For statistical graphics enthusiasts and John Tukey admirers in particular,
for the first time in my lifetime, I have seen a reference to Tukey's
"box-and-whisker
plots" in a public non-technical forum -- a National Weather Service report
for the San Francisco Bay Area:
"Looking at box and whisker
To add to Rolf's comments:
1. What you do with your data may depend on what you want to do with it
*after* you have fiddled with its structure. Treating numerical categories
as numerics is often a bad idea for many kinds of analyses or even graphics;
2. Note that na.omit(d) when d is a data frame
Just wanted to point out that:
1. range() is a generic function and the default version works for
character objects also:
> range("Joe", "Q", "Public")
[1] "Joe" "Q"
## but
> diff(range("Joe", "Q", "Public"))
Error in r[i1] - r[-length(r):-(length(r) - lag + 1L)] :
non-numeric argument to bin
mns in a data
frame/table for processing
If I've misunderstood, of course just ignore the above.
-- Bert
On Wed, Jun 18, 2025 at 7:48 AM Bert Gunter wrote:
> If I understand correctly, see ?"[" : data tables inherit from data
> frames, and data frame columns can be i
If I understand correctly, see ?"[" : data tables inherit from data frames,
and data frame columns can be indexed by their names as character strings.
The following should give you the idea:
d <- data.frame(a1 = 1:3, a2 = 4:6)
x <- vector("numeric",2)
for(i in 1:2){
nm <- paste0("a",i)
x[i]
mple).
Context matters!
Cheers,
Bert
On Thu, Jun 5, 2025 at 10:09 PM Bert Gunter wrote:
> ... and here is a a simple 2-liner without a sort that I think is linear
> in time and space (but please correct if this is wrong):
>
> x <- cumsum(runif(10))
> x/x[10] *
tics? I believe yes, but someone more statistically competent
than me needs to verify or correct this.
Cheers,
Bert
On Thu, Jun 5, 2025 at 5:19 AM Bert Gunter wrote:
>
> Richard:
>
> "The "use an upper bound of 100 - (n+1)*5" and then "add back
> cumsum(
Richard:
"The "use an upper bound of 100 - (n+1)*5" and then "add back
cumsum(rep(5,n)) at the end" (or equivalent) trick ensures the gaps
are right but does nothing about the distribution.."
If I understand you correctly, I think the above is wrong. Here is a
one-line version of Peter's code for
can I extend this to draw only integer
> mid-point between 0-100 while maintaining minimum difference as 5?
>
> Also, are all generated points are equally likely?
>
> Thanks for your time and suggestions.
>
> Thanks and regards,
>
> On Wed, 4 Jun 2025 at 17:13, Bert Gunte
To answer my own question, yes they are different.
Peter's code can generate the solution 1 20 26 32 38 44 50 56 62 68. Mine
cannot.
So, again, context?
-- Bert
On Wed, Jun 4, 2025 at 4:43 AM Bert Gunter wrote:
> Is Peter's solution different then:
>
> diffs <- cumsum(r
Is Peter's solution different then:
diffs <- cumsum(runif(9, 5, 100/9))
x <-runif(1,0,100-diffs[9])
c(x, x+diffs)
I ask because:
1. If yes, this is why more context is needed;
2. If no, the above avoids a sort.
Cheers,
Bert
On Tue, Jun 3, 2025 at 2:15 PM peter dalgaard wrote:
> Can't you j
s.
> [Here, "k_(i)" is the usual notation for order statistics.]
>
> Is that the task?
>
> If so, are there are other requirements on the k_i ?
>
> The word "simulate" suggests the k_i are supposed to be realizations of
> random variables. If so, what sort
If this is a real problem and not homework, can you tell us the
context? It is not at all clear (to me) what you mean by "simulate",
i.e. what your target distribution is, which may depend on/be defined
by the context.
Bert
"An educated person is one who can entertain new ideas, entertain
others,
Oh, and for your last example, what you actually do is plot 25
separate axes with one label each, one on top of the other.
Again, correction requested if this is wrong.
-- Bert
"An educated person is one who can entertain new ideas, entertain
others, and entertain herself."
On Wed, May 28, 2025
I believe the following is the reason, but please correct is this is wrong:
from ?axis:
"The code tries hard not to draw overlapping tick labels, and so will
omit labels where they would abut or overlap previously drawn labels.
