van=function(a,x,n){
A-matrix(c(x[1]^2,x[1],1,x[2]^2,x[2],1,x[3]^2,x[3],1),byrow=T,nrow=3)
for(i in 1:3){
o-x[i]
for(j in sort(1:n+1,decreasing=T)){
B[i][j]-(o)^(n+1-j)
}
}
y=vector(length=3)
y-B%*%a
p=vector(length=3)
p-solve(A,y)
return(p)
}
this is my function and I am getting an error
Dear all,
I want to measure the goodness of prediction of my linear model. That's why
I was thinking about the area under roc curve.
I'm trying the following, but I don't know how to avoid the error. Any help
would be appreciated.
library(ROCR)
model.lm -
Hi there,
I am having a similar problem with reading in a large text file with around
550.000 observations with each 10 to 100 lines of description. I am trying
to parse it in R but I have troubles with the size of the file. It seems
like it is slowing down dramatically at some point. I would be
as of right now
x = function(a) print(a)
attr(x, srcref)
returns NULL in 2.13, am I doing something wrong?
(also, should I post this to a new thread, or the development thread?)
About me: I like long walks on the beach, and this is my current version of
R:
t(as.data.frame(R.Version()))
Hi dear list,
I want to compare the amount of computation of two functions. For example,
by using this algorithm;
data - rnorm(n=100, mean=10, sd=3)
output1 - list ()
for(i in 1:100) {
data1 - sample(100, 100, replace = TRUE)
statistic1 - mean(data1)
output1 - c(output1, list(statistic1))
}
Dear All:
I do see I have weeks of codes in my console when I check with my arrow up
keys. I have been clearing them with Control L but it seems to clear it
clear the screen temporally. I do see the previous codes again when I open R
the next day, after quitting the session!
Q:
How do I clear
Hi,
I have a number of genes (columns) for which I want to examine pairwise
associations of genotypes (each row is an individual)...For example (see
data below), I would like to compare M1 to M2, M2 to M3, and M1 to M3 (i.e.
does ac from M1 tend to be found with bc from M2 more often than
Hello,
I am using the following model
model1=lmer(PairFrequency~MatingPair+(1|DrugPair)+(1|DrugPair:MatingPair),
data=MateChoice, REML=F)
1. After reading around through the R help, I have learned that the above
code is the right way to analyze a mixed model with the MatingPair as the
fixed
Hi,
I have a problem with an external library on a previous R version.
We've created our own package containing a mixture of C++ as well as R
code which works fine under R 2.12.1. However, trying to install the
very same package ZIP file on R 2.11.1 will issue an error when loading
a library:
Hi,
I have a number of genes (columns) for which I want to examine pairwise
associations of genotypes (each row is an individual)...For example (see
data below), I would like to compare M1 to M2, M2 to M3, and M1 to M3 (i.e.
does ac from M1 tend to be found with bc from M2 more often than
hi , this can be done easily if (cond) expr
ex:
for (i in 1: 4)+ {+ if(i==2) print(a)+ if(i==2) print(b)+ }
output : [1] a[1] b
but i want this
if (cond) expr1 expr 2
i tried this :
for (i in 1: 4)+ {+ if(i==2) (print(b) print(a))+ }
output : [1] bError in print(b) print(a) : invalid 'x'
Hi,
I am experimenting with the function predict() in two versions of R and the R
extension package survival.
library(survival)
set.seed(123)
testdat=data.frame(otime=rexp(10),event=rep(0:1,each=5),x=rnorm(10))
testfm=as.formula('Surv(otime,event)~x')
testfun=function(dat,fm)
{
Hello, I have a following time series data
head(mend.dat)
ID PARAM Year Month Day Value SYM
1 15052500 1 1965 5 15 128 A
2 15052500 1 1965 5 16 135 A
3 15052500 1 1965 5 17 157 A
4 15052500 1 1965 5 18 176 A
5 15052500 1 1965 5 19
Dear R experts
I was using kinship package to fit mixed model with kinship matrix.
