Thanks Rolf,
Sorry, I was trying to be coherent, but obviously failed abysmally!
I am new to this list, I will give a reproducible example next time, and
more of the relevant script.
I managed to do what I wanted to with subset:
leveneTest(subset(results,ADHES == 'Poly'|ADHES == 'Cryst')[,5],
Thanks William. I've tried the first code, ( with f0() ), but still for
n=25, m=15 , I got this:
> s<-f0(25,15)
Error in rep.int(rep.int(seq_len(nx), rep.int(rep.fac, nx)), orep) :
invalid 'times' value
In addition: Warning message:
In rep.int(rep.int(seq_len(nx), rep.int(rep.fac, nx)), orep) :
Hi
What about using the second function f1 which William suggested?
Cheers
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Maram
> SAlem
> Sent: Tuesday, October 20, 2015 11:06 AM
> To: William Dunlap; r-help@r-project.org
> Subject: Re: [R] E
Hello,
This must be a bloody R-beginner question but I just can't find an
answer in my beginner books:
I have a dataframe like this:
myframe <- data.frame (Timestamp=c("24.09.2012 09:00:00", "24.09.2012
10:00:00", "24.09.2012 11:00:00"), Event=c("0.1","0.5","1.2") )
myframe$Timestamp <- as.POS
Thanks for the hint Petr.
I'm just a little bit confused with the function f1(). could you please
help me and insert comments within f1() to be able to relate it with f0()?
Thanks a lot.
Maram Salem
On 20 October 2015 at 11:29, PIKAL Petr wrote:
> Hi
>
> What about using the second function f
Jim's solution works.
Thank you
Carol
On Monday, October 19, 2015 11:53 PM, Jim Lemon
wrote:
I think what you may have done is simply changed x.init= to x=x.init. x.init
may or may not be there when the function is called, and that is what the
warning is saying. While you have sa
Hello
Please could you help me to select the most appropriate/fastest function to use
for the following constraint optimisation issue?
Objective function:
Min: Sum( (X[i] - S[i] )^2)
Subject to constraint :
Sum (B[i] x X[i]) =0
where i=1��n and S[i] and B[i] are real numbers
Need to
Hi
Why are you confused?
> f0(5,3)
[,1] [,2]
[1,]00
[2,]10
[3,]20
[4,]01
[5,]11
[6,]02
> f1(5,3)
[,1] [,2]
[1,]00
[2,]10
[3,]20
[4,]01
[5,]11
[6,]02
>
seems to give the same result.
> res0<
Hi
Thanks for the data.
There is function na.locf in zoo package.
> library(zoo)
Attaching package: ‘zoo’
The following objects are masked from ‘package:base’:
as.Date, as.Date.numeric
new$Event <- na.locf(new$Event)
[1] 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0
On 20/10/2015 6:58 AM, Andy Yuan wrote:
> Hello
>
> Please could you help me to select the most appropriate/fastest function to
> use for the following constraint optimisation issue?
Just project S into the space orthogonal to B, i.e. compute the
residuals when you regress S on B (with no inter
> On 20 Oct 2015, at 15:35, Duncan Murdoch wrote:
>
> On 20/10/2015 6:58 AM, Andy Yuan wrote:
>> Hello
>>
>> Please could you help me to select the most appropriate/fastest function to
>> use for the following constraint optimisation issue?
>
> Just project S into the space orthogonal to B, i
Thanks a lot Petr.
I did exactly what you did and found that f1() works for n=25 and m=15. But
I just wanted to figure out how f1() works, as I used its output for this
larger code and It has been running for almost 5 hours now.
s<-f1(25,15)
simpfun<- function (x,n,m,p,alpha,beta)
{
a<-fac
Hello All,
I am trying to create an S4 class with a slot that is an array. Below is the
code I tried but I have missed something. Any suggestions are appreciated.
setClass("CashFlow",
representation(
CashFlowArray = "array"))
setMethod("initialize",
signature = "CashFlow",
function(.Object,
f0 is essentially your original code put into a function, so
expect it to fail in the same way your original code did.
f1 should give the same answer as f0, but it should use
less memory and time.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Oct 20, 2015 at 2:05 AM, Maram SAlem wrote:
>
I am running r-markdown from r-studio and can't work out how to keep
the figures.