This can result in, for example, every other tick being labelled. The
project.org/articles/RN-2001-021/RN-2001-021.pdf
>
> Bert is correct about the 20+ year time frame… :-)
>
> Regards,
>
> Marc Schwartz
>
>
> On May 22, 2025, at 10:32 PM, Bert Gunter wrote:
>
> John:
> 1. You might get a better response if you could give at
John:
1. You might get a better response if you could give at least a couple of
use case examples of what you mean.
2. I have a hazy memory of this being actually discussed on this list ca ~
20 years ago. As I recall, some sort of macro facility may have been
implemented in a package using R's com
5, 21:58 Bert Gunter wrote:
> I am out of the country and will reply more fully to you (privately) when
> I return. But briefly, and subject to my possible
> misunderstanding/misinterpretation of your specification, I would say both
> your examples illustrate exactly what I said. In th
I am out of the country and will reply more fully to you (privately) when I
return. But briefly, and subject to my possible
misunderstanding/misinterpretation of your specification, I would say both
your examples illustrate exactly what I said. In the first, the clea
On Tue, May 6, 2025, 14:23 Kev
Heh. I suspect you'll get some interesting responses, but I won't try to
answer your questions. Instead, I'll just say:
(All just imo, so caveat emptor)
1. What you have been taught is mostly useless for addressing "real"
statistical issues;
2. Most of my 40 or so years of statistical practice i
ote:
> On 2025-04-24 3:25 p.m., Bert Gunter wrote:
> > Folks:
> >
> > Unless my wee old brain errs (always a danger), uniform sampling from an
> > n-vector for which 0 <= ai <= xi <= bi and SUM(xi) = k, a constant,
> > where to ensure that the constraints are
Folks:
Unless my wee old brain errs (always a danger), uniform sampling from an
n-vector for which 0 <= ai <= xi <= bi and SUM(xi) = k, a constant,
where to ensure that the constraints are consistent (and nontrivial),
SUM(ai)< k and SUM(bi) > k, is a simple linear transformation (details left
to
ot considered.
> I only worked with the vector. If the vector is part of a data frame where
> I need to know that "this zero" came from Bob, and the other zero came from
> Samantha, then the code will not work as written. My solution would be to
> ask ChatGPT more questions an
Just a comment:
You wish to draw subsets of size n with or without replacement -- and I
suspect without replacement is simpler than with -- from a set of positive
integers that sum to a fixed value T. This sounds related to the so-called
subset sum problem in computational complexity theory: Given
My *opinion*:
1. As it stands, this "opinion survey" seems pretty useless;
2. It *might* be more useful if you explained what the package will do, who
it's intended users are, etc.
-- Bert
"An educated person is one who can entertain new ideas, entertain others,
and entertain herself."
On Mon
. Heiberger wrote:
> > somenames <- c("Heigh", "Ho", "Silver", "Away")
> > as.formula(paste("~(",paste(somenames, collapse="+"),")^2"))
> ~(Heigh + Ho + Silver + Away)^2
> >
>
> &g
You might do better posting here:
https://stat.ethz.ch/mailman/listinfo/r-sig-finance
Cheers,
Bert
"An educated person is one who can entertain new ideas, entertain others,
and entertain herself."
On Wed, Apr 2, 2025 at 4:39 PM Arnaud Gaboury
wrote:
> I will try to be more precised. Here is
The R-help listserv deletes most attachments (as security risks) and your
attachments (I assume) did not come through. Provide code for a minimal
reprex or a site from which those interested in helping can get your plots
instead if no one who already knows about this issue responds. You might
also
;
On Sun, Mar 30, 2025 at 6:11 AM Gabor Grothendieck
wrote:
> Another solution. reformulate + substitute + as.formula:
>
> substitute(~ (.)^2, list(. = reformulate(somenames)[[2]])) |> as.formula()
>
> On Sat, Mar 29, 2025 at 5:31 PM Bert Gunter
> wrote:
> >
&
eling system without generating a
> formula (a string) that gets parsed.
>
> -Original Message-
> From: R-help On Behalf Of Richard M.