The package looks like lme4, but I could find a way to extract p-value
out of it. I need to extract is as I need to analyse large number of
variables ( 1).
Please help me:
require(kinship)
Generating random
Dear all again:
I have been trying to clear weeks of codes in my R console, even after
quitting and coming back another day!. One of you out there graciously
suggested this rm(list=ls(all=TRUE)). I tried this but no dice.
Codes tried.
Ctrl +L , clears for the time... and the the latest i
Can you recommend an R library that will help me create a diagram of a
phylogenetic tree on which specific images are placed at appropriate nodes of
the tree?
For example, I have specific image files associated with each member of the
phylogenetic tree, and I would like to automate the display
I have a matrix M
dim(M)
[1] 30380 561
I have another list L contains , that contains some row names of matrix M
str (L)
chr [1:21037]
Now I want to extract the submatrix subM (21037 , 561) from the matrix M
by matching the rownames (M) to the 21037 rownames o f L
How do I do that
click in console window and type rm(list=ls(all=TRUE)),
everything will be gone
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Hello all,
I have arranged my data as per Dennis's suggestion in this post
http://www.mail-archive.com/r-help@r-project.org/msg107156.html.
the posted code works fine but when I try to apply it to my data, i get
u2 - ddply(xxm, .(plateid, cytokine), as.data.frame.function(f))
Error in
Hello fellow R-users,
I have a data set that includes times in the hh:mm format. I am uncertain
how to tell R that it is time data. Currently when I try to calculate a mean
with this time subset, I get an error message:
mean(mar2011stats$TT)
[1] NA
Warning message:
In
Cameron and I are pleased to announce version 0.6.0 of the tikzDevice
package which should be available shortly at your local CRAN mirror!
The tikzDevice makes it possible to export R graphics as LaTeX code
that can be included in other documents or compiled into stand alone
figures. The full
To improve the efficiency of a process I am writing I would like to cache
results. So I would like a data structure like a hash table.
So if I call Z - f(Y) I can cache Z associated with Y: CACHE[Y] - Z
I am stumped. I expected to be able to use a list for this but I cannot
figure how
On 14/04/11 15:21, YAddo wrote:
Dear all again:
I have been trying to clear weeks of codes in my R console, even after
quitting and coming back another day!. One of you out there graciously
suggested this rm(list=ls(all=TRUE)). I tried this but no dice.
Codes tried.
Ctrl +L , clears for the
On Wed, Apr 13, 2011 at 04:12:39PM -0700, helin_susam wrote:
Hi dear list,
I want to compare the amount of computation of two functions. For example,
by using this algorithm;
data - rnorm(n=100, mean=10, sd=3)
output1 - list ()
for(i in 1:100) {
data1 - sample(100, 100, replace =
HI Berend,
Thank you for your reply.
2011/4/13 Berend Hasselman b...@xs4all.nl
Questions:
1. why are you defining Bo within a loop?
2. Why are you doing library(nleqslv) within the loop?
Yes, I see what you mean. There's no reason for defining that within the
loop.
Doing both those
On Apr 14, 2011, at 01:29 , (Ted Harding) wrote:
On 13-Apr-11 17:40:53, Jim Silverton wrote:
I have a matrix say,
1 4
23 30
and I want to find the previously attainable fisher's exact test
p-value. Is there a way to do this in R?
--
Thanks,
Jim.
I do not understand what you
Hi Louis,
You pose an interesting question.
Bellow is my take on a solution.
It is a function named compare.all.column.pairs which gets as input the
original data (as a data.frame or a matrix, with all the columns you'd want
to compare).
And also the function you would like to use on each pair
I have a categorical variable with a nested structure. For example,
region: a country is split into parts, which in turn contain
provinces, which contain municipalities:
Part - Province - Municipality
North
Province A
Municipality 1
Municipality 2
Municipality 3
I use the very nice expm functions available from the expm and Matrix
packages. My understanding is that for large sparse matrices the
currently best methods available are Krylov subspace methods, but
they are as far as I can tell not implemented in either of the packages
mentioned, nor in any
Look at ExpoKit (http://www.maths.uq.edu.au/expokit/)
It implement Krylov methods and is really fast.