I mean I have a few figures in the document and would like to have
them as separate pdf's too as I have been used to have them when using
Sweave.
best regards
Witold
--
Witold Eryk Wolski
__
Hi all,
I want to plot two graphs and I use this :
par(mar=c(10,6,6,6))
matplot(Simulation,type="l")
abline(h = 0, lwd = 2, col = "black")
fhist<-hist(Simulation,plot=FALSE)
par(mar=c(6,0,6,6))
barplot(fhist$counts,axes=FALSE, space=0,horiz=TRUE,col="lightgray")
The question is, that the legt
Yes Indeed William. f1() works perfectly well and took only 30 secs to
execute f1(25,15), but I wonder if there is anyway to speed up the
execution of the rest of my code (almost seven hours now) ?
Thanks for helping.
Maram Salem
On 20 October 2015 at 18:11, William Dunlap wrote:
> f0 is essen
On Tue, Oct 20, 2015 at 11:58 AM, Andy Yuan wrote:
>
> Please could you help me to select the most appropriate/fastest function to
> use for the following constraint optimisation issue?
>
> Objective function:
>
> Min: Sum( (X[i] - S[i] )^2)
>
> Subject to constraint :
>
> Sum (B[i] x X[i]) =0
>
Yes, it's the projection of S onto the subspace orthogonal to B which is:
X <- S - B%*%B / sum(B*B)
and is also implied by Duncan's solution since that is what the residuals
of linear regression are.
On Tue, Oct 20, 2015 at 1:00 PM, Paul Smith wrote:
> On Tue, Oct 20, 2015 at 11:58 AM, Andy Yu
Correction.
Yes, it's the projection of S onto the subspace orthogonal to B which is:
X <- S - (B%o%B) %*% S/ sum(B*B)
and is also implied by Duncan's solution since that is what the residuals
of linear regression are.
On Tue, Oct 20, 2015 at 1:11 PM, Gabor Grothendieck wrote:
> Yes, it's the
Hello!
Recently I am trying to transfer a large 3-dimensional array to a matrix. For
example, a array like:
, , 1
[,1] [,2]
[1,]14
[2,]25
[3,]36
, , 2
[,1] [,2]
[1,]7 10
[2,]8 11
[3,]9 12
, , 3
[,1] [,2]
[1,] 13 16
[2,] 14 17
[3,]
> x <- array(1:18, dim=c(3, 2, 3))
> x
, , 1
[,1] [,2]
[1,]14
[2,]25
[3,]36
, , 2
[,1] [,2]
[1,]7 10
[2,]8 11
[3,]9 12
, , 3
[,1] [,2]
[1,] 13 16
[2,] 14 17
[3,] 15 18
> apply(x, 3, t)
[,1] [,2] [,3]
[1,]17 13
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Glenn
> Schultz
> Sent: Tuesday, October 20, 2015 10:36 AM
> To: R Help R
> Subject: [R] Creating S4 class with a slot as an array
>
> Hello All,
>
> I am trying to create an S4 class with a slot that
Or use aperm() (array index permuation):
> array(aperm(x, c(2,1,3)), c(6,3))
[,1] [,2] [,3]
[1,]17 13
[2,]4 10 16
[3,]28 14
[4,]5 11 17
[5,]39 15
[6,]6 12 18
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Oct 20, 20
Hi,
Bill was faster than me in suggesting aperm() instead of apply(),
however, his solution is still suboptimal. Try to avoid array(), and
set the dimensions directly if possible.
fn1 <- function(x) {
apply(x, 3, t)
}
fn2 <- function(x) {
array(aperm(x, c(2, 1, 3)), c(prod(dim(
Hi bgnumis,
I'm too lazy to try to work out what "Simulation" contains, but try this:
Simulation<-sin(seq(0,6*pi,length.out=144))*5000+
2000*runif(144)+seq(8000,5000,length.out=144)
png("bb.png",width=800,height=400)
par(mfrow=c(1,2))
plot(Simulation,type="l",ylim=c(0,2))
abline(h = 0, lwd =
Hi All,
Thank you very much for your suggestion to use ddply function for my data
set. I used the function like the following
mydata <- read.table(header=TRUE, text='
cntry state city gender
1 1 1 1
1 1
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