> Heiberger
> Sent: Saturday, March 29, 2025 8:47 PM
> To: Bert Gunter
> Cc: R-help
> Subject: Re: [R] [External] Creating model
Note: I am almost certain that this has been asked and answered here
before, so my apologies for the redundant query.
I also know that there are several packages that will do this, but I wish
to do it using base R functions only (see below).
The query: Suppose I have a character vector of names l
(I neglected to cc the list)
-- Forwarded message -
From: Bert Gunter
Date: Mon, Mar 24, 2025 at 11:00 AM
Subject: Re: [R] how to create model matrix of second order terms
To: Stephen Bond
In view of the fact that your claim that:
"The current setup for formulas doe
Full disclosure: I did not attempt to decipher your code.
But ~(A+B +C)^2 - (A + B + C)
gives all 2nd order interactions whether the terms are factors or numeric.
~I(A^2) + I(B^2) gives quadratics in A and B, which must be numeric, not
factors, of course
You can combine these as necessary to get
Bravo for your unrequired R efforts.
You misunderstand the nested call. sample() is called only once,
producing 1 sample of 10 with replacement. Since your matrix call
needs 50 values, ?matrix tells you (in details):
"If there are too few elements in data to fill the matrix, then the
elements in d
As has now been explained, there is a lot going on under the hood
here. I would just note that the Rmpfr package can do arbitrary
precision arithmetic; and so can the Ryacas package, which extends
these capabilities to e.g. arbitrary precision linear algebra.
(I am just parroting what I found via
Full disclosure: I have never plotted time series using this xyplot method.
However, ?xyplot.ts says:
"screens
factor (or coerced to factor) whose levels specify which panel each
series is to be plotted in. screens = c(1, 2, 1) would plot series 1,
2 and 3 in panels 1, 2 and 1. May also be a nam
Hi Ben:
I realize that for the OP whether it takes 1/2 second or 1 microsecond to
do what he wants may be irrelevant, but just for fun I thought I'd time the
condense() function you found vs. the compr() function I worked out, which
are similar in their approach.
compr <- function(x, sep ="-")
{
Well, as I predicted, my initial suggestions were, ... ummm, rather dumb.
Also, Rui's suggestions are probably preferable to the below. However, it
*is* a very simple, bare-boned approach to converting a sequence of
increasing integers to a character representation using interval notation.
The code
Cute exercise!
"Simple" is in the eyes of the beholder, of course. There is probably a
package out there that can do this in a trice .
But my first thought -- so caveat emptor!, as my thoughts, first or last,
are often not so, um... thoughtful -- is to diff() the sequence (as
numerics) -- so that
It sounds like your first course of action should be to contact the package
maintainer. See ?maintainer or consult the Description file for the package
for their email.
Cheers,
Bert
"An educated person is one who can entertain new ideas, entertain others,
and entertain herself."
On Fri, Feb 21
Again, **Completely Off Topic,** but I hope of interest to at least some on
this forum.
https://www.science.org/content/article/renowned-scientific-integrity-investigator-endows-fund-support-fellow-sleuths
Please do not reply. This is just FYI.
Best to all,
Bert
"An educated person is one who c
Possibly because:
panel.hist is not an existing R function -- you have to first create
it so pairs() can use it. ?pairs shows you how in the Help examples,
i.e.
panel.hist <- function(x, ...)
{
usr <- par("usr")
par(usr = c(usr[1:2], 0, 1.5) )
h <- hist(x, plot = FALSE)
breaks <- h
I think there may be some confusion here.
4) x which is a 844x7 matrix having value "Good", "Moderate", etc.
Correct.
**
To be clear:
x is a matrix of **integers** (presumably population counts) with 7 columns
with names "Good", "Moderate",
See ?"::"
Many packages hide some of their internal objects, because they are not
meant to be directly called by users. As the Help file referenced above
explains, they can be referenced by the triple colon operatorm in this case:
MASS:::boxcox.lm
See also ?methods and
methods(boxplot)
for furt
dget() should be dput(), of course. But you don't need dput() either, as
assigning the structure() call suffices.
-- Bert
On Mon, Jan 20, 2025 at 6:18 PM Sorkin, John
wrote:
> I have used ggplot to create a graph on which the y-axis is on the log
> scale. (see data and code, below.) I would lik
AM Ian Farm wrote:
> I might add that there seems to be a subtle difference between using
> `...elt()` and `match.call()`, which is that the former causes `a` itself
> to be evaluated while the latter doesn't:
> ```
> # Some approaches that have been suggested:
>
> #
gt; c(one, two, three(...))
