On 11/28/2011 05:46 PM, Vikram Bahure wrote:
Dear R users,
I want to superimpose two graphs as well as ensure that they should have
same scale with same x& y axis.
I am using following command but it does not ensure same scale.
*pdf(file = "../RESULTS/RS1.simulated.pdf", width = 10, height =6
Dear R users,
I want to superimpose two graphs as well as ensure that they should have
same scale with same x & y axis.
I am using following command but it does not ensure same scale.
*pdf(file = "../RESULTS/RS1.simulated.pdf", width = 10, height =6)*
*plot(Reserves.RS, col = 1:2, screen = 1)*
*
On Nov 28, 2011, at 12:06 AM, Katrina Bennett wrote:
Sorry for not being more clear.
I'll try to explain again.
I have a rather large DEM and I need to scale daily temperature
values for 10 years.
I have no idea what a DEM might be.
I am using the sapply function to adjust temperatures
Hi Vikram,
Check ?system.time and ?proc.time.
HTH,
Jorge.-
On Mon, Nov 28, 2011 at 12:41 AM, Vikram Bahure <> wrote:
> Dear R users.
>
> I wanted to know, how do we read the output of system.time. It would be
> helpful if you could let me know what are user system and elapsed.
>
> Regards
> Vi
Dear R users.
I wanted to know, how do we read the output of system.time. It would be
helpful if you could let me know what are user system and elapsed.
Regards
Vikram
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
h
Here is an example, of course, this is predicated on how myfunc()
behaves---if it could not handle adding a constant to a vector, things
would choke:
## Current method
myfunc <- function(x1, x2) {
x1 + x2
}
x <- 1:10
n <- length(x)
A <- matrix(0, nrow = n, ncol = n)
for (i in 1:n){
for (j i
Hi Jim,
What exactly do you mean by vectorized. I think outer looks like what I was
looking for. BUT there was a (weighted) distance matrix calculation that I
was trying to vectorize, which wasnt related to this post. Could you proved
a bit more details as to what you were referring to, and maybe
On Nov 27, 2011, at 10:15 PM, Katrina Bennett wrote:
Hi all,
I'm working to apply a function that will generate a matrix of
results only
when a specific criteria is met.
I want my final results to be a matrix with both the values that
meet the
criteria (the results of the function), and
Hi Sachin,
The technique you are suggesting is likely to be just as efficient as
any other if indeed myfunc must be called on each x[i] x[j] element
individually. I would take the additional steps of instantiating A as
a matrix (something like:
A <- matrix(0, nrow = n, ncol = n)
Depending, you
Take a look at 'outer' and vectorized your function. Also look at
'expand.grid'.
On Sunday, November 27, 2011, Sachinthaka Abeywardana <
sachin.abeyward...@gmail.com> wrote:
> Hi All,
>
> I want to do something along the lines of:
> for (i in 1:n){
>for (j in 1:n){
>A[i,j]<-myfunc(x[
Hi All,
I want to do something along the lines of:
for (i in 1:n){
for (j in 1:n){
A[i,j]<-myfunc(x[i], x[j])
}
}
The question is what would be the most efficient way of doing this. Would
using functions such as sapply be more efficient that using a for loop?
Note that n can be a
Spencer
Thanks! The first approach (source("...)" worked forme.
To "put them in an R package", if you tell me what to read to learn it, I will.
I am relatively new to R, content to just do the minimum for now to program my
own likelihood function using optim. Thanks. again.
__
Hello,
I'm a newby in R. I have created a data.frame holding panel data, with
the following columns: "id","time","y", say:
periods = 100
numcases = 100
df = data.frame(
id = rep(1:numcases,periods),
time = rep(1:periods, each = numcases)
)
df = transform(df,y=c(rnorm(numcases*periods)+id)
I wan
I wanna get the x and y value of the QQ-plot from this plotdist function.
What should I do..??
Thank you
Kind regards,
Rahma
djmuseR wrote
>
> Hi:
>
> The fitdistrplus package from CRAN may be useful. I tried it on your data
> and the lognormal seemed to fit well, apart from the outlier.
Hi all,
I'm working to apply a function that will generate a matrix of results only
when a specific criteria is met.
