Thank you both for your help. If I have any further questions after I give
this a go I'll get back to you. This was much appreciated.
M.
--
View this message in context:
http://r.789695.n4.nabble.com/Randomized-Points-in-space-saving-model-results-tp3922612p3954834.html
Sent from the R help ma
Ok, to michael
Looking over what you posted I'm thinking this is the way forward but my
lack of experience using functions is getting in the way of accomplishing
the task. Examining the outputs into the functions and putting in my own
info is giving me a headache. Here's my thoughts at each step
I am trying to compute the approximate numerical integration of the following
expression using the integrate function:
> (integrate(function(x) {log(1+x^2)*(1+x^2)^(-20.04543)},low,Inf)$val)
Error in integrate(function(x) { : the integral is probably divergent
which gives me an error. If
This post has NOT been accepted by the mailing list yet.
Hi,
I'm passing the input parameters and calculating the linear regression.
some time i'm getting the error
Error in dim(data) <- dim : attempt to set an attribute on NULL.
But when i restart my R studio, then it runs.
Please help to res
Hello,
I have a for loop that generates data in R. With the IML program, I would
like to analyze data in SAS from each iteration of the for loop in R. It
would be helpful if someone could explain to me how to analyze data this
way.
Thanks
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Hi Smart Guy,
You can try use.missings = TRUE. I am not sure how well this will
work. If you really need it and use.missings = TRUE does not work, I
would just read it in as a list, and then maybe create a new data
frame based on the list. See ?as.data.frame for a way to do this.
Then you could
Hi All,
Here is my problem.
I need to get data frame (and not a list) while reading spss file using
read.spss. For which I can put one parameter of read.spss as
to.data.frame=TRUE.
Now in doing so, I dont get missing values.
Is there a way to get missings values along with data frame?
T
Rainer wrote:
* On Thu, Oct 20, 2011 at 12:22 AM, Norm Matloff
* wrote:
*
* >
* > I've developed a new R debugging tool, debugR, available at
* > http://heather.cs.ucdavis.edu/debugR.html
* >
* > This basically replaces my edtdbg, which I will no longer be
* > supporting.
* > The new to
try this:
> M <- matrix(c("1","2","3","2","4","5","5","3","2","1","3","2","4","4"),
> ncol=2)
> M
[,1] [,2]
[1,] "1" "3"
[2,] "2" "2"
[3,] "3" "1"
[4,] "2" "3"
[5,] "4" "2"
[6,] "5" "4"
[7,] "5" "4"
> # not the most efficient
> M.sorted <- apply(M, 1, function(x) paste(sort(x), co
Dear list,
Suppose I have the following matrix:
> M <-
matrix(c("1","2","3","2","4","5","5","3","2","1","3","2","4","4"), ncol=2)
> M
[,1] [,2]
[1,] "1" "3"
[2,] "2" "2"
[3,] "3" "1"
[4,] "2" "3"
[5,] "4" "2"
[6,] "5" "4"
[7,] "5" "4"
In this matrix, row 1 contains elements "1" an
dear R experts. about once a year, I wonder where GPU computing is. this
is of course a quickly moving field, so I am hoping to elicit some advice
from those that have followed the developments more actively. fortunately,
unlike my many other cries for help, this post is not critical. it is mor
Hi,
Regarding, ..., you literally just pass For example:
f <- function(d, ...) {
lapply(d, mean, ...)
}
f(1:10) # but you could pass other arguments to lapply.
Regarding saving, my thought would be to just switch which device is started
based on path. I'm sure thre are other, possibly
One option is to use table to get the number of repeats
tab<-table(tr1$time)==3
time3<-names(tab)[tab]
tr1[tr1$time%in%time3,]
Weidong Gu
On Sun, Oct 30, 2011 at 5:10 PM, Zheng Lu wrote:
>
>
>
>
> With the following dataset, trl =1, 2,3, how to extract data with the time
> repeated 3 times. Fo
Good point indeed,
actually I want to have numericVector==0 (get that points).
What is the difference between though
!numericVector==0 and
-numericVector==0
when used inside a matrix to pinpoint matrix entries?
B.R
Alex
From: William Dunlap
Cc: "R-hel
Duncan Murdoch-2 wrote:
>
> On 11-10-30 2:52 PM, Alaios wrote:
>> Dear all,
>>
>> Could you please explain me why
>>
>>> OverloadsTesT
>> [1] 1 0 1 0 0 0 0 0 0 0
>>> a<-matrix(data=seq(1,10),nrow=10)
>>> a
>> [,1]
>> [1,]1
>> [2,]2
>> [3,]3
>> [4,]4
>> [5,]5
>> [6,]6
>> [7
Szűcs Ákos wrote:
>
> Hi!
> I got a loop where i print out the results of Jarque Bera tests, but I
> have to put, the p-values in a vector. Can you help me how to do it in
> an effective way and not just typing in the results to a vector? Thanks
> a lot, here is the code:
> for(i in 1:60){
> p
With the following dataset, trl =1, 2,3, how to extract data with the time
repeated 3 times. For example, I am only interested in the data records with
time repeated 3 times. time=0.1 only has 1 time, we don't want it, time=0.3
only has two times, we don't want it, and so on. Thanks a lot.