> }
>
> three <- function(a, ...) {
> a
> }
>
> f1(a = 1, b = 2, c = 3)
> #> [1] 1 1 1
>
>
> On Sun, Jan 5, 2025 at 12:00 PM Bert Gunter
> wrote:
>
>> Consider:
>>
>> f1 <- function(...){
>> one <
Biostatistics and Informatics Core, University of Maryland School of
> Medicine Claude D. Pepper Older Americans Independence Center;
> >Senior Statistician University of Maryland Center for Vascular Research;
> >
> >Division of Gerontology and Paliative Care,
> >10 North
have I entirely misunderstood you?!
NB. Terminology: Function "arguments" not "parameters". "Parameters"
actually means something different in R.
Cheers,
Bert
On Mon, Jan 6, 2025 at 3:26 PM Sorkin, John
wrote:
> Bert and other on this Chain,
>
> The origina
e way you want in that situation. I leave such delights to
wiser heads, as well as any corrections or refinements to anything that
I've said here.
Cheers,
Bert
On Mon, Jan 6, 2025 at 9:55 AM Jorgen Harmse wrote:
> I think Bert Gunter is right, but do you want partial matches (not f
Thanks, Iris.
That is what I suspected, but it wasn't clear to me from the docs.
Best,
Bert
On Sun, Jan 5, 2025 at 10:16 AM Iris Simmons wrote:
>
> I would use two because it does not force the evaluation of the other
> arguments in the ... list.
>
>
>
> On Sun, Jan
Consider:
f1 <- function(...){
one <- list(...)[['a']]
two <- ...elt(match('a', ...names()))
c(one, two)
}
## Here "..." is an argument list with "a" somewhere in it, but in an
unknown position.
> f1(b=5, a = 2, c=7)
[1] 2 2
Which is better for extracting a specific named argument, one<- o
... but do note:
glm(lot1 ~ log(u), data = clotting, family = gaussian)
is a plain old *linear model*, which is of course a specific type of
glm, but not one that requires the machinery of glm() to fit. That
is, the above is exactly the same as:
lm(lot1 ~ log(u), data = clotting)
and gives exac
?deviance ?anova
Bert
On Tue, Dec 24, 2024 at 6:22 AM Christofer Bogaso
wrote:
>
> I think vcov() gives estimates of VCV for coefficients.
>
> I want estimate of SD for residuals
>
> On Tue, Dec 24, 2024 at 7:24 PM Ben Bolker wrote:
> >
> > vcov(). ?
> >
> >
> > On Tue, Dec 24, 2024, 8:45 AM C
choose.files is only for MS Windows (the Help file says).
-- Bert
On Wed, Dec 18, 2024 at 8:06 AM Gabor Grothendieck
wrote:
>
> Try choose.files
>
> choose.files(default = file.path(mydir, "*.*"), multi = FALSE)
>
> On Wed, Dec 18, 2024 at 10:33 AM J C Nash wrote:
> >
> > I've been working
Please look at what you wrote.
get.symbols vs. getSymbols.
-- Bert
On Wed, Dec 18, 2024 at 7:56 AM Phil Smith via R-help
wrote:
>
> Hello r-project:
>
> I want to load and use the tiny quant libary.
>
> Hello R-project:
>
> get.symbols doesn't work this morning.
>
> I use this code:
>
>
Bruce,
You failed to post (some of) your data using dput() as was requested
by John Kane. The reason that this is important is that it would tell
us exactly what your data consist of -- i.e date-times, character,
factors, etc. -- which your use of cut and paste does not.
So if for some reason you
ributes including names.
> Character strings ‘c("T", "TRUE", "True", "true")’ are regarded as
> true, ‘c("F", "FALSE", "False", "false")’ as false, and all others
> as ‘NA’.
>
>
> On
Ivo, et al.:
--IMHO only ... and with apologies for verbosity
Defining, let alone enforcing, "consistent behavior" can be a
philosophical conundrum: what one person deems "consistent" behavior
for a function across different data structures and circumstances may
not be the same as another's. While
Thanks, Ivan. Exactly my reaction.
