> I wanted to know the codes of Fourier Transformation of time series
> data.
The location of complete source code for all versions of R is given at
https://cran.r-project.org/sources.html
fft is in the stats package and is implemented as a call to C code, so you'll
need to look in the C source
> In doing meta-analysis of diagnostic accuracy I produce ellipses of confidence
> and prediction intervals in two dimensions. How can I calculate the area of
> the ellipse in ggplot2 or base R?
There are established formulae for ellipse area, but I am curious: in a 2-d
ellipse with different qu
Kevin,
I didn't have the data set (you might want to post a link to a downloadable
file instead), but on a quick look at the code your foo1 function looks as if
it is not guaranteed to return an array of the same length each time. It's
testing for nonzero coefs in a fitted model and then droppi
> Whether or not you agree with my reason for doing so, my question was
> how to contact the creator of the "colors" function. If you do not have advice
> on this, please refrain from weighing in.
... which is precisely what Prof Gunter said.
However, it is worth consulting the documentation
> Patrick (Malone Quantitative)
> There's only one version and it's free.
You're forgetting Microsoft R, formerly Revolution R
(https://mran.revolutionanalytics.com/download), which comes in open and
enterprise flavours, both with multithreading and the latter with beefed up
database access ser
Ivan Krylov [krylov.r...@gmail.com] said:
> Instead you can either close over X:
>
> X <- c(...)
> holly <- function(p) (p$a * X^2) / (p:b^2 + X^2)
> # function holly now "contains" the vector X
That would not be an accurate statement as written.
The function only contains an unevaluated call re
You have four character vectors in your data frame; those return non-finite
ranges.
plot(test[sapply(test, class)!="character"])
plots the non-character columns.
S Ellison
From: R-help [r-help-boun...@r-project.org] on behalf of Rasmus Liland
[j...@pos
Define a wrapper function for the t test that only returns the p-value?
by(mydata$WtBaseline,mydata$Group, function(...) t.test(...)$p.value)
S Ellison
From: R-help [r-help-boun...@r-project.org] on behalf of Sorkin, John
[jsor...@som.umaryland.edu]
Sent:
work...
____
From: PIKAL Petr [petr.pi...@precheza.cz]
Sent: 10 June 2020 07:59
To: Stephen Ellison; r-help@r-project.org
Subject: RE: [R] almost logistic data evaluation
Hi
External heating. Normally I would use TA instrumentation but for technical
reasons
> Actually „y“ is growing temperature, which, at some point, rise more rapidly
> due to exothermic reaction.
> This reaction starts and ends and proceed with some speed (hopefully
> different in each material).
Are you applying external heating or is it solely due to reaction kinetics?
Steve
lass="histogram")
}
plot(freqhist(freqs, breaks=c(4*0:5, 10*3:5, 70, 100, 140)) )
#Also works equidistant with default 0:length(counts) breaks:
f2 <- c(30, 39, 31, 29, 10, 6, 3, 1, 0, 1)
plot(freqhist(f2))
Steve E
> -Original Message-
> From: Stephen Ellison
> Se
> I have a vector like say 73,53,42,67,41,50 where these numbers are the
> number of occurrences of the data values 1,2,3,4,5,6 - so in essence I have
> the frequency bit from the hist() function. I can't see an elegant way (there
> are clearly messy workarounds like generating a vector of 73 1's,
> Sure, it's a silly example, but it makes about as much sense as using
> "slave" to mean "quiet".
It doesn't. It's a set of options chosen for when R is called as a slave
process from a controlling process, and in that it is a reasonable description
of the circumstance.
--quiet is a separate c
> When I had breaks = 18, I get total number of cells as 16, which is
> same when I put breaks = 20
>
> In the 2nd case I was expecting total number of cells (i.e. bars) as
> 20 i.e. if I understand the documentation correctly I should expect
> total number of cells (bars) should be same as breaks
Google "levionson-Durbin in R
The first reference is the levinson function in the signal package.