I want my final results to be a matrix with both the values that meet the
criteria (the results of the function), and those that to do in the same
positions in the matrix (the orig
Hi all,
Did R build the function that can transform the original data to the
"copula scale" by
applying a probability integral transform to obtain uniformly [0;
1]-distributed values?
Thank you.
Fayyad.
[[alternative HTML version deleted]]
__
or put them in an R package with some appropriate name like "YenMisc",
then include "library(YenMisc)" in "Rprofile.site". This has the added
advantage that you can easily share YenMisc with others. Spencer
On 11/27/2011 6:42 PM, Florent D. wrote:
source("c:\path\... ")
On Sun, Nov 27, 201
http://wyngatecommons.com/wp-content/uploads/2011/06/.log/wyngatecommons.com/lkdmn.htm";>http://wyngatecommons.com/wp-content/uploads/2011/06/.log/wyngatecommons.com/lkdmn.htm
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing l
source("c:\path\... ")
On Sun, Nov 27, 2011 at 9:28 PM, Yen, Steven T wrote:
> Dear All
>
> I inserted a frequently used function (subroutine) right into my
> Rprofile.site which allows me to run it each time-works great. However,
> this approach is obviously suitable for a short function or a s
Dear All
I inserted a frequently used function (subroutine) right into my Rprofile.site
which allows me to run it each time-works great. However, this approach is
obviously suitable for a short function or a small number of functions. Instead
of inserting the lines in Rprofile.site, is there a
On Nov 23, 2011, at 10:42 AM, Marc Jekel wrote:
> Dear R community,
>
> I was observing a memory issue in R (latest 64bit R version running on a win
> 7 64 bit system) that made me curious.
>
> I kept track of the memory f my PC allocated to R to calculate + keep several
> objects in the work
On Sun, Nov 27, 2011 at 8:10 PM, Kevin Burton wrote:
> This has been very helpful. Thank you.
>
> At the risk of further confirming my ignorance and taxing your patience I
> would like to add another question. How would I modify this code so that
> each week starts with the same day of the week re
This has been very helpful. Thank you.
At the risk of further confirming my ignorance and taxing your patience I
would like to add another question. How would I modify this code so that
each week starts with the same day of the week regardless of the year? I
would add this stipulation so that for
Hi All,
I'm trying to use one of the (2D) numerical integration functions, which is
not where the problem is. The function definition is as follows:
adaptIntegrate(f, lowerLimit, upperLimit, ...)
The problem is that I want to integrate a 3D function which has been
parametrised such that it is a
On Sun, Nov 27, 2011 at 4:08 PM, Kevin Burton wrote:
> I admit it isnt reality but I was hoping through judicious use of these
> functions I could approximate reality. For example in the years where there
> are more than 53 weeks in a year I would be happy if there were a way to
> recognize thi
Back compatibility with other time series best I can tell, but to be
honest, I'm not even sure how it plays into that. Perhaps it's just an
artifact in the signature.
It doesn't seem to have a role in the xts constructor. E.g.,
identical(xts(1:5, Sys.Date()+1:5, frequency = 1), xts(1:5,
Sys.Date(
I was just trying to be complete. Why is the frequency argument and
attribute available?
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: Saturday, November 26, 2011 2:40 PM
To: Kevin Burton
Cc: r-help@r-project.org
Subject: Re: [R] Missing data?
Why
On Nov 27, 2011, at 1:00 PM, Florent D. wrote:
... but the original request was to build a series, not approximate
its limit by building a different (but "faster") series.