And as a Turing complete language, R allows one to do anything in R --
including writing an email package that does email directly from R:
see package emayili .
Indeed, I think one would find it difficult to find *any* software
that does not interact with the in
Just a slight technical note -- Ben gave you a good answer already, imo.
The note is: R is Turing complete, which mean that *anything* any
language can do, R could be programmed to do also. The point is what
can be done well in R and what can be done (often much) better with
other tools, as Ben ex
Please read **and follow** the posting guide linked below to learn how
to ask for help on this list.
Cheers,
Bert
On Thu, Dec 5, 2024 at 2:16 PM Figueiredo, Carlos via R-help
wrote:
>
> Hi there
>
>
> I want to create a script in Rstudio and load in the reagent dataset ensuring
> that the diffe
tps://www.uni-giessen.de/math/eichner
> -
>
> Am 04.12.2024 um 14:38 schrieb Bert Gunter:
> > matrices are vectors with a "dim" attribute.
> > So what I think is happening is:
> >
> >> A &
matrices are vectors with a "dim" attribute.
So what I think is happening is:
> A <- matrix(1:25, nrow = 5, ncol = 5)
> diag(A[-1,]) <- 0
> A
[,1] [,2] [,3] [,4] [,5]
[1,]16 11 16 21
[2,]07 12 17 22
[3,]30 13 18 23
[4,]490 19 24
[5,]
riatrics Research, Education, and Clinical Center;
> PI Biostatistics and Informatics Core, University of Maryland School of
> Medicine Claude D. Pepper Older Americans Independence Center;
> Senior Statistician University of Maryland Center for Vascular Research;
>
> Division of Geron
Rui:
"f these two, diff is faster. But of all the solutions posted so far,
Ben Bolker's is the fastest."
But the explicit version of diff is still considerably faster:
> D <- c(rep(1,10),rep(2,6),rep(3,2))
> microbenchmark(c(1L,diff(D)), times = 1000L)
Unit: microseconds
expr min
May I ask *why* you want to do this?
It sounds to me like like you're using SAS-like strategies for your
data analysis rather than R-like.
-- Bert
-- Bert
On Sat, Nov 30, 2024 at 6:27 PM Sorkin, John wrote:
>
> Dear R help folks,
>
> First my apologizes for sending several related questions to
I assume that the responses that John already received to his recent
post met his needs. However, when I read it, I had a slightly
different interpretation. So feel free to ignore the rest of this post
if you like, but here's my interpretation and a simple solution to it.
An example to help explai
The grouping solutions offered seem to be the obvious way to do this and
may even be more efficient in R then what follows below. However, note
that they are to some extent doing unnecessary work, since the ordering in
the data frame already implicitly provides the grouping, and the hashing or
wha
If you haven't already done so, this might be better posted on
Bioconductor:
https://www.bioconductor.org/
Cheers,
Bert
On Fri, Nov 15, 2024 at 2:53 AM Clark Jeremy
wrote:
> Dear All,
>
> The following code extracts from NCBI very nice output for ONE allele of a
> SNP (often the allele with th
Is the problem reading the file in or processing it after it has been read
in?
Bert
On Fri, Nov 8, 2024 at 5:13 PM Jeff Newmiller via R-help <
r-help@r-project.org> wrote:
> Can you tell us what is wrong with the "chunked" package which comes up
> when you Google "r read large file in chunks"?
>
Sorry, wrong language.
"through the function's closures" in my email should be:
through the function's chain of environments. (A function in R *is* a
closure).
-- Bert
On Wed, Nov 6, 2024 at 12:46 PM Bert Gunter wrote:
> "It seems therefore that there
0 1990 1990 1990 1990 1990 1990 1990 1990 ...
> $ DOY : num 1 2 3 4 5 6 7 8 9 10 ...
> $ Ta: num -2.67 -2.77 -2.23 -2.21 -0.98 0.82 0.49 -1.02 -2.31 -3.36 ...
> $ Tmin : num -3.5 -3.7 -4.26 -2.87 -2.98 0.3 -0.83 -1.27 -3 -3.82 ...
> $ Tmax : num -1.13 -0.15 -0.13 -0.45 1
Not quite sure if I understand you.
list.files() simply returns a character vector(not a list). You can simply
use a vector index to select whatever file you wish to read. So if your
desired filename is the 5th element of filelist above, something like
myfile <- read.table(filename[5], ...)