Cem Girit-2 wrote:
>
> Hello,
>
> I am interested in parametric multi comparison tests such
> as
> Dunnett, Duncan, Tukey, Newman-Keuls, Bonferonni, Scheffe, and
> non-parametric tests such as Kruskal-Wallis, and Mann-Whitney U. Are
> there
> packages that include most of these
I am replying to myself as regardless of the save I described below I also want
to do something like the following
do_analysis_for_data_sets <- function(AnalysisFunction, DataSets, ...) {
lapply(DataSets,AnalysisFunction, "how to pass the ... to lapply")
}
which means that I want to p
You are absolutely right it is better. This was just an example how one
can convert the positives to TRUE and zeros to FALSE as it was asked.
Andrija
On Sun, Oct 30, 2011 at 9:57 PM, William Dunlap wrote:
> I like to use numericVector != 0 instead of
> is.logical(numericVector) because the for
I like to use numericVector != 0 instead of
is.logical(numericVector) because the former
more directly indicates what you want to happen
instead of relying on knowledge that numeric 0
maps to logical FALSE.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> Fro
myvec != 0
does the same as
ifelse(myvec == 0, FALSE, TRUE)
but more quickly
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of Sarah Goslee
> Sent: Sunday, October 30,
I think this does the work
return(m[!as.logical(data)])
I am not sure though if this is the same with
return(m[!as.logical(data)])
From: andrija djurovic
Cc: "R-help@r-project.org"
Sent: Sunday, October 30, 2011 9:58 PM
Subject: Re: [R] why the a[-indx] do
Hi:
Here are a few ways (untested, so caveat emptor):
# plyr package
library('plyr')
ddply(df, .(Plant, Tissue, Gene), summarise, ntest =
shapiro.test(ExpressionLevel))
# data.table package
library('data.table')
dt <- data.table(df, key = 'Plant, Tissue, Gene')
dt[, list(ntest = shapiro.test(Exp
Well, you need to use the current version, then.
-thomas
On Fri, Oct 28, 2011 at 10:09 PM, amitava wrote:
> I am using Package survey version 3.20 and R 2.11.1.
>
> --Amitava
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Problem-with-svyvar-in-survey-package-tp39328
Hi:
I'd double check the form of the upper bound of the sequence. In the
first case (i = 1), it uses the subvector loghozamok[1:220]; when i =
2, it uses the subvector loghozamok[21:260], etc. In the last case, it
uses loghozamok[1181:1420]. If instead you want to split the vector
into 60 groups o
> as.logical(c(1,0,1,1))
[1] TRUE FALSE TRUE TRUE
?as.logical
On Sun, Oct 30, 2011 at 8:50 PM, Alaios wrote:
> probably you mean
>
>
>
>
> For [-indexing only: i, j, ... can be logical
> vectors, indicating elements/slices to select. Such vectors
> are recycled if necessary to mat
ifelse(myvec == 0, FALSE, TRUE) # set 0 to FALSE, other values to TRUE
On Sun, Oct 30, 2011 at 3:50 PM, Alaios wrote:
> probably you mean
>
>
>
>
> For ‘[’-indexing only: ‘i’, ‘j’, ‘...’ can be logical
> vectors, indicating elements/slices to select. Such vectors
> are recycled if necessary to
I suppose I wasn't clear enough.
For some purposes, I want to label the horizontal axis like the first
label-line below.
It separates the longer periods better even though it squashes the shorter
ones.
The real scale for the corresponding plotted data is the second label-line.
The same-range linear
probably you mean
For â[â-indexing only: âiâ, âjâ, â...â can be logical
vectors, indicating elements/slices to select. Such vectors
are recycled if necessary to match the corresponding extent.
âiâ, âjâ, â...â can also be negative integers, indicating
elements/slices
Hi,
Is it possible to calculate confidence intervals for support vector
regression?
In the "ksvm{kernlab}" manual, it says that it supports confidence
intervals for regression.
However, i couldn't find any information about how to calculate confidence
interval.
Do you have any documents or example
Dear R users,
I have a data frame in the form below, on which I would like to make normality
tests on the values in the ExpressionLevel column.
> head(df)
ID Plant Tissue Gene ExpressionLevel
1 1 p1 t1 g1 366.53
2 2 p1 t1 g2 0.57
3 3 p1 t1 g311.81
4 4
> a[overLoadTesT==0]
[1] 2 4 5 6 7 8 9 10
Look into help('[') or help('Subscript') to see
how integer and logical (Boolean) subscripts differ.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun..
Dear all,
I was reading for the nice ... parameter option in R, which if I got it right
it is used to pass other parameter to some other function inside your function
for example
plot_duty_cycle<- function(data, ...) {
barplot(duty_cycle_per_bin(data), ...)