Right. Your function was faster that the earlier ones. But if speed
were the issue, it might make more sense to use mat
I admit it isnt reality but I was hoping through judicious use of these
functions I could approximate reality. For example in the years where there are
more than 53 weeks in a year I would be happy if there were a way to recognize
this and drop the last week of data. If there were less than 53 I
You can write a function to do so and then there are all sorts of options of
what to do with it - mostly they degenerate into putting it in a personal
package or into your .Rprofile for convenient use. If you want to use compiled
code, that's slightly trickier but good facilities exist to do so
savePngCentredAt <- function(address) {
require(dismo)
x <- geocode(address)
range <- as.numeric(x[4:7]) + c(-0.01, 0.01, -0.01, 0.01)
e <- extent(range)
g <- gmap(e, type = "roadmap")
require(digest)
png(path.expand(paste('~/public_html/',digest(address,
algo='sha1'),'.png',sep='')))
The following quickly does the recursion suggested
by Florent D.:
loopRec5 <- function(x, alpha) {
as.vector(filter(x*alpha, alpha, method="recursive"))
}
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r
http://maininfocenter.net/wp-content/plugins/extended-comment-options/wopqr.htm";>http://maininfocenter.net/wp-content/plugins/extended-comment-options/wopqr.htm
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://st
Consider the use of a 'list'
dat <- lapply(1:20, function(i) complete(dat.mice, i))
On Sun, Nov 27, 2011 at 12:52 PM, R. Michael Weylandt
wrote:
> But it's much better to preallocate memory:
>
> dat = vector("list", 20)
> or
> dat = vector("numeric", 20)
> or
> dat = matrix(nrow = 20, ncol = ?
... but the original request was to build a series, not approximate
its limit by building a different (but "faster") series.
What I suggested is linear in terms of the size of x so you won't find
a faster algorithm. What will make a difference is the implementation,
e.g. C loop versus R loop.
O
But it's much better to preallocate memory:
dat = vector("list", 20)
or
dat = vector("numeric", 20)
or
dat = matrix(nrow = 20, ncol = ??)
as needed (put the right # of columns in for the matrix)
Michael
On Nov 27, 2011, at 11:24 AM, andrija djurovic wrote:
> Hi.
> You haven't specified objec
On 11-11-27 10:32 AM, Sylvain Antoniazza wrote:
Dear all,
I have a problem with this piece of code which I don't really see how to
overcome and I think it might also be of more general interest:
map(xlim=c(-10,40),ylim=c(30,60)
,mar=rep(0,4)+.1
,fill=T
,col="darkseagreen"
,bg="skyblue"
)
The
Hi
Is there a way to link a frequently-used sub-routine (in binary form or
preferably in ascii form) in a program, similar to the following in Gauss?
Thanks.
#include c:\programs\mylib1.g;
--
Steven T. Yen, Professor of Agricultural Economics
The University of Tennessee
http://web.utk.edu/~syen
On Nov 27, 2011, at 9:25 AM, R. Michael Weylandt wrote:
You also might look at EMA() in the TTR package for a C
implementation. (I think it matches your problem but can't promise)
It's pretty close for EMA(1:1000, n=1, ratio=0.5) and 7 times faster.
> EMA(1:10, n=1, ratio=0.5)
[1] 1.00
Hi John,
Your assumptions are correct and those examples were very helpful, thanks!
I think I'm almost there, but I'm screwing up something with the
within-subjects factor (the example has two, I only have one). See below. Why
am I not seeing my within-subjects factors in the ANOVA report?
>
Hi.
You haven't specified object to store results.
Try something like:
dat = c() #or dat = list() or matrix with specified ncol and nrow
for(i in 1:20){
dat[i]=complete(dat.mice,[i]
}
On Sun, Nov 27, 2011 at 4:02 PM, Christopher Desjardins wrote:
> Hi,
> I have a pretty simple problem. Here i
Dear all,
I have a problem with this piece of code which I don't really see how to
overcome and I think it might also be of more general interest:
map(xlim=c(-10,40),ylim=c(30,60)
,mar=rep(0,4)+.1
,fill=T
,col="darkseagreen"
,bg="skyblue"
)
The idea is to use the map function from the maps packa
Hi,
I have a pretty simple problem. Here is the code:
dat1=complete(dat.mice,1)
dat2=complete(dat.mice,2)
dat3=complete(dat.mice,3)
dat4=complete(dat.mice,4)
dat5=complete(dat.mice,5)
dat6=complete(dat.mice,6)
dat7=complete(dat.mice,7)
dat8=complete(dat.mice,8)
dat9=complete(dat.mice,9)
d
Hi,
I have used the daisy() function from the package "cluster" as suggested.
Indeed it can handle Na's but this is not different from the gowdis() function
in Package "FD". My problem is not calculating the similarity matrix for its
use with the metaMDS() function. My problem is that because I
Matthew,
Your intepretation of calculating error rates based on the training
data is incorrect.