You
But note:
> zip("hello.zip", "hello.txt")
updating: hello.txt (stored 0%)
> readChar(unz("hello.zip","hello.txt"),100)
[1] "hello"
I leave it to you and other wiser heads to figure out.
Cheers,
Bert
On Thu, Oct 24, 2024 at 8:57 AM Iris Simmons wrote:
> Hi Mikko,
>
>
> I tried running a few di
Thomas:
1. I have no knowledge of JASP
2. I am pretty sure that the answer strongly depends on what sort of
statistics one needs to do. I am also certain that no GUI out there can
come close to R's breadth and depth of capabilities.
3. There are many GUIs for R. Here's one fairly recent discussi
(only of interest -- maybe! -- to those who followed this thread of a
couple of weeks ago)
Just for the heckuva it, I compared the timing of Deepayan's unsplit(x,f)
solution to my as.vector(do.call(rbind, x)) approach to the query for a
list of 3 vectors each of length 1000 (the original toy examp
:33 AM Deepayan Sarkar
> > > wrote:
> > >>
> > >> > unsplit(x, f)
> > >> [1] 7 2 6 13 5 9 1 14 15 4 8 12 10 11 3
> > >>
> > >> Is more general (works if the subgroups are imbalanced), and hopefully
> > more
> > >> se
... And, in fact, I just realized that
c(do.call(rbind, x))
is even better.
-- Bert
On Thu, Sep 26, 2024 at 9:26 PM Bert Gunter wrote:
> Sorry, hit send by accident.
> The 2-line version is:
>
> x <- do.call(rbind, x)
> dim(x) <- NULL
>
> Cheers,
> Bert
>
Sorry, hit send by accident.
The 2-line version is:
x <- do.call(rbind, x)
dim(x) <- NULL
Cheers,
Bert
On Thu, Sep 26, 2024 at 9:23 PM Bert Gunter wrote:
> How about:
> as.vector(do.call(rbind,x))
>
> Cheers,
> Bert
>
>
>
>
> However, I much prefer a 2 li
How about:
as.vector(do.call(rbind,x))
Cheers,
Bert
However, I much prefer a 2 line version:
On Thu, Sep 26, 2024 at 8:56 PM Rolf Turner wrote:
>
> I have (toy example):
>
> x <- list(`1` = c(7, 13, 1, 4, 10),
> `2` = c(2, 5, 14, 8, 11),
> `3` = c(6, 9, 15, 12, 3))
> an
This is off topic and only tangentially related to statistics or R
(through "HARK"ing -- Hypothesizing After Results are Known). But
given the research interests of many on this list, I thought others
would enjoy it. My apologies if I have overstepped (please let me know
if so). Also, PLEASE DON'T
You might try posting on r-sig-geo if you don't get a satisfactory
response here. I assume there's a lot of expertise there on handling
raster-type data.
Cheers,
Bert
On Mon, Sep 23, 2024 at 11:31 PM javad bayat wrote:
>
> Dear R users;
> I have downloaded a grib file format (Met.grib) and I wan
"If you can get a vector with all the package names (I do not know how
to do this)..."
See ?available.packages
grep-ing the rownames of the available.packages() result appropriately
should give you a vector of the desired package names to install,
which can then be given to install.packages.
...
Well, you may have good reasons to do things this way -- and you
certainly do not have to explain them here.
But you might wish to consider using R's poly() function and a basic
nested list structure to do something quite similar that seems much
simpler to me, anyway:
x <- rnorm(20)
df <- data.fr
Hmmm... typos and thinkos ?
Maybe:
mean_narm<- function(x) {
m <- mean(x, na.rm = T)
if (is.nan (m)) NA else m
}
-- Bert
On Mon, Sep 16, 2024 at 4:40 PM CALUM POLWART wrote:
>
> Rui's solution is good.
>
> Bert's suggestion is also good!
>
> For Berts suggestion you'd make the list bit
>
need to go there and need help.
Cheers,
Bert
On Mon, Sep 16, 2024 at 11:02 AM Bert Gunter wrote:
>
> It's NA *not* Na. Details matter.
