Hi
see par.setting in ?xyplot
or
trellis.par.get() # beware it is verbose
or for starters
names(trellis.par.get() )
eg
par.settings = list(axis.text = list(cex = 0.75),
par.xlab.text = list(cex = 0.85),
par.ylab.text = list(cex = 0.85)),
Duncan Mackay
Department of Agronomy and Soil Scienc
On Thu, 26 Apr 2012, Christopher Desjardins wrote:
Hi,
I am trying to replicate Lambert (1992)'s simulation with zero-inflated
Poisson models. The citation is here:
@article{lambert1992zero,
Author = {Lambert, D.},
Journal = {Technometrics},
Pages = {1--14},
Publisher = {JSTOR},
Title = {Zero-i
On 27/04/2012 06:24, arun wrote:
Dear R'ers,
I have trouble installing tikzDevice in Ubuntu. When I use
install.packages("tikzDevice"), it gives error message:
ERROR: dependency ‘filehash’ is not available for package ‘tikzDevice’
* removing ‘/usr/local/lib/R/site-library/tikzDevic
Thanks a lot for your reply.
I have managed to do it but still something that I do not know how to do
that.This code below will read the first row in D and write it to a new
file.I want to make a another loop to read all rows and write each result of
every row to a file.
library(Matrix)
M <- Matr
Dear R'ers,
I have trouble installing tikzDevice in Ubuntu. When I use
install.packages("tikzDevice"), it gives error message:
ERROR: dependency âfilehashâ is not available for package âtikzDeviceâ
* removing â/usr/local/lib/R/site-library/tikzDeviceâ
Then I tried filehash instal
Dear all,
I am having trouble with the following problem. Suppose we have the fourth
order ODE with boundary conditions:
http://r.789695.n4.nabble.com/file/n4591748/problem.jpg problem.jpg
where q(t) is a known function.
Note here the lambda parameter is changing, so essentially we have a seri
Dear R'ers,
I was not able to change the font size of axis label using cex. Tried R
archive solutions, though not solved yet.
My code is:
print(dotplot(ss ~ Response | bb*sr*tr, pp, pch = 21,
strip = FALSE, strip.left = TRUE,layout = c(3,8),
scales = list(y = lis
Hi,
How can I extract the phone numbers( consecutive 10 digits or in a pattern
like 3digits-3digits-4digits) from a verbatim?
Thanks in advance for any help...
Antony
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See http://r.789695.n4.nabble.com/reorder-a-matrix-td4588839.html#a4588845
--JIV
On Thu, Apr 26, 2012 at 10:31 PM, Rebecca <> wrote:
> Hi,
> Here are part of my data,
>
> > pr16.5
>origin dest ldco time.slot distortion
> 1 111 14.3
> 2 111
Hi,
Here are part of my data,
> pr16.5
origin dest ldco time.slot distortion
1 1 1 1 1 4.3
2 1 1 1 2 4.7
3 1 1 1 3 5.6
4 1 1 1 4 7.7
5 1 1 2 1 6.8
6
A new version of glmnet is uploaded to CRAN.
This should take care of the problem on PCs that
caused it to fail there. Many thanks to B. Narasimhan
for his stoic efforts in debugging this problem, which was
a real nasty idiosyncrasy in the gfortran compiler that exists
on windows but not on linux
On 04/26/2012 03:21 PM, Kamil Slowikowski wrote:
My goal is simple: calcuate GC content of each sequence in a list of
nucleotide
sequences. I have figured out how to vectorize, but all my attempts at
memoization failed.
Can you show me how to properly memoize my function?