Unfortunately no R wrapper that I know yet.
If you plan to implement it, I can provide my testing code.
Ciao!
mario
On 14-Apr-11 10:16, Niels Richard Hansen wrote:
I use
On Wed, 13 Apr 2011, Katrina Bennett wrote:
Hello, I have a following time series data
head(mend.dat)
ID PARAM Year Month Day Value SYM
1 15052500 1 1965 5 15 128 A
2 15052500 1 1965 5 16 135 A
3 15052500 1 1965 5 17 157 A
4 15052500 1 1965 5 18
I do see I have weeks of codes in my console when I check with my arrow up
keys. I have been clearing them with Control L but it seems to clear it
clear the screen temporally.
CTRL-L simply clears the screen and not the history.
I do see the previous codes again when I open R
the next day,
On Apr 13, 2011, at 21:32 , Louis Plough wrote:
Hi,
I have a number of genes (columns) for which I want to examine pairwise
associations of genotypes (each row is an individual)...For example (see
data below), I would like to compare M1 to M2, M2 to M3, and M1 to M3 (i.e.
does ac from M1
Hi Petr,
Your idea looks like logically. So, can we say this with your idea; the
expected number of computation in unique(sample(...)) is fewer than
sample(...). Because, the expected length is 63.39677 in unique case, while
the expected length is 100 in non-unique case ?
Thanks for reply,
Hi,
I am trying to use autoarima function to make prediction of more than
1 items ( each has 100 values so it is 1x100 dimension matrix)
with frequency 52(weekly forecast). The problem is that some of fitted
arimas have drift and some not. It gives this error and stop processing
Best
Hi list, I would like to use the following data.frame to generate matrices
over a 3 year moving window:
DF = data.frame(read.table(textConnection( A B C
80 8025 1995
80 8026 1995
80 8029 1995
81 8026 1996
82 8025 1997
82 8026 1997
83 8025 1997
83 8027 1997
84 8026 1999
84
Did you mean this?
http://rgm2.lab.nig.ac.jp/RGM2/func.php?rd_id=latticeExtra:c.trellis
In fact I'm already using latticeExtra package because my xyplot is little bit
complicated...
So, I'm tring, thanks fro tricks baptiste!
Francesco
Date: Thu, 14 Apr 2011 08:52:45 +1200
Subject: Re: [R]
On 04/14/2011 03:30 AM, Michael Friendly wrote:
I have a diagram to be included in latex, where all my figures are .eps
graphics (so pdflatex is not an
option) and I want to achieve something
like the following: three concentric filled circles varying in lightness
or saturation. It is easiest to
On Thu, Apr 14, 2011 at 06:44:53PM +1200, Worik R wrote:
To improve the efficiency of a process I am writing I would like to cache
results. So I would like a data structure like a hash table.
So if I call Z - f(Y) I can cache Z associated with Y: CACHE[Y] - Z
I am stumped. I expected to
I am running some simulations in R involving reading in several
hundred datasets, performing some statistics and outputting those
statistics to file. I have noticed that it seems that the time it
takes to process of a dataset (or, say, a set of 100 datasets) seems
to take longer as the
On Apr 14, 2011, at 10:16 , Niels Richard Hansen wrote:
A third question. There exists a Fortran implementation called Expokit
with an ad hoc license stating that
Permission to use, copy, modify, and distribute EXPOKIT and its
supporting documentation for non-commercial purposes, is
Please reply to the list, so the OP and otheres following the thread
can see your contributions. I'm taking this back to r-help.
