA, NA, "gray")) to indicate that "al", "ben" and "chris" have lines, and "dave" has a gray shading. Many thanks,

>>>>> "JonR" == Jonathan Rougier <J.C.Rougier@durham.ac.uk> writes: JonR> ... By the way, I have `solve' JonR> methods for triangular matrices and variance matrices -- would JonR> you be interested? { Jonathan, I hope it's okay if I CC this to R-devel; this must be of a wider interest } Ye.e..s; for triangular ones,

Hi there, I hope this question is not as stupid as the one before ... I tried to shorten my histogram (because the distribution is quite skewed and I simply don't want to see the long tail but still use the histogram plot). How can I do something like this? (The example does not work but I don't know why...) data <- rnorm(100) # as example, of course this is not skewed... h <-

Can someone please tell me to make a density histogram? hist makes one with count or relative frequency on the y-axis. I want the density, which is (rel freq)/(bin width) In the help I see: intensities: values f^(x[i]), as estimated density values. If `all(diff(breaks) == 1)', they are the relative frequencies `counts/n' and in general satisfy

Hello, all; I get an unexpected result when trying to plot a probability histogram with R1.9.1 on windows xp: #with the following code: > x <- runif(100,0,1) > hist(x) > hist(x, freq=F) > h <- hist(x, freq=F) > summary(h) # Length Class Mode #breaks 11 -none- numeric #counts 10 -none- numeric #intensities 10 -none- numeric #density 10

I have two large datasets (156K and 2.06M records). Each row has the hour that an event happened, represented by an integer from 0 to 23. R's histogram is combining some data. Here's the command I ran to get the histogram: > histinfo <- hist(crashes$hour, right=FALSE) Here's histinfo: > histinfo $breaks ?[1] ?0 ?1 ?2 ?3 ?4 ?5 ?6 ?7 ?8 ?9 10 11 12 13 14 15 16 17 18 19 20 21

Hi everyone, Is there any reason why we should not automatically coerce a factor supplied as an argument to col in a plotting function? The following modification (to R-1.6.1) seems pretty harmless > plot.xy function (xy, type, pch = 1, lty = "solid", col = par("fg"), bg = NA, cex = 1, ...) { if (is.factor(col)) col <- codes(col)

Dear all, I have done further modifications on the histogram functions that I reported earlier this year, and I hope this can be of use and perhaps included in the distribution. I have been using this stuff a couple of months myself, and while it is nothing sophisticated, it has it's applications. :-) I did a few small modifications today to make it a bit more compact. I have modified the

Hi Everyone, At the moment it is not possible to replicate complex lists, but only simple ones: > rep(list(fred = 1:10), 10) # works fine > rep(list(fred = 1:10, happy = "squash"), 10) Error in rep(list(fred = 1:10, happy = "squash"), 10) : Unimplemented feature in rep There is nothing in ?rep that suggests that the latter should not work, and I think it would

I was surprised to find that the variance of a scalar, using var(), is NA. Surely this should be zero? Cheers, Jonathan. --please do not edit the information below-- Version: platform = sparc-sun-solaris2.7 arch = sparc os = solaris2.7 system = sparc, solaris2.7 status = Patched major = 0 minor = 99.0 year = 2000 month = February day = 9 language = R Search Path: .GlobalEnv,

I find this a bit puzzling ... > solve(matrix(c(5, 2, 3, 1), 2, 2), c(Inf, 3)) [1] NaN Inf > solve(matrix(c(5, 2, 3, 1), 2, 2)) %*% c(Inf, 3) [,1] [1,] -Inf [2,] Inf I would expect the answer to be c(-Inf, Inf), so why has the -Inf been replaced by NaN in solve? Cheers, Jonathan. --please do not edit the information below-- Version: platform = sparc-sun-solaris2.7 arch = sparc

Hi everyone, There is no mention in ?new.env (R-1.6.0) of what the effect of setting the hash argument of new.env() actually does. What does it mean in performance terms to say that "the environment will be hashed"? Thanks, Jonathan. -- Jonathan Rougier Science Laboratories Department of Mathematical Sciences South Road University of Durham

The following commands z <- matrix(seq(0, 1, len=1000)) image(z, 1, z, zlim=c(0, 1), col=gray(1:10/10)) appear to violate the image functionality as described on the help page. As I understand it, by specifying a zlim of c(0, 1) and a colour vector of length 10, the colours should correspond to the intervals (0, 0.1), (0.1, 0.2), ..., (0.9, 1.0), this being 10 equally-spaced intervals on (0,