>
> Ah, but note:
> > mean(c(NA,NA), na.rm = TRUE)
> [1] NaN
>
> So if that might happen, you'll have to write your own mean fun
ncesca wrote:
>
> All' Na Is Na.
>
>
> Il lun 16 set 2024, 16:29 Bert Gunter ha scritto:
>>
>> See the na.rm argument of ?mean
>>
>> But what happens if all values are NA?
>>
>> -- Bert
>>
>>
>> On Mon, Sep 16, 2024 at 7:24 AM Fr
See the na.rm argument of ?mean
But what happens if all values are NA?
-- Bert
On Mon, Sep 16, 2024 at 7:24 AM Francesca wrote:
>
> Sorry for posting a non understandable code. In my screen the dataset
> looked correctly.
>
>
> I recreated my dataset, folllowing your example:
>
> test<-data.fr
Oh, and for correct syntax with R's |> operator, "data = ." , should
be "data = _" ; although the former seemed to work.
-- Bert
On Fri, Sep 13, 2024 at 10:45 AM Bert Gunter wrote:
>
> "Why does key = list(points = 16:17) not work? "
>
>
"Why does key = list(points = 16:17) not work? "
Because, from the "key" section of ?xyplot
" The contents of the key are determined by (possibly repeated)
components named "rectangles", "lines", "points" or "text". Each of
these must be **lists** with relevant graphical parameters (see later)
co
Ha, the proper answer!
Thanks for this, Iris. I followed up by consulting the Wikipedia "byte
order mark" entry and learned something I knew nothing about.
FWIW, if I had simply searched on t it would have immediately led
me to BOMs.
Best,
Bert
On Sat, Sep 7, 2024 at 1:30 PM Iris Simmons wrote:
Well, this is frankly an unsatisfactory answer, as it does not try to deal
properly with the issues that you experienced, which I also did. However,
it's simple and works. As this is a small text file,
simply copy it in your browser to the clipboard, and then use:
thefile <- read.table(text =
"", h
efined
> -> 0 + 0i + Inf i + NaN * i^2
> -> 0 + 0i + Inf i - NaN
> -> NaN + Inf i
>
> I am not sure how complex arithmetic could arrive at another answer.
>
> I advise against messing with infinities... use atan2() if you don't
> actually need complex arithmetic.
l division/multiplication on a
> complex number. Which R actually does for these very particular cases; but
> not when only Im(x) is Inf.
>
> Sincerely,
>
> Leonard
>
> --
> *From:* Bert Gunter
> *Sent:* Friday, September 6, 2024 1:12 AM
Perhaps
> Inf*1i
[1] NaN+Infi
clarifies why it is *not* a bug.
(Boy, did that jog some long dusty math memories :-) )
-- Bert
On Thu, Sep 5, 2024 at 2:48 PM Duncan Murdoch
wrote:
> On 2024-09-05 4:23 p.m., Leo Mada via R-help wrote:
> > Dear R Users,
> >
> > Is this desired behaviour?
> > I
What version of R are you using and on what platform?
I get:
> atan(1i)
[1] 0.7853982+Infi
> atan(1i)/5
[1] NaN+Infi
on:
R version 4.4.1 (2024-06-14)
Platform: aarch64-apple-darwin20
Running under: macOS Sonoma 14.6.1
-- Bert
On Thu, Sep 5, 2024 at 1:23 PM Leo Mada via R-help
wrote:
> Dear R
Do the "at" and "labels" components of the "scales" list argument to xyplot
not do what you want?
Cheers,
Bert
On Thu, Sep 5, 2024 at 4:05 AM Gerrit Draisma wrote:
> Dear R-helpers,
>
> In the plot below I would like to have labels at positions 2^(3*(0:10)),
> and keep the labels in the exponen
I have no clue, but I did note that you are using different versions of
BLAS/LAPACK on the different platforms. Could that be (part) of the issue?
Cheers,
Bert
On Tue, Sep 3, 2024 at 10:24 PM Iago Giné Vázquez wrote:
> Hi all,
>
> I build a dataset processing in the same way the same data in Wi
n.
>
> On August 29, 2024 2:29:16 PM PDT, Bert Gunter
> wrote:
> >Petr et.al:
> >
> >I think using merge is a very nice idea! (note that the email omitted the
> >last rows of the result, though your code of course produced them)
> >
> >The only minor
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