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Faheem
> Jan via R-help
> Sent: 04 September 2019 11:29
> To: R-help Mailing List
> Subject: [R] Durbin-
> I want to sort a DF, temp, on two columns, patid and time. I have searched
> the internet and found code that I was able to modify to get my data sorted.
> Unfortunately I don't understand how the code works. I would appreciate it
> if someone could explain to me how the code works. Among other
> However, having installed "Action of the Toes" (are R releases named
> by Culture AIs, by any chance?)
Peanuts cartoons.
https://twitter.com/snoopy/status/269963551135891456
Steve E
***
This email and any attachments are confid
> This corresponds with the info found in my global environment for the
> object indicated. Now, how do I go about determining if the dataset is a
> matrix?
Consider ?is.matrix ?
S Ellison
***
This email and any attachments are co
> Thanks for the suggestion but I'm not sure that it answers my original
> question.I need to know how many samples I need to collect to collect in
> order to estimate the sample size needed to achieve a specific margin of
> error for confidence intervals for the population variance. I'm not sure
> Thomas Subia
Could you please explain the difference between:
> any differences in variance between factors.
and
> variance differences between factors
?
***
This email and any attachments are confidential. Any use...{{drop
If you want to pass a data frame (not just its name) plus (some) column names
to your function, the easiest way is to put the _quoted_ names in a vector, as
previously posted. For example
f1 <- function(dfrm, colnames) {
print(dfrm[,colnames]) #or, if you like, for(nn in colnames)
prin
> > 3) Is there a way to assign the numbers to the treatment group that they
> were
> > originally in? For example, "1" in the result should be "3", indicating that
> > that was the temperature the experiment took place at. Pairwise.prop.test
> > doesn't seem to accept matrices with more than 2 col
May be a kludge, but it might be simpler to write your own density function for
a few specified points.
For example
my.density <- function(x, bw = 'nrd0', at) {
x<-na.omit(x)
#
#Borrowed from density.default for compatibility
if (is.character(bw)) {
if (leng
>From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On Behalf Of
>Ann [trivialesc...@gmail.com]
>Subject: [R] reading the results of a within subject test
>
>Hi, I ran the following in R (on item means):
>
> aov(VAR ~(a*b)+Error(item/(a*b)), data = item)
>
>
> I'm confused...
Depending on how critical the problem is, you might also want to look at the
literature on bootstrap CI's, perhaps starting from the references in boot.ci
in the boot package. The simple quantiles are not necessarily the most
appropriate. For example I seem to recall that BCa intervals were the
Not according to the help page.
Bit kludgy,but try something like
y <- rnorm(73, 17.2)
nbreaks <- function(x, n) {
r <- range(pretty(y))
seq(r[1], r[2], length.out=n+1)
}
hist(y, breaks=nbreaks(y,13))
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-he
Dear Uwe,
> If you think the relevant section is not sufficient, it's
> your turn to provide a short and precise suggestion for a
> change, preferable a diff against the R-devel sources. That
> way you may be able to convince an R Core member to work on it.
I would be more than happy to rearra
> -Original Message-
>
>
> You can install to any library, even a temporary one that
> does not need to be in .libPaths() of your regular R
> installation. So you won't do any harm to your sytsem.
>
The principal point I was making is that "the manuals" are not very informative
on
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Uwe Ligges
> Sent: 17 June 2011 12:04
> >> See the manual. It tells you
> >>
> >> R CMD INSTALL --build
> >>
> >> will generate a binary package.
> >
...
>> it seems I am not
You can plot explicitly over a range of x; for example
x<-seq(10, 15, 0.1)
#for mean 12.5, sd 0.6
plot(x, pnorm(x, 12.5, 0.6), type="l", ylim=c(0,1)
Or you can try the default plot for a univariate function (see ?curve)
plot(function(x) pnorm(x, 12.5, 0.6), xlim=c(10,15))
#Note use of t
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