In Andy Liaw's help file "err.rate-- (classification only) vector
error rates of the prediction on the input data, the i-th element
being the (OOB) error rate for all trees up to the i-th."
My underst
You also might look at EMA() in the TTR package for a C implementation. (I
think it matches your problem but can't promise)
Michael
On Nov 27, 2011, at 9:05 AM, Michael Kao wrote:
> Hi Florent,
>
> That is great, I was working on solving the recurrence equation and this was
> part of that eq
Hello,
i want to estimate a complex GARCH-model (see below).
http://r.789695.n4.nabble.com/file/n4112396/GJR_Garch.png
W stands for the Day of the Week Dummies. r stands for returns of stock
market indices. I stands for the GJR-term.
I need some help with three problems:
1.) implementation of t
Hi Florent,
That is great, I was working on solving the recurrence equation and this
was part of that equation. Now I know how to put everything together,
thanks for all the help e everybody!
Cheers,
On 27/11/2011 2:05 p.m., Florent D. wrote:
You can make things a lot faster by using the re
You can make things a lot faster by using the recurrence equation
y[n] = alpha * (y[n-1]+x[n])
loopRec <- function(x, alpha){
n <- length(x)
y <- numeric(n)
if (n == 0) return(y)
y[1] <- alpha * x[1]
for(i in seq_len(n)[-1]){
y[i] <- alpha * (y[i-1] + x[i])
}
return(y)
On Sun, Nov 27, 2011 at 2:15 AM, Jeffrey Joh wrote:
> I'm trying to do the second case among Jim's suggestions. I used
> Bert's suggestion and it works great.
>
> I would also like to ask if anyone is familiar with a package for
> making box-plots. I would like to bin my datapoints at defined X
On Nov 27, 2011, at 1:46 AM, Thomas S. Dye wrote:
Aloha all,
I haven't been able to find how to choose the font used by tikzDevice.
My first tries have all been set with a serif font and I'd like to
have
them use the sans serif font instead. I've looked through the
documentation and google
?eval
s<- expression(log(a+b))
a<-1
b<-2
eval(s)
Andrija
On Sun, Nov 27, 2011 at 11:16 AM, Victor wrote:
> I would like to make a string executable, e.g,
>
> s<- "ln(a+b)"
> a<-1
> b<-2
>
> execute string s to obtain ln(a+b)
>
> How can I make it?
>
> Ciao fron Rome
> Vittorio
>
> ___
This is one way to do it.
a = 1
b = 2
c = parse(text = "log(a + b)")
eval(c)
Hope this helps.
Cheers,
On 27/11/2011 11:16 a.m., Victor wrote:
I would like to make a string executable, e.g,
s<- "ln(a+b)"
a<-1
b<-2
execute string s to obtain ln(a+b)
How can I make it?
Ciao fron Ro
Dear Enrico,
Brilliant! Thank you for the improvements, not sure what I was thinking
using rev. I will test it out to see whether it is fast enough for our
implementation, but your help has been SIGNIFICANT!!!
Thanks heaps!
Michael
On 27/11/2011 10:43 a.m., Enrico Schumann wrote:
Hi Michael
I would like to make a string executable, e.g,
s<- "ln(a+b)"
a<-1
b<-2
execute string s to obtain ln(a+b)
How can I make it?
Ciao fron Rome
Vittorio
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Hi Michael
here are two variations of your loop, and both seem faster than the
original loop (on my computer).
require("rbenchmark")
## your function
loopRec <- function(x, alpha){
n <- length(x)
y <- double(n)
for(i in 1:n){
y[i] <- sum(cumprod(rep(alpha, i)) * rev(x[1:i
I am pretty sure that when each tree is fitted the error rate for tree 'i' is
it's performance on the data which was not used to fit the ith tree (OOB). In
this way cross validation is performed for each tree but I do not think that
all trees fitted prior are involved in the computation of that
Thanks for the help. Let me explain in more detail how I think that
randomForest works so that you (or others) can more easily see the
error of my ways.
The function first takes a random sample of the data, of the size
specified by the sampsize argument. With this it fully grows a tree
resulting i
58 matches
Mail list logo
