Thanks in advance R
users
I have time series data
and I need to estimate the parameters involved in three different models for
generalized extreme values
Model 1: a, b,
c are constants.
Model 2: a(t)=B0+B1 t,
but b, c are constants
Model 3: c(t)=
Exp(B0+B1 t) but a, b are
Well, then, you had better get busy and stop posting here. To learn why, read
the Posting Guide. Some pointers:
a) No homework help here.
b) No posting in HTML.
c) This list is for questions about R, not statements about your needs.
I am no HPC expert, but I have been computing for awhile.
There are already many CPU-specific optimizations built into most compilers
used to compile the R source code. Anyone sincerely interested in getting work
done today should get on with their work and hope that most of the power of new
Hi all,
I am having trouble loading a ff object previously saved in a different
computer. I have both files .ffData and .RData, and the first of them is
13Mb large from which I know the data is therein. But when I try to ffload
it,
checkdir error: cannot create /home/_myUser_
Hello,
I would like to get an advice on how the notorious eval(parse()) construct
could possibly
be avoided in the following example. I have an array x, which can have
different number of dimensions,
but I am only interested in extracting, say, the first element of the first
dimension.
1. Please always cc. the list; do not reply just to me.
2. OK, I see. I ERRED. Had you cc'ed the list, someone might have
pointed this out. The correct example reproduces what you saw.
z- sample(1:10,30,rep=TRUE)
table(z)
w - data.frame(table(z))
w
z Freq
1 12
2 23
3 31
Mike:
On Sat, May 25, 2013 at 7:24 PM, C W tmrs...@gmail.com wrote:
Thomas, thanks for the cool trick. I always thought browser() was the
only thing existed, apparently not.
Which you would have known had you read the docs!
See section 9 on Debugging in the R Language Definition Manual
--
Le dimanche 26 mai 2013 à 13:53 +0200, Djordje Bajic a écrit :
Hi all,
I am having trouble loading a ff object previously saved in a different
computer. I have both files .ffData and .RData, and the first of them is
13Mb large from which I know the data is therein. But when I try to ffload
On 26-05-2013, at 15:56, Martin Ivanov tra...@abv.bg wrote:
Hello,
I would like to get an advice on how the notorious eval(parse()) construct
could possibly
be avoided in the following example. I have an array x, which can have
different number of dimensions,
but I am only interested in
Martin:
Well, assuming I understand, one approach would be to first get the
dim attribute of the array and then create the appropriate call using
that:
z - array(1:24,dim=2:4)
d - dim(z)
ix -lapply(d[-c(1,1)],seq_len)
do.call([, c(list(z),1,1,ix))
[1] 1 7 13 19
Is that what you want?
--
library(abind)
asub(x, 1, 1)
On Sun, May 26, 2013 at 10:43 AM, Bert Gunter gunter.ber...@gene.com wrote:
Martin:
Well, assuming I understand, one approach would be to first get the
dim attribute of the array and then create the appropriate call using
that:
z - array(1:24,dim=2:4)
d -
Hi,
Is it possible to :
[1] set a default location to plot graphs in png format with specific
dimensions resolution. I want to plot to a directory which is a shared on
the network (samba share), so as to view the plots from a different machine.
[2] call dev.off() 'automagically' after a call
the code returns Error:
do.call([, c(list(z),1,1,ix))
#Error in 1:24[1, 1, 1:3, 1:4] : incorrect number of dimensions
May be something is missing.
A.K.
