Here's a tip for the original poster.
> ?numeric
and then follow the link it suggests
> ?double
which says amongst other things
All R platforms are required to work with values conforming to the
IEC 60559 (also known as IEEE 754) standard. This basically works
with a precision of 53
> R by default uses floating-point arithmetic, which
> is subject to problems described in [*].
Yes.
I want to note that both graphics and modern statistics, require
efficient floating point arithmetic.
So, R does what it's designed to do...
__
R-help@
Oh, that's a nice option, too. Thanks!
-Original Message-
From: Richard M. Heiberger
Sent: Tuesday, September 17, 2019 6:54 PM
To: Sabatier, Jennifer F. (CDC/DDPHSIS/CGH/DGHP)
Cc: r-help@r-project.org
Subject: Re: [R] bi-directional bar chart with a central axis
I would use the liker
I would use the likert function in the HH package
> library(HH)
> likert(my.dta)
> as.pyramidLikert(likert(my.dta))
>
See the demo
demo("likert-paper", package="HH", ask=FALSE)
for more complex examples, including the population pyramid.
We can also get the multiple coloring that your posted e
Thanks, Jim (author of plotrix!), that's a real easy way to do it!
I'll play around with options.
Jen
-Original Message-
From: Jim Lemon
Sent: Tuesday, September 17, 2019 6:43 PM
To: Sabatier, Jennifer F. (CDC/DDPHSIS/CGH/DGHP) ; r-help mailing
list
Subject: Re: [R] bi-directional ba
Hi Jennifer,
This is one way:
library(plotrix)
pyramid.plot(my.dta[,1],my.dta[,2],
labels=c("Statement 1","Statement 2","Statement 3",
"Statement 4","Statement 5","Statement 6",
"Statement 7","Statement 8","Statement 9",
"Statement 10","Statement 11","Statement 12","Statement 13"),
top.labels
Oh, I guess I just wasn't thinking of them as population pyramids since the
variables A-M are actually responses to unrelated survey questions and not age
bins. But that's silo'd thinking because why would that matter?
Thanks for the brain nudge. I'll check out the plotrix link.
Jen
-Ori
On 9/17/19 3:14 PM, Sabatier, Jennifer F. (CDC/DDPHSIS/CGH/DGHP) via
R-help wrote:
Hi R-help,
I have this data:
my.dta <-data.frame(matrix(c(
26.3, 21.4,
20.1, 13.4,
7.9,3.9,
16.5, 14.6,
5.3,3.6,
38.6, 25.6,
34.4, 21.6,
77.4, 79.5,
58.2, 56.1,
80.5, 84,
37.7, 31.9,
Your numbers are 70 bits long, R double precision numbers are 53 bits long.
You need Rmpfr to get the higher precision.
> log(569936821221962380720, 2)
[1] 68.94936
> print(569936821221962380720, digits=22)
[1] 569936821221962350592
> library(Rmpfr)
> mpfr("569936821221962380720", 70)
1 'mpfr' nu
On Wed, 18 Sep 2019 00:02:47 +0200
Martin Møller Skarbiniks Pedersen wrote:
> I know I can use gmp and R will do it correctly.
Which is equivalent to what Python does: it uses so-called long
arithmetic, allowing scalar variables with as many digits as it fits in
the computer memory. R by default
Hi R-help,
I have this data:
my.dta <-data.frame(matrix(c(
26.3, 21.4,
20.1, 13.4,
7.9,3.9,
16.5, 14.6,
5.3,3.6,
38.6, 25.6,
34.4, 21.6,
77.4, 79.5,
58.2, 56.1,
80.5, 84,
37.7, 31.9,
19.9, 28.1,
6.2,5.9
), nrow=13, ncol=2, byrow=T,
dimnames=list(c('A','B'
On 17/09/2019 6:02 p.m., Martin Møller Skarbiniks Pedersen wrote:
Hi,
I don't understand why R computes this wrong.
This is pretty well documented. R uses double precision floating point
values for these expressions, which have about 15 digit precision. I
believe for whole numbers Python
Hi,
I don't understand why R computes this wrong. I know I can use gmp and
R will do it correctly.
$ echo '569936821221962380720^3 + (-569936821113563493509)^3 +
(-472715493453327032)^3' | Rscript - [1] -4.373553e+46
Correct answer is 3 and Python can do it:
$ echo
'pow(569936821221962380720,3)
On 9/17/19 2:08 PM, David Winsemius wrote:
On 9/17/19 1:35 PM, varin sacha wrote:
Many thanks David, it perfectly works.
Now, one last think.
If I want my R code here below to run let's say B=500 times and at
the end I want to get the average for the MSE_GAM and for the
MSE_MARS. How can I
On 9/17/19 1:35 PM, varin sacha wrote:
Many thanks David, it perfectly works.
Now, one last think.