On Thu, Apr 14, 2011 at 01:43:31AM -0700, Mohammad Tanvir Ahamed wrote:
you can try it ...
rm(list=ls())
No - this has been suggested before and saying it gain
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
I have 10.000 simulations for a sensitivity analysis. I have done a few
sensitivity analysis for different response variables already,
but now, as most of the simulations (if not all) show some cyclic
behaviour, see how the independent input
Dear R-users,
After upgrading to the recent version of R, I'm having a problem updating task
views:
sessionInfo()
R version 2.13.0 (2011-04-13)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)
locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1]
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 14/04/11 11:57, Mike Marchywka wrote:
Date: Thu, 14 Apr 2011 11:29:23 +0200
From: r.m.k...@gmail.com
To: r-help@r-project.org
Subject: [R] Identify period length of time series
On 11-04-14 4:48 AM, Philipp Pagel wrote:
On Thu, Apr 14, 2011 at 06:44:53PM +1200, Worik R wrote:
To improve the efficiency of a process I am writing I would like to cache
results. So I would like a data structure like a hash table.
So if I call Z- f(Y) I can cache Z associated with Y:
Hi R-users,
This is a generic question, is there a way to plot a line graph for the output
from crosstable function? one of the inputs to the crosstab function is
categorical.
Taby --
[[alternative HTML version deleted]]
__
I have seen a couple of posts about this, but no solutions. The problem is
fitting the Basic Structural Model (BSM) to the AirPassengers time series using
StructTS. For this particular time series, if the length is reduced below 140
months, the BSM fits are bad.
The following illustrates the
It depends on how you want to work with the data. Here is an example
of converting both to POSIXct and a numeric value:
x - c('1:24', '13:46')
# convert to POSIXct
x.POSIXct - as.POSIXct(x, format = '%H:%M')
x.POSIXct
[1] 2011-04-14 01:24:00 EDT 2011-04-14 13:46:00 EDT
# to get a numeric
Dear all,
I am trying to implement the following in a loop:
g - cbind(c(1, 2, 3), c(1, 2, 3))
h - cbind(c(1, 2, 3), c(1, 2, 3))
c - cbind(g[,1]*h[,1], g[,2]*h[,2])
g-rowSums(c)
My attempt looks like this but does not produce the desired results as
above.
for (i in 1:2) {g-
Dear All,
#I have the following data:
primate-log(c(100,95,90,85,80,75,60,40,20,10))
rank-log(1:10)
#from which I can compute a linear regression, and obtain the confidence
interval for the slope coefficient as follows:
res-lm(primate~rank)
confint(res,level=0.9)[2,]
#Now, what I would like
Thank you Phillip for your post. I am reading in:
1. a 3 x 100 item parameter file (floating point and integer data)
2. a 100 x 1000 item response file (integer data)
3. a 6 x 1000 person parameter file (contains simulation condition information,
person measures)
4. I am then computing several
Hi Tanvir,
Line breaks matter in R. If I understand correctly you want
for (i in 1: 4){
if(i == 2){
print(a)
print(b)
}
}
Best,
Ista
On Wed, Apr 13, 2011 at 10:07 PM, Mohammad Tanvir Ahamed
mashra...@yahoo.com wrote:
hi , this can be done easily if (cond) expr
ex:
for (i in 1:
The matrix is generated by print.dist function ( getS3method(print, dist) ).
You could try this:
d - as.matrix(dist(matrix(rnorm(27), 9), diag = TRUE, upper = TRUE))
colnames(d) - letters[1:9]
On Wed, Apr 13, 2011 at 1:29 PM, Radhouane Aniba arad...@gmail.com wrote:
Hi,
I have a vector V
On Thu, Apr 14, 2011 at 3:51 AM, mathijsdevaan mathijsdev...@gmail.com wrote:
Hi list, I would like to use the following data.frame to generate matrices
over a 3 year moving window:
DF = data.frame(read.table(textConnection( A B C
80 8025 1995
80 8026 1995
80 8029 1995
81 8026
Hi:
Here's an example:
m - as.data.frame(matrix(rnorm(1), nrow = 100))
dim(m)
[1] 100 100
# Assign rownames to rows of m
rownames(m) - paste('A', 1:100, sep = '')
# Select a random subset of the rownames
vn - sample(rownames(m), 50)
# Subset m by selecting the rownames from vn
m2 -
--- begin included message ---
I am experimenting with the function predict() in two versions of R and
the R extension package survival.
library(survival)
set.seed(123)
testdat=data.frame(otime=rexp(10),event=rep(0:1,each=5),x=rnorm(10))
testfm=as.formula('Surv(otime,event)~x')
Many thanks for all the quick replies...
