> Personally I much prefer backwards compatibility to political correctness.
I agree with Rolf, here.
And as someone that's planning to write a Linux Terminal Emulator, in
the medium-term future, I *strongly* defend this approach.
And to the original poster.
Haven't you seen The Matrix?
(Second
On 18/09/19 6:00 PM, Benjamin Lang wrote:
Dear R project,
I have a very simple question:
How, in late 2019, is there an option called "--slave" to "make R run as
quietly as possible"?
Let me reiterate that it is 2019, i.e. "The Future", rather than 1970 when
R was presumably developed,
I think there is no confusion except in the minds of those with nothing better
to do. I agree with Antirez, quoted in [1], which nevertheless indicates that
perspective lost the debate.
Any accurate alternative notation will have similar connotations because in
fact the "slave" side of that
Hi Edward,
Say your "data frame" is named "epdat". This may do it:
epmat<-matrix(epdat[10:289],nrow=28)
colnames(epmat)<-sub("1","",names(epdat[10:289])[seq(1,270,by=28)])
This one looks like the Sorceror's Apprentice tangled with one of
those experimental schedule scripting programs.
Jim
On
Hi R Help,
How would I convert the data below so that I have it formatted with trials
along the rows and then each type of measure separately? e.g.,
Subject RT OnOff Feedback
Trial_1
Trial_2
Trial_3
Trial_4
Thanks!
Edward
structure(list(TAP_ID = "967372 ", TAP_Date =
I have recently installed R 3.6.1 on my MacBook running Mac OS Mojave 10.14.5
in order to run a custom package called MH1823 (probability of detection).
This package requires me to load the following packages from CRAN, tcltk,
tcltk2, survival and xlsx.
I have installed survival, xlsx and
Ar ! What a pity, I have not seen it ! Many thanks !
Le mercredi 18 septembre 2019 à 19:07:42 UTC+2, peter dalgaard
a écrit :
Redefining n is probably not a good idea...
[...snip...]
> m <-runif(n, 0, 5)
> n <-rnorm(n, 2, 3)
Oops! n is now a vector of length 2000.
[...snip...]
For what it's worth, this is an ongoing conversation in computer
science and engineering. And has been so for decades.
Not R, but related to this it's only in the past few months that a
fork of the photo-manipulation software GIMP (slur for handicapped)
renames it (GLIMPSE).
Note, I am not
Dear R project,
I have a very simple question:
How, in late 2019, is there an option called "--slave" to "make R run as
quietly as possible"?
Let me reiterate that it is 2019, i.e. "The Future", rather than 1970 when
R was presumably developed, based on its atrocious syntax, documentation
and
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 <- c( rnorm(n*0.9, y_model, 0.1), rnorm(n*0.1, y_model, 0.5) )
then y_obs:
> length(y_obs)
[1] 2000
De: varin
Redefining n is probably not a good idea...
[...snip...]
> m <-runif(n, 0, 5)
> n <-rnorm(n, 2, 3)
Oops! n is now a vector of length 2000.
[...snip...]
> y_obs <- y_model +c( rnorm(n*0.97, 0, 0.1), rnorm(n*0.03, 0, 0.5) )
now length(n*0.97) == 2000 > 1, so rnorm(n*0.97, ...) gets you a
Dear Peter Dalgaard,
Really appreciated, but my code does not work. There is still a problem ! Here
below the reproducible example with 20 variables
library(mgcv)
library(earth)
n<-2000
x<-runif(n, 0, 5)
z <- rnorm(n, 2, 3)
a <- runif(n, 0, 5)
b <- rnorm(n, 2, 3)
c <- runif(n, 0, 5)
d
That worked! Thanks for the explanation.
On Wed, Sep 18, 2019 at 10:06 AM Duncan Murdoch
wrote:
>
> On 18/09/2019 8:43 a.m., Huzefa Khalil wrote:
> > Hello R-users,
> >
> > I have been running a script which produces objects based on the
> > column names of a data.frame. The column names are of
On 18/09/2019 8:43 a.m., Huzefa Khalil wrote:
Hello R-users,
I have been running a script which produces objects based on the
column names of a data.frame. The column names are of the form CB_1-1,
CB_1-2, etc. Now this calculation was rather long and memory
intensive, so I would rather not have
Hello R-users,
I have been running a script which produces objects based on the
column names of a data.frame. The column names are of the form CB_1-1,
CB_1-2, etc. Now this calculation was rather long and memory
intensive, so I would rather not have to do it again after fixing the
column names
A little note on quoting in regular expressions.
I find writing \\. when I want a quoted . somewhat confusing,
so I would use the pattern "_w_.*[.]csv$".
Better still, if you want to match file names,
there is a function glob2rx that converts shell ("glob")
patterns into regular expression
> 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
Hi Cristofer,
If you just ask for a number of breaks, you will get what "hist"
thinks you should. Try this or something similar:
hist(x,breaks=seq(min(x),max(x),length.out=21))
Jim
On Wed, Sep 18, 2019 at 8:55 PM Christofer Bogaso
wrote:
>
> Hi,
>
> I have a numerical vector as below
>
> x =
Hi,
I have a numerical vector as below
x = c(92958.2014593977, -379826.025677203, 881937.411562002, 25761.5278163719,
-11837.158273897, 48450.8089746788, -415505.62910869, -168462.98512054,
328504.255373387, -298966.051027528, 237133.794811816, -49610.1148173768,
-92459.1170329526,
Um, I think not... The mean of the last 200 observation won't line up with the
x and z.
Possibly, if what you want is the last 200 obs to have a different variance,
y_obs <- y_model + c(rnorm(0.9 * n, 0, 0.1), rnorm(0.1 * n, 0, 0.5))
or
y_obs <- rnorm(n, y_model, rep(c(0.1, 0.5), c(.9 * n,
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
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