There is a StackOverfl
Yes, you can see that in the new model the slope is now 1- the old
slope, so it is measuring difference from 1. Since it is significant
that means the slope is significantly different from 1. To test the
null that slope = 0.5 just change the offset to
"offset(0.5*log(data$SIZE,10))"
To save your
I have tried this but its not the same as in the links.
x=seq(-3,3,length.out=100)
y=dnorm(x)
plot(x,y,type="l")
y3=y*2
x3=x+1
plot(range(c(-4,4)),range(c(0,3)),xlab="x",ylab="y",col="white" )
lines(x,y4,type="l")
x2=x*0.5
lines(x2,y,type="l")
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How can I write a R-code so I can get these diagrams? (The skewness and
Kurtosis for a normal distribution)
http://www.google.dk/imgres?um=1&hl=da&biw=1280&bih=605&tbm=isch&tbnid=aWXNm22wHhXWWM:&imgrefurl=http://en.wikipedia.org/wiki/Skewness&docid=klEV7X9U6qwTRM&imgurl=http://upload.wikimedia.org
# dput() example
# lets say you have data called y, like this:
> y
sp1 sp2 sp3 sp4
d 0 0 0 0
e 0 0 0 0
f 0 0 0 0
# ok, so do this:
> dput(y)
structure(list(sp1 = c(0, 0, 0), sp2 = c(0, 0, 0), sp3 = c(0,
0, 0), sp4 = c(0, 0, 0)), .Names = c("sp1", "sp2", "sp3", "sp4"
)
My goal is simple: calcuate GC content of each sequence in a list of
nucleotide
sequences. I have figured out how to vectorize, but all my attempts at
memoization failed.
Can you show me how to properly memoize my function?
There is a StackOverflow post on the subject of memoization, but it does
Hi everyone,
I'm using the HoltWinters() function to do a time series analysis. The
function only returns the back fitted values ($fitted) after the first year
of data, which is my case, is a little more than half. However, when I use
the plot() function, it plots the back fit for almost the ent
Sorry , i am really tiered i misread you're message.
Date: Thu, 26 Apr 2012 10:59:32 -0700
From: ml-node+s789695n4590509...@n4.nabble.com
To: guar...@hotmail.com
Subject: Re: ErrError in f(x, ...) : object 'g.' not found
Just what it says:
You define g but refer to a variable g. in
Hello,
Try ?is.na.
In the example below I've changed your first row name from NA to NA.0
x <- read.table(text="
ID Category Sex Beak Head
NA.0NANA NA
NA.1 NANA NA
NA.2 NANA NA
NA.3 NANA NA
NA.4 NA
On 27/04/12 05:49, Greg Snow wrote:
The phrase "does not work" is not very helpful, it can mean quit a few
things including:
* Your computer exploded.
* No explosion, but smoke is pouring out the back and microsoft's
"NoSmoke" utility is not compatible with your power supply.
* The computer stop
Hi,
try also grid.table() in gridExtra
b.
On 27 April 2012 01:26, statquant2 wrote:
> Hello,
> I would like to be able to plot an array on a plot, something like:
> |arg1 | arg2 | arg3
> val1| 0.9 | 1.1 | 2.4
> val2| 0.33 | 0.23 | -1.4
> val3| hello| stop | test
> I know Rwave is
Hi,
I am trying to replicate Lambert (1992)'s simulation with zero-inflated
Poisson models. The citation is here:
@article{lambert1992zero,
Author = {Lambert, D.},
Journal = {Technometrics},
Pages = {1--14},
Publisher = {JSTOR},
Title = {Zero-inflated {P}oisson regression, with an application to d
Hi Greg and others,
Thanks for your replies. Okay, I'm convinced that the offset is the best
approach and wonder if you might have a quick look at what I did.
Here's the original model containing the slope (0.56) that I'd like to test
if it's different from 1.0
>model1 <- glm(log(data$AB.obs+1,1
If it helps... R is not Matlab. There is no '.*' operator in R.
/H
On Thu, Apr 26, 2012 at 2:37 PM, Sarah Goslee wrote:
> On Thu, Apr 26, 2012 at 5:34 PM, Guaramy _ wrote:
>> Thanks for your answer, but how can i correct that, the mathematical
>> expression is correct that way with that g. Do
Hi,
I'm trying to do some constrained non-linear optimisation, but my
function does not have second order derivatives everywhere.
To be a little more specific (the actual function is huge and
horrible, so it would probably be better to just describe it) my model
has four variables and I'm using op
On Thu, Apr 26, 2012 at 5:34 PM, Guaramy _ wrote:
> Thanks for your answer, but how can i correct that, the mathematical
> expression is correct that way with that g. Do you know how can i program it
> ?