On 14/04/2011 13:37, Hao Wu wrote:
you can use this instead, apply(g*h,1,sum)
On Thu, Apr 14, 2011 at 7:28 PM, Thomas thomas.tri...@cantab.net
mailto:thomas.tri...@cantab.net wrote:
Dear all,
I am trying to implement the following in a
On Thu, Apr 14, 2011 at 06:50:56AM -0500, Barth B. Riley wrote:
Thank you Phillip for your post. I am reading in:
1. a 3 x 100 item parameter file (floating point and integer data)
2. a 100 x 1000 item response file (integer data)
3. a 6 x 1000 person parameter file (contains simulation
On Thu, Apr 14, 2011 at 12:40:53AM -0700, helin_susam wrote:
Hi Petr,
Your idea looks like logically. So, can we say this with your idea; the
expected number of computation in unique(sample(...)) is fewer than
sample(...). Because, the expected length is 63.39677 in unique case, while
the
Hi:
It's hard to diagnose the problem without an illustrative example. Perhaps
the following might help:
(1) When writing a function to use in ddply(), make a generic data frame the
input argument
to the function and refer to the variables within the function either
with the $ notation
On 04/13/2011 02:55 PM, Barth B. Riley wrote:
Dear list
I am running some simulations in R involving reading in several
hundred datasets, performing some statistics and outputting those
statistics to file. I have noticed that it seems that the time it
takes to process of a dataset (or, say, a
On 14-04-2011, at 09:00, Kristian Lind wrote:
HI Berend,
Thank you for your reply.
..
Finally the likelihood function at the end of your code
#Maximum likelihood estimation using mle package
library(stats4)
#defining loglikelighood function
#T - length(v)
#minuslogLik -
Dear list members,
I need to fit a regression with two inequality constraints. I look up in
the literature and R archive and found some ways to do it, for example
using the ic.infer package.
However I do not know how to write in R my constraints. In the package
help, it is told that I
On Apr 14, 2011, at 11:45 , Michael Kubovy wrote:
Dear R-users,
After upgrading to the recent version of R, I'm having a problem updating
task views:
Yes, a bug in the ctv package needs fixing...
sessionInfo()
R version 2.13.0 (2011-04-13)
Platform: x86_64-apple-darwin9.8.0/x86_64
What Ted and Peter did were Fisher's exact test, To get the previous
attainable p-value, what you do is the the fisher exact test p-values of
ALL the possible tables with margins fixed and choose the p-value that is
just below the one for fisher's exact test of the original table.
n Thu, Apr 14,
Hi,
I have to one set of inputs and their observed true values of each
input. Now I have a model takes the input and predict a value. I only
consider the ranks based on either the observed true values or the
predicted values. My question is how do I compare this two rank in R?
That is, how close
Hi,
Just wanted to mention that this worked perfectly.
Many thanks again and best wishes,
Ranjan
On Tue, 12 Apr 2011 13:04:14 -0500 Jonathan P Daily jda...@usgs.gov
wrote:
try:
options(help_type = 'text')
?options
If this works, you can create a site profile (A default is created
Which one is more efficient?
x2=c()
for (i in 1:length(x)) {
x2=c(x2,func(x[i]))
}
or
x2=x
for (i in 1:length(x)) {
x2=func(x[i])
}
where func is any function?
Dirk
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Hello,
I have a contingency table showing relation between two datasets. I tried
to see association among them with the assocplot, but it shows
error. mosaicplot of the same data worked perfectly.
Can anyone please help me.