}
I can put some param
On 11-10-30 2:52 PM, Alaios wrote:
Dear all,
Could you please explain me why
OverloadsTesT
[1] 1 0 1 0 0 0 0 0 0 0
a<-matrix(data=seq(1,10),nrow=10)
a
[,1]
[1,]1
[2,]2
[3,]3
[4,]4
[5,]5
[6,]6
[7,]7
[8,]8
[9,]9
[10,] 10
a[-OverloadsTesT]
[1] 2 3 4
On 11-10-24 7:16 PM, Hadley Wickham wrote:
On Mon, Oct 24, 2011 at 5:39 AM, Duncan Murdoch
wrote:
Suppose I have data like this:
A<- sample(letters[1:3], 1000, replace=TRUE)
B<- sample(LETTERS[1:2], 1000, replace=TRUE)
x<- rnorm(1000)
I can get a table of means via
tapply(x, list(A, B), mea
Dear all,
Could you please explain me why
> OverloadsTesT
[1] 1 0 1 0 0 0 0 0 0 0
> a<-matrix(data=seq(1,10),nrow=10)
> a
[,1]
[1,]1
[2,]2
[3,]3
[4,]4
[5,]5
[6,]6
[7,]7
[8,]8
[9,]9
[10,] 10
> a[-OverloadsTesT]
[1] 2 3 4 5 6 7 8 9 10
the last line d
Hi!
I got a loop where i print out the results of Jarque Bera tests, but I
have to put, the p-values in a vector. Can you help me how to do it in
an effective way and not just typing in the results to a vector? Thanks
a lot, here is the code:
for(i in 1:60){
print(jarque.bera.test(loghozamok[(
On Tue, Oct 25, 2011 at 12:04 PM, Eliano wrote:
> hi people,
>
> I'm trying to maximize this function:
>
> fn= function (x) {x[1]^2+5*x[2]^2}
>
> with this restriction
> fn1 = function (x) {x[1]+x[2] <=5}
>
> Can someone help me how to procedure this?
>
> I tried in the alabama and genoud package
Dear users,
I'm using rpart for classification trees, but my code isn't working when I
try to use all the variables in my data frame. This data frame was created
from a data frame with 1775 variables, but I choose only 13.
arv13<-rpart(iv~.,data=gn,method="class",parms=list(split="information"))
Hello list.
I need help (e.g., a reference, code, package, etc.) in calculating the
joint entropy of many variables (some sure highly mutually-informative and
some not).
Is there anyone here who knows a computationally-efficient solution (such
as an R package)? I appreciate you help ...
Best, Reza
On 30.10.2011 04:51, Wendy wrote:
Hi,
I have a vector and a matrix. For example,
A = [
12
3
4];
B = [
4 13
10 2
4 8];
I am comparing A to each column of B using A>B[,ii], so the expected result
is
C = [
10
01
00];
This list is about R rather than Matlab dial
The recycling rule should apply here (see 'An Introduction to R', Sec.
5.4.1; and ?Comparison, under 'Value').
x <- -5:5
A <- cbind(x, x, x)
vec <- numeric(length(x))
A > vec ### recycling
apply(A,2,`>`,vec) ### using apply
vec <- numeric(11) + 3; vec[1] <- -6
A > vec
Hi,
To compare row wise is merely to compare column wise using the transpose matrix:
t(B) < A
or
t(t(B) wrote:
> Given that you want to compare
> columns, you can just do:
>
> A > B
>
> If you wanted to compare rows, then
> it is more troublesome. One approach
> would be:
>
> rep(A, each=nrow(
Given that you want to compare
columns, you can just do:
A > B
If you wanted to compare rows, then
it is more troublesome. One approach
would be:
rep(A, each=nrow(B)) > B
On 30/10/2011 03:51, Wendy wrote:
Hi,
I have a vector and a matrix. For example,
A = [
12
3
4];
B = [
4 13
10
Hi Wendy,
Most of the binary operators can deal with matrices and vectors natively:
A<-c(12,3,4)
B<-matrix(c(4,10,4,13,2,8),3,2)
B
[,1] [,2]
[1,]4 13
[2,] 102
[3,]48
B wrote:
> On 10/30/2011 02:51 PM, Wendy wrote:
>>
>> Hi,
>>
>> I have a vector and a matrix. For exampl
On 10/30/2011 02:51 PM, Wendy wrote:
Hi,
I have a vector and a matrix. For example,
A = [
12
3
4];
B = [
4 13
10 2
4 8];
I am comparing A to each column of B using A>B[,ii], so the expected result
is
C = [
10
01
00];
I am looking for a way to do this quickly ins
uqtr.ca> writes:
>
> Hi Guys,
>
> First, English is not my native language so sorry if the question is
> too difficult to understand. I can rephrase it if necessary.
>
> I noticed the following code to explain as clearly as possible the
> problems encountered.
>
> I am not able to add
> "SG" == Spencer Graves writes:
SG> If you know nothing about the black box except that its domain is
SG> bounded, then I would random sample uniformly from the domain. If the
SG> function is monotonically increasing in all variables, then you only
SG> need to test the two extreme points.
Hi,
I have a vector and a matrix. For example,
A = [
12
3
4];
B = [
4 13
10 2
4 8];
I am comparing A to each column of B using A>B[,ii], so the expected result
is
C = [
10
01
00];
I am looking for a way to do this quickly instead of going through the for
loop
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