Then you need to define g. and how it differs from g alone.
g. and g are two separate object
Assuming everything else is good, the "all" or "all.x" or "all.y"
arguments to merge() should do what I think you're asking for. You did
read the help page for merge, right?
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
O
I am attempting to calibrate a forest growth model using FME. I have
several spatially separate plots with time series growth data for
each(eg, height, basal area, etc.). Thus 2 independent variables: time
and plot.
Is there a way that I can calculate a model cost that takes into account
thes
Hi there,
Thanks for your responses. I haven't used/heard of dput() before. I'm
looking it up & understanding how it works.
Thanks!
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As Bert Gunter points out, the statistic R-squared should not be given any
grand meaning.
However, (also in the archives) I use it in my nonlinear least squares codes as
a "sanity
check" -- and not more than that. Since it is computed as
{1 - (residual SS)/(SS from mean)},
do you really want t
Hello, I'm making a simple plot using xYplot in the Hmisc library and having
problems with labeling the values on the x-axis. Using the reproducible example
below, how can I have the text (jan, feb,mar, etc.) in place of 1:12.
Thanks, AB
x <- c(seq(0,0.5,by=0.1),seq(0.5,0,by=-0.1))
ci <- rnor
Look at the 'addtable2plot' function in the 'plotrix' package or the
'textplot' function in the 'gplots' package.
On Thu, Apr 26, 2012 at 7:26 AM, statquant2 wrote:
> Hello,
> I would like to be able to plot an array on a plot, something like:
> |arg1 | arg2 | arg3
> val1| 0.9 | 1.1
Hi,
As Sarah reiterated -- it'd *really* be helpful if you give us data we
can actually work with.
That having been said:
On Thu, Apr 26, 2012 at 4:12 PM, RHelpPlease wrote:
> Hi again,
> I tried the sample code like this:
>
>> merged_clmno <- subset(bestPartAreadmin, !CLAIM_NO %in% hrc78_clm_n
You'd get better help if you actually did as Steve requested and
provided sample data (a reproducible example!) using dput().
But since you didn't:
> fakedata <- data.frame(a = 1:5, b=11:15, c=c(1,1,1,2,2))
> fakedata
a b c
1 1 11 1
2 2 12 1
3 3 13 1
4 4 14 2
5 5 15 2
> notb <- c(12, 14, 15)
>
On Thu, Apr 26, 2012 at 02:49:08PM -0500, cassie jones wrote:
> Thanks Berend, you are right. The break command would not work here. But
> the while loop is taking time to generate the desired.
>
> On Thu, Apr 26, 2012 at 2:39 PM, Berend Hasselman wrote:
[...]
> > You need a while loop for this.
Hi again,
I tried the sample code like this:
> merged_clmno <- subset(bestPartAreadmin, !CLAIM_NO %in% hrc78_clm_no)
> dim(merged_clmno)
[1] 1306893
Note that:
> dim(bestPartAreadmin)
[1] 1306893
So, no change between the original data.frame (bestPartAreadmin) & the
(should be) less-row
I second the proposition !
Thanks to let me discover that absolutely essential library !! :-)
Arnaud
Le 26 avril 2012 15:30, Marc Schwartz a écrit :
>
> On Apr 26, 2012, at 12:49 PM, Greg Snow wrote:
>
> > The phrase "does not work" is not very helpful, it can mean quit a few
> > things includi
On Apr 26, 2012, at 3:40 PM, michaelyb wrote:
David -
My question to you may sound (actually, it really is) silly, but
please do
take your time to answer it.
What is the difference between:
fac<-function(x){a<-1
for (i in 1:x){
a<-a*i
}a}
and
Dear R-community,
I am using R (V 2.14.1) on Windows 7. I have a dataset which consists of 19
variables for 91 individuals or rows. Two of my variables are Age
(adult/chick, with no NA values) and Sex (0 for females/1 for females, with
quite a few NA values). The sex of many adult birds is unknow
Hi Steve,
Thanks for replying. Here's a small piece of the data.frame:
> bestPartAreadmin[1:5,1:6]
DESY_SORT_KEY PRVDR_NUM CLM_THRU_DT CLAIM_NO
NCH_NEAR_LINE_REC_IDEN_CD NCH_CLM_TYPE_CD
1 10193 290003 20090323 20
On Thu, Apr 26, 2012 at 02:30:09PM -0500, cassie jones wrote:
> Hello R-users,
>
> I am having a problem with the 'break' command in R. I am wondering if
> anyone can help me out with this. My program is similar to the following.