Con.table=as.matrix(Con.table)
dim(Con.table)
[1] 27 27
thank you very much for your response!
But I have now another problem: optim( ) gives me the exactly same initial
values for the 4 parameters. I have tried many initial values, and it always
didnt change them.
opt$convergence = 1. I increased the number of iterations, but the problem is
Date: Thu, 14 Apr 2011 11:29:23 +0200
From: r.m.k...@gmail.com
To: r-help@r-project.org
Subject: [R] Identify period length of time series automatically?
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Hi
I have 10.000 simulations for a
Date: Wed, 13 Apr 2011 10:57:58 -0700
From: frederikl...@gmail.com
To: r-help@r-project.org
Subject: Re: [R] Incremental ReadLines
Hi there,
I am having a similar problem with reading in a large text file with around
550.000 observations with
Hi all,
I'm new to R and Rcpp, and I'm trying to learn Rcpp with the simplest code
possible. My goal is to be able to call R functions from C++.
The code I'm trying to run is:
#include
#include
#include
#include
int main(int argc, char* argv[])
{
Rcpp::NumericVector v(1);
Hi
I'm trying to plot several return level plots on the same axes to compare the
return levels from different data sets. To do this I am having to manually plot
the return levels for return periods. The problem is when I try to plot the
automatically calculated confidence intervals as the
Dear list,
I am running a fama macbeth regression using the lm.fit function and am
having problems extracting the rsquareds from each of the cross sectional
regressions (second stage) I have estimated.
I can obtain the coefficients using cfit = lm.fit(t(bhat[-1,]),
t(q))$coefficients
However I
I knew that the NA's in my data were the root cause of the trouble, but did
not find out how to get rid of them. Untill I found your another post
mentioning to use 'na.omit' to remove the lines containing 'NA's and the
problem got fixed after that.
Thanks for the help and all the trouble you have
Hi R-Users
I need to estimate Lyapunov exponent of my time series. After reading
description of all functions available I still don't know how to determine
time delay. My time series length is 4200. Is it possible to determine time
delay with other function's output or I can choose any random
Thanks, I thought that removing the list would take care of it. the question
is I do not see a .Rhistory file in my current working directory, so where
it is stored. it is not visible in C:\Program files\R either. Serarching the
C;\ and D:\ drives shows some old .Rhistory files but not the recent
thank you very much for your response!
But I have now another problem: optim( ) gives me the exactly same initial
values for the 4 parameters. I have tried many initial values, and it always
didnt change them.
opt$convergence = 1. I increased the number of iterations, but the problem is
Dear Dr. Therneau,
Thank you for your response. Just to point out that we didn't experience any
problems with the lm() function under R version 2.12.2 (2011-02-25):
set.seed(123)
testdat=data.frame(y=rexp(10),event=rep(0:1,each=5),x=rnorm(10))
testfm=as.formula('y~x')
If you don't want your history, why not just do this.
q()
Save workspace image? [y/n/c]:
n
??
Am I missing something?
1Rnwb wrote:
Thanks, I thought that removing the list would take care of it. the
question is I do not see a .Rhistory file in my current working directory,
so where it
Dear Pert,
Many thanks to your reply. Fully you are right!
Best wishes,
Helin.
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Dear Peter, thank you for your reply.
It works very well!
I have an additional question: what about the Wilcoxon test in case of
unequal paired samples with ties?
thank you!
net
--
View this message in context:
Hi Dirk,
On Thu, Apr 14, 2011 at 4:59 AM, dirknbr dirk...@gmail.com wrote:
Which one is more efficient?
x2=c()
for (i in 1:length(x)) {
x2=c(x2,func(x[i]))
}
or
x2=x
for (i in 1:length(x)) {
x2=func(x[i])
}
where func is any function?
This one. Creating a
On Apr 14, 2011, at 16:55 , Jim Silverton wrote:
What Ted and Peter did were Fisher's exact test, To get the previous
attainable p-value, what you do is the the fisher exact test p-values of ALL
the possible tables with margins fixed and choose the p-value that is just
below the one for
Thanks for the clarification, Jim. The terminology previous
was not self-explanatory!