- Original Message -
From: Bert Gunter gunter.ber...@gene.com
To: Martin Ivanov tra...@abv.bg
Cc: r-help@r-project.org
Sent:
Hi,
You could use:
library(abind)
#using Berend's and Bert's example
x1 - array(runif(9),dim=c(3,3))
x2 - array(runif(8),dim=c(2,2,2))
z - array(1:24,dim=2:4)
#applying your code:
eval(parse(text=paste0(x1[1,paste(rep(,,length(dim(x1))-1),collapse=),])))
#[1] 0.6439062 0.7139397 0.6017418
---BeginMessage---
Dear Mr Gunter,
with a slight correction:
z - array(1:24,dim=2:4)
d - dim(z)
ix - lapply(d[-1], seq_len)
do.call([, c(list(z),1,ix))
[,1] [,2] [,3] [,4]
[1,]17 13 19
[2,]39 15 21
[3,]5 11 17 23
your suggestion worked and is exactly what
I think this is mostly but not fully correct.
most users are better off with double precision most of the time...but
not all of the time if the speedup and memory savings are 4 and 2,
respectively.
algorithm inefficiency may well be true, too---but if I spend one week
of my time (or even 3 days)
Dear Arun,
Thank You very much, your suggestion also works and seems to also be more
convenient.
I think though, that Mr Gunter's suggestion should be more efficient, as it
uses directly the Extract operator.
Thank You all very much for Your responsiveness. R does have a wonderful
Hi,
Another way would be:
library(arrayhelpers)
slice(aperm(x1,c(2,1)),j=1)
#[1] 0.6439062 0.7139397 0.6017418
slice(aperm(x2,c(2,1,3)),j=1)
# [,1] [,2]
#[1,] 0.026671344 0.2116831
#[2,] 0.003903368 0.1551140
slice(aperm(z,c(2,1,3)),j=1)
# [,1] [,2] [,3] [,4]
#[1,] 1
colnames(dd)
#[1] col1 colb
null_vector- colnames(dd)
sapply(null_vector,makeNull,dd)
# col1 colb
#[1,] NA 4
#[2,] 2 NA
#[3,] 3 2
#[4,] 4 NA
#[5,] 1 4
#[6,] NA 5
#[7,] 1 6
A.K.
I am trying to make a column value in a dataframe = NA if there is a 0
or
Hi,
see the R.devices package
[http://cran.r-project.org/web/packages/R.devices/]. FYI, there is a
vignette [R.devices-overview.pdf], but for some reason it's hard find.
However it is there: help.start() - R.devices - 'User guides,
package vignettes and other documentation.' -
Yes, for the record, the typo in my earlier post is corrected below.
(Martin's previous correction both corrected and slightly changed
what I provided).
-- Bert
On Sun, May 26, 2013 at 7:43 AM, Bert Gunter bgun...@gene.com wrote:
Martin:
Well, assuming I understand, one approach would be to
Sorry, forgot to add: Hi, [a somewhat different approach but] see the
R.devices package /Henrik
On Sun, May 26, 2013 at 10:14 AM, Henrik Bengtsson h...@biostat.ucsf.edu
wrote:
Hi,
see the R.devices package
[http://cran.r-project.org/web/packages/R.devices/]. FYI, there is a
vignette
Hello Bert.
I didn't reply to the list because i forgot. I hit reply instead of reply
all
Thanks for your example.
I understood now that i was trying to do something that didn't made sense and
that was why it failed.
I should have used an histogram do do a graph of the frequency of each
Patrick Likongwe patricklikongwe at yahoo.co.uk writes:
Dear Team,
Help me out here. I have managed to run a Geographically Weighted
Regression in R with all results coming up. The problem now comes in
mapping the parameter estimates and the t values that are significant in
the model. My
i *SOLVED*
Thanks Milan, I have receivd some feedback externally to the list and
managed to solve the issue.
I saved the document as follows:
x.Arr - ff(NA, dim=rep(ngen,3), vmode=double)
ffsave (x.Arr, file=x.Arr)
finalizer(x.Arr) - delete
The problem was related to the rootpath argument. As
How's this:
big.gap = diff(test) 1
cbind(test[c(TRUE, big.gap)], test[c(big.gap, TRUE)])
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Lizzy Wilbanks
Sent: Tuesday, 14 May 2013 1:18p
To: r-help@r-project.org
Subject: [R]
26 matches
Mail list logo