>
> a=rep(NA,5)
> a[1]=0
> for(i in 2:5)
> {
> a[i]=a[i-1]+runi
David -
My question to you may sound (actually, it really is) silly, but please do
take your time to answer it.
What is the difference between:
fac<-function(x){a<-1
for (i in 1:x){
a<-a*i
}a}
and:
fac<-function(x){a<-1
for
Thanks Berend, you are right. The break command would not work here. But
the while loop is taking time to generate the desired.
On Thu, Apr 26, 2012 at 2:39 PM, Berend Hasselman wrote:
>
> On 26-04-2012, at 21:30, cassie jones wrote:
>
> > Hello R-users,
> >
> > I am having a problem with the 'b
Hi,
To increase the chances of you getting help on this one, please give
example data (a small data.frame, a small list) that you are trying to
do this on, and also show the desired output. Whip these variables up
in your R workspace and paste the output of `dput` for each into your
follow up emai
On 26-04-2012, at 21:30, cassie jones wrote:
> Hello R-users,
>
> I am having a problem with the 'break' command in R. I am wondering if
> anyone can help me out with this. My program is similar to the following.
>
> a=rep(NA,5)
> a[1]=0
> for(i in 2:5)
> {
>a[i]=a[i-1]+runif(1,0,3)
>if
Hi there,
I wish to merge a common variable between a list and a data.frame & return
rows via the data.frame where there is NO match. Here are some details:
The list, where the variable/col.name = CLAIM_NO
CLAIM_NO
20
83
1440
4439
7002
...
> dim(hrc78_clm_no)
[1] 66781
The data.frame, where
On Apr 26, 2012, at 12:49 PM, Greg Snow wrote:
> The phrase "does not work" is not very helpful, it can mean quit a few
> things including:
>
> * Your computer exploded.
> * No explosion, but smoke is pouring out the back and microsoft's
> "NoSmoke" utility is not compatible with your power supp
Hello R-users,
I am having a problem with the 'break' command in R. I am wondering if
anyone can help me out with this. My program is similar to the following.
a=rep(NA,5)
a[1]=0
for(i in 2:5)
{
a[i]=a[i-1]+runif(1,0,3)
if(a[i]>5)
{
i=2
break
}
}
What exactly I am
Dear R users.
I'm trying to know how to create a function. I have developed a big code that
use a file and makes many process and then creates a table that is exported as
txt.
The point is to make something like a function. I need to make it easy for
others, so they can run it. This has a p
Nice, thank you !
I particularly appreciated the list of possible explanations for "it does
not work" ! :-)
Arnaud
Le 26 avril 2012 13:49, Greg Snow <538...@gmail.com> a écrit :
> The phrase "does not work" is not very helpful, it can mean quit a few
> things including:
>
> * Your computer explo
The latest version of fortunes (from R-forge, not sure about CRAN)
have fortunes up to number 317 which is from just a couple of days ago
(and 312 is the quoted one from February). For some reason some of
the instances of R on my computer stop at 291, others go up to 317
(running a new instance of
Hello,
>
> gives you 120, but you cannot access it after the end of execution.
>
Because you're just printing the final value of 'a', not returning it.
fac <- function(x){
a <- 1
for(i in 1:x) a <- a*i
a
}
The return value must be the last instruction in a function.
The
Sorry I took so long getting back to this, but the paying job needs to
take priority.
The regular expression "(? wrote:
> Hi Greg,
>
> This is quite helpful. Not so good yet with regular expressions in general or
> Perl-like regular expressions. Found the help page though, and think I was
> able
the solution is to write functions that return the data you want changed, and
let the caller of the function decide where to put the answer. If you want to
return multiple "answers", collect them in a vector or list and return that.