If I want my R code here below to run let's say B=500 times and at the end I
want to get the average for the MSE_GAM and for the MSE_MARS. How can I do that
?
The `replicate` function is desi
Many thanks Ana, it perfectly works.
Le mardi 17 septembre 2019 à 22:12:35 UTC+2, Ana PGG a
écrit :
Dear Varin Sacha,
My guess to try to help you is the following:
I think you may want to change this:
y_obs <- rnorm(n*0.9, y_model, 0.1) + rnorm(n*0.1, y_model, 0.5)
for:
y_obs <
Many thanks David, it perfectly works.
Now, one last think.
If I want my R code here below to run let's say B=500 times and at the end I
want to get the average for the MSE_GAM and for the MSE_MARS. How can I do that
?
library(mgcv)
library(earth)
n<-2000
x<-runif(n, 0, 5)
z <- runif(n, 0,
It depends on what you want to do, which is likely not what you do do
You might be looking for
y_obs <- ifelse(runif(n) < .9, rnorm(n, y_model, 0.1), rnorm(n, y_model, 0.5))
-pd
> On 17 Sep 2019, at 21:48 , varin sacha via R-help
> wrote:
>
> Dear R-helpers,
>
> Doing dput(x) and dput
On 9/17/19 12:48 PM, varin sacha via R-help wrote:
Dear R-helpers,
Doing dput(x) and dput(y_obs), the 2 vectors are not the same length (1800 for
y_obs and 2000 for x)
How can I solve the problem ?
Here is the reproducible R code
# # # # # # # # # #
library(mgcv)
library(earth
Dear R-helpers,
Doing dput(x) and dput(y_obs), the 2 vectors are not the same length (1800 for
y_obs and 2000 for x)
How can I solve the problem ?
Here is the reproducible R code
# # # # # # # # # #
library(mgcv)
library(earth)
n<-2000
x<-runif(n, 0, 5)
y_model<- 0.1*x^3 -
Dear all
I am trying to find one breakpoint in my data. I use segmented package.
Here is my data
> dput(temp)
temp <- structure(list(spotreba = 0:40, sqsp = c(0, 1, 1.4142135623731,
1.73205080756888, 2, 2.23606797749979, 2.44948974278318, 2.64575131106459,
2.82842712474619, 3, 3.16227766016838
?regexp ## Search the text on "backreference" .(or websearch it: "regular
expression backreference")
-- Bert
On Tue, Sep 17, 2019 at 7:52 AM Ivan Calandra wrote:
> Thank you Bert.
> That's more like what I was looking for.
>
> Could you please tell me where I can find information on the "\\1
Thank you Bert.
That's more like what I was looking for.
Could you please tell me where I can find information on the "\\1"? This
is the part I still don't get.
Ivan
--
Dr. Ivan Calandra
TraCEr, laboratory for Traceology and Controlled Experiments
MONREPOS Archaeological Research Centre and
Mus
Thanks Jeff!
It does indeed make sense that there is no "AND" corresponding to the "|".
Ivan
--
Dr. Ivan Calandra
TraCEr, laboratory for Traceology and Controlled Experiments
MONREPOS Archaeological Research Centre and
Museum for Human Behavioural Evolution
Schloss Monrepos
56567 Neuwied, German
(For the units)
Why not simply:
sub(".*\\[(.+)\\]","\\1", headers)
Cheers,
Bert
On Tue, Sep 17, 2019 at 6:40 AM Ivan Calandra wrote:
> Thank you Ivan for your help!
>
> Your solution for the first problem is so simple I didn't even think
> about it!
> What I find weird is that "_w_|\\.csv$"
https://stackoverflow.com/questions/3041320/regex-and-operator/37692545
On September 17, 2019 6:39:13 AM PDT, Ivan Calandra wrote:
>Thank you Ivan for your help!
>
>Your solution for the first problem is so simple I didn't even think
>about it!
>What I find weird is that "_w_|\\.csv$" works as e
Thank you Ivan for your help!
Your solution for the first problem is so simple I didn't even think
about it!
What I find weird is that "_w_|\\.csv$" works as expected ("OR"), but is
there no way to combine two patterns with an "AND"?
Your solution to the second problem is actually unfortunate
On Tue, 17 Sep 2019 10:14:24 +0300
Ivan Krylov wrote:
> '\\[.*\\]'
Sorry, I forgot to take it into account that you don't want the [] in
your units, either. That's still doable, but requires so-called
look-around assertions in the regular expression:
'(?<=\\[).*(?=\\])'
This should match any c
On Tue, 17 Sep 2019 08:48:43 +0200
Ivan Calandra wrote:
> CSVs <- list.files(path=..., pattern="\\.csv$")
> w.files <- CSVs[grep(pattern="_w_", CSVs)]
>
> Of course, what I would like to do is list only the interesting files
> from the beginning, rather than subsetting the whole list of files.
29 matches
Mail list logo