The following implements (in a somewhat crude way, but explicit)
a solution to your question:
M - matrix(c(1, 4, 23, 30), byrow=TRUE, ncol=2)
M
# [,1] [,2]
# [1,]14
# [2,] 23 30
ok ?c.trellis works well.
But I still have a problem.
One of my plot is a combination of two xyplot on different scales:
a-xyplot(NDVI_P10~dek_num
| Year, type=a, data=data, xlim=c(1,37), ylim=c(0.1,0.8), as.table = TRUE,
layout =
c(13,1), aspect = 2, col=darkgrey,
col.axis=black, lwd=3,
Neither is the fastest method. The best way would be to vectorize func so
that it accepts and returns a vector. Many builtin R functions do this and
say so in their documentation.
The slower way would be to use one of the apply functions, such as:
?lapply
x2 - lapply(x, func)
If you must use
Date: Thu, 14 Apr 2011 12:42:28 +0200
From: r.m.k...@gmail.com
To: marchy...@hotmail.com
CC: r-help@r-project.org
Subject: Re: [R] Identify period length of time series automatically?
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On
Jurgens de Bruin debruinjj at gmail.com writes:
Hi,
I do not have much R experience just the basics, so please excuse
any obvious questions.
I would like to create bubble plot that have Categorical data on the x and y
axis and then the diameter if the bubble the value related to x and
Thanks Martin, this is very helpful.
Barth
-Original Message-
From: Martin Morgan [mailto:mtmor...@fhcrc.org]
Sent: Thursday, April 14, 2011 11:04 AM
To: Barth B. Riley
Subject: Re: [R] for loop performance
On 04/14/2011 07:12 AM, Barth B. Riley wrote:
Hi Martin
Question--when
Dear R users,
Hi. I want know R code of a function: predict.cv.glmnet (which is
included in glmnet package).
Could you let me know how I can see the R code of the function?
Thank you,
Soyeon Kim
__
R-help@r-project.org mailing list
Dear all
I've just tried the brand new 'grdevice' option in Sweave but couldn't
make it work. When I declare
\SweaveOpts{grdevice=pdf}
or
\SweaveOpts{grdevice=cairo_pdf}
trying to plot something simple
fig=T, echo=T=
plot(1:10,1:10,main='Some title')
@
would result in an Sweave error:
I have two suggestions to speed up your code, if you
must use a loop.
First, don't grow your output dataset at each iteration.
Instead of
cases - 0
output - numeric(cases)
while(length(line - readLines(input, n=1))==1) {
cases - cases + 1
output[cases] -
See help for getAnywhere()
Kevin
On Thu, Apr 14, 2011 at 11:24 AM, Soyeon Kim yunni0...@gmail.com wrote:
Dear R users,
Hi. I want know R code of a function: predict.cv.glmnet (which is
included in glmnet package).
Could you let me know how I can see the R code of the function?
Thank
You could also just download the source package from CRAN and look through it.
The thing is, the predict.cv.glmnet function isn't exported by the
package (via its namespace file) -- so it's somehow protected. You
could still see it using `:::`, like so:
R glmnet:::predict.cv.glmnet
Hi-
Tabular data have been provided to me within .csv files. I need to transform
the data from tabular format into a dataframe with three
columns. The columns need to be the table row id, table column id, and the
tabulated variable. An example dataset can be downloaded here:
Dear Community,
this is my first programming in R and I am stuck with a problem. I
have the following code which automatically calculates Granger
causalities from a variable, say e.g. bs as below, to all other
variables in the data frame:
log.returns-as.data.frame( lapply(daten, function(x)
Hello all,
First off, I am using R version 2.13.0 in Ubuntu.
I have read previous posts in the R mailing list on saving models for
later use, and the responses indicate using the R save() function to
save a model and then using load() to load it is one way to go.
However, when I use the save()
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