---
Dear Rainer
On Wed, Apr 25, 2012 at 5:34 PM, Rainer Schuermann
wrote:
> <>=
>
I like the 'eval=FALSE' trick.
> SweaveInput( "setup.Rnw" )
>
> and from here, I can suse the named chunks almost like function calls, as you
> you describe below. The advantage (for me) is that I have only one place
In R, the preferred method is to assign the result to a new object:
fac <- function(x) {
a<-1
for(i in 1:x){
a<-a*i
print(a)
}
a # need to explicitly state what the function should return
}
myresult <- fac(5)
myresult
Sarah
On Thu, Apr 26, 2012 at 2:16 PM, michaelyb wrote:
>
On Apr 26, 2012, at 2:16 PM, michaelyb wrote:
Ista,
Since you seem to know your stuff very well, how would you get 120
out of a
function that gives you the factorial of 5, without using
factorial(5)?
Meanwhile, look at this example instead:
fac<-function(x){a<-1
for(i in
Simply return it like a function is supposed to:
fac <- function(x, loud = TRUE){
a <- 1
for(i in seq_len(x)) { # seq_len is faster and more robust
a <- a * i
if(loud) print(a)
}
return(a)
}
fac3 <- fac(3)
print(fac3) # As desired
Michael
On Thu, Apr 26, 2012 at
Ista,
Since you seem to know your stuff very well, how would you get 120 out of a
function that gives you the factorial of 5, without using factorial(5)?
> Meanwhile, look at this example instead:
> fac<-function(x){a<-1
> for(i in 1:x){
> a<-a*i
>
On Thu, Apr 26, 2012 at 1:32 PM, michaelyb wrote:
> Any solution for that type of problem?
Solution for what type of problem?
> I did read the ?"<<-", and seems very similar to the "assign" function, if I
> am not mistaken
Bottom line: don't use <<- until you know what you are doing. By the
Just what it says:
You define g but refer to a variable g. in the next line.
Just get rid of the typo.
Sarah
On Thu, Apr 26, 2012 at 1:43 PM, Guaramy wrote:
> Hi , R is a new language for me so sorry in advance if this error is to basic
> for posting. I have tried the R manual and search onli
Yes, don't do it.
I know this can seem appealing if you're coming from another language,
but playing fast and loose with globals, non-local effects, and
superassignment is almost-never a good idea.
Michael
On Thu, Apr 26, 2012 at 1:32 PM, michaelyb wrote:
> Any solution for that type of problem
The phrase "does not work" is not very helpful, it can mean quit a few
things including:
* Your computer exploded.
* No explosion, but smoke is pouring out the back and microsoft's
"NoSmoke" utility is not compatible with your power supply.
* The computer stopped working.
* The computer sits aroun
Hi , R is a new language for me so sorry in advance if this error is to basic
for posting. I have tried the R manual and search online for quite a few, if
anyone could help i would be very thankful.
Here is my code.
kappa = 1.1
theta = 0.1
sigma = 0.4
rho = -0.6
v0 = 0.2
r = 0.05
T = 0.5
s0 = 1
K
Hello,
I am using constrOptim to maximize a likelihood function (the values of my
parameter vector must be between zero and one and must sum up to <=1). I am
getting the error 'initial value in 'vmmin' is not finite'. I've tracked it
down to a problem in the function 'R' defined within the cons
Hello,
ok for xtable, but how will I print the latex code generated in a plot ?
Cheers
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_
Any solution for that type of problem?
I did read the ?"<<-", and seems very similar to the "assign" function, if I
am not mistaken
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On 25/04/12 14:02, Mabille, Geraldine wrote:
Hi, I am working with gam models in the mgcv library. My response
variable (Y) is binary (0/1), and my dataset contains repeated
measures over 110 individuals (same number of 0/1 within a given
individual: e.g. 345-zero and 345-one for individual A, 22
You could also use the grconvertX function to convert from the middle
of the plotting region (from='npc') to user coordinates (to='user'),
or many other combinations.
On Wed, Apr 25, 2012 at 5:53 AM, Ramon Ovelar wrote:
> Many many thanks for the tip and for authoring this function!
>
> The final
At 00:55 26/04/2012, Steven Orzack wrote:
I am looking for an R package with which one can calculate an effect
size for a set of contrasts given an omnibus chi-square test
statistic (more than 1 degree of freedom). Is there such a package?
Presumably, it would implement the procedure (or someth
Dear All,
I am recently working on neural network using nnet package. The network has
4 hidden layers and 1 output layer, the target output 1 or 0.
The model I use is as follows:
nn<-nnet(target~f1+f2+f3+f4+f5+f6+f7+f8+f9+f10,data=train,size=4,linout=FALSE,decay=0.025,maxit=800)
It works well
Thanks a lot, totally forgot cran there.
Hm.. so they're multiplying some specifically computed Kernelmatrix with the
pcv's.. interesting.. too tired to check the math there, guess i'll just accept
its possible and go to sleep.
Am 26.04.2012 um 18:10 schrieb Steve Lianoglou:
> Hi Jessica,
>
>
Hello R users,
Hope everyone is doing great.
I have a dataset that is in .csv format and consists of two columns: one
named Period (which contains dates in the format _mm) and goes from
1995_10 to 2007_09 and the second column named pcumsdry which is a
volumetric measure and has been formatt
On Apr 26, 2012, at 9:26 AM, statquant2 wrote:
Hello,
I would like to be able to plot an array on a plot, something like:
|arg1 | arg2 | arg3
val1| 0.9| 1.1| 2.4
val2| 0.33 | 0.23 | -1.4
val3| hello| stop | test
I know Rwave is good to report but don't want to use it.
? Is the
Dear all,
I have a series of variables that looks roughly like the sample data below and
I'm trying to conduct a factor analysis. I've omitted cases with missing
values for the factor analysis, but now I'd like to use the scores on each
component as new variables in the *original* data set for
Hello,
Your example code has a bug, there's no fata.frame called 'AB'. It's
corrected below.
pannigh wrote
>
> Dear list,
> I get the ifelse function to work on a data frame but don't know how to do
> something similar (only more conditions) with the combination of if and
> else like in the exa
Hi Jessica,
On Thu, Apr 26, 2012 at 11:59 AM, Jessica Streicher
wrote:
> Hi!
>
> how do i get to the source code of kpca or even better predict.kpca(which it
> tells me doesn't exist but should) ?
Probably you have to do kernlab:::predict.kpca from your R workspace,
but why not just download th
I guess there is not a steep learning curve if you want to use cache
in knitr or cacheSweave -- just use the chunk option cache=TRUE and
you are all set.
<>=
# time-consuming computation here
@
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa
Hi!
how do i get to the source code of kpca or even better predict.kpca(which it
tells me doesn't exist but should) ?
(And if anyone has too much time:
Now if i got that right, the @pcv attribute consists of the principal
components, and for kpca, these are defined as projections of some rando
Dear list,
I get the ifelse function to work on a data frame but don't know how to do
something similar (only more conditions) with the combination of if and else
like in the example:
A <- c("a","a","b","b","c","c")
B <- c(rep(2,6))
dat <- data.frame(A,B)
dat$C <- if(AB$A=="a") {AB$B^2} else
The easiest way probably is to put the code that takes so long to execute, in a
separate file, such as "graph1.Rnw", and get it into your master file via
SweaveInput( "graph1.Rnw" ).
Once you are happy with your graph1, you can comment this line out and only
compile the stuff that keeps changing.
Hello Rui,
For the write.table, it's OK!
And for the second one (for the 2nd best correlation) seems to work great!
You're too strong ^^
I have to check a bit more to be sure, but it seems to do it!
If you come in the Alps, it will be more liqueurs such as Chartreuse or
Génépi (from mountain plan
> Did you read ?'<<-' as
> Peter suggested?
Better yet, do not read ?'<<-' and do not use <<-.
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 Ista Zahn
> Sent: Thur
Hello everybody out there using Sweave,
There are some complicated SQL queries and laborous calculations against large
data included as R code chunks using Sweave in my LaTeX document.
These code chunks create graphs that do not change most of the time, but they
are of course recompiled every ti
Perhaps PerformanceAnalytics and the CAPM.* functions.
Michael
On Thu, Apr 26, 2012 at 8:46 AM, and_mue wrote:
> Hi all
>
> I am looking for an add-in. I am currently working on something and I use
> daily data of closing stock prices. As not all companies are traded daily
> (e.g. on monday, the
On Apr 26, 2012, at 9:29 AM, Neil Davis wrote:
Hi,
I have a data.frame which contains timeseries from several different
locations, which I want to compare against each other for example
calculating RMSE, or normalized mean bias of each location against
the others. An example of this is t
On Apr 26, 2012, at 7:31 AM, Jim Lemon wrote:
On 04/26/2012 08:02 PM, Manish Gupta wrote:
Hi,
I have my data in below format.
position var1var2
2 .1 10
3 .29 89
12.56
Hi,
you can iterate over the vertices, but it'll be probably slow. The
best solution is to create the graph directly from the data frame(s)
containing all structure and attribute data. See the
graph.data.frame() function for this.
Btw. it might be worth to post your igraph related questions to t
Thanks!
-Steve
-Original Message-
From: Duncan Murdoch [mailto:murdoch.dun...@gmail.com]
Sent: Thursday, April 26, 2012 9:28 AM
To: Steven Wolf
Cc: r-help@r-project.org
Subject: Re: [R] Heatmap fidelity
On 26/04/2012 9:01 AM, Steven Wolf wrote:
> I'm having a problem when using heatmap.
On Thu, Apr 26, 2012 at 8:56 AM, michaelyb wrote:
> Peter, your solution is actually very interesting. I have never seen or heard
> of before. I will look into it.
>
> Meanwhile, look at this example instead:
> fac<-function(x){a<-1
> for(i in 1:x){
> a<-a*i
>
Hello,
I would like to be able to plot an array on a plot, something like:
|arg1 | arg2 | arg3
val1| 0.9| 1.1| 2.4
val2| 0.33 | 0.23 | -1.4
val3| hello| stop | test
I know Rwave is good to report but don't want to use it.
? Is there a package that allow quick and dirty plot of da
Hi,
I have a data.frame which contains timeseries from several different
locations, which I want to compare against each other for example
calculating RMSE, or normalized mean bias of each location against the
others. An example of this is the cor function where I can put in a
data.frame and
On 26/04/2012 9:01 AM, Steven Wolf wrote:
I'm having a problem when using heatmap. Even though the diagonal of my
matrix is all the same value, the diagonal of my heatmap is not all the same
color. Any suggestions?
heatmap() rescales the matrix by default. Use
heatmap(abs(psim), scale="none
Hi all
I am looking for an add-in. I am currently working on something and I use
daily data of closing stock prices. As not all companies are traded daily
(e.g. on monday, then on thursday etc) at the stock exchange, there is
satistically a problem. There are some papers which explain the approach
i found one clue in our forum, but it can give position for whole matrix,
>n<-read.table(file.choose(),header=T)
>y<-n[,c("1","2","3")]
> my.number=1.12270420185886
> z<-abs(y-my.number)==min(abs(y-my.number))
> which(z)
[1] 19
i have uploaded f
Peter, your solution is actually very interesting. I have never seen or heard
of before. I will look into it.
Meanwhile, look at this example instead:
fac<-function(x){a<-1
for(i in 1:x){
a<-a*i
print(a)}}
The result is :
> fac(5)
[1] 1
[1] 2
[1]
hHllo,
I'm looking for an algroithm to transform an existing toeplitz matrix
(autocorrelation matrix) to the nearest positive semidefinite toeplitz
matrix.
I merely found an algorithm to transform an correlation matrix via the
function nearcor() based on the algorithm of Higham.
But as I exam
Dear R users,
I was just wondering if anyone would be
able to advice me on how to create a confidence envelope on the graph when
plotting a pair correlation function (or mark correlation function).
Is the command qqenvl() or envl.plot() what I need or is it something
completely dif
I'm having a problem when using heatmap. Even though the diagonal of my
matrix is all the same value, the diagonal of my heatmap is not all the same
color. Any suggestions?
Here is some reproducible code:
#
# Get data
nba <- read.csv("http://datasets
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