Hello,
I'm trying to install ncdf4 on RHEL 5.8, R version 2.15.1.
Previously installed is netcdf 3.6.2 from Red Hat, so I've compiled and
installed netcdf 4.2.1.1 (with hdf5 and zlib as per install instructions, and
also set --enable-netcdf4 option) into /usr/local.
When attempting to
I am wondering how to obtain the model/equation at each level automatically
in a regression model with a few factors
without looking at summary of the lm model. For example, consider
lm.factors - lm(y ~ x1 + factor(x2)*factor(x3)+x4*factor(x5))
The coefficients of lm.factors in
I am sorry I have confused you, the logs are all base e:
ln(a) = 1347
ln(b) = 1351
And I am trying to solve this expression:
exp( ln(a) ) - exp( ln(0.1) + ln(b) )
Thank you.
2013/2/4 francesca casalino francy.casal...@gmail.com:
Dear R experts,
I have the logarithms of 2 values:
log(a)
I have the same problem as Daniel Westphal, packages are always installed with
umask 022, I'd like 002.
Any solution so far?
Thank you.
Francois
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
Hi all,
I've written a function for a simulation which will - in general operation
without being specific to the simulation scenario, duplicate or delete columns
from a matrix based on two values which determine how many as a proportion of
the matrix: the two values are always between 0.01 and
Hi guys,
I have a problem with the placement of my gwindow widget. When I start the
gwindow widget it is displayed near the middle of my screen but I want that
the widget is on the upper left side of my screen. Is it possible to define
where my gwindow pops up? And additionally, I want to focus
I am having a puzzling problem with bloomberg connection. When i run from R
prompt some code that has
library(Rbbg)
conn - blpConnect(throw.ticker.errors = FALSE)
print(connected)
...
I establish connection every time and then proceed to get data when i run this
code from R prompt.
Hi there,
Thanks for the quick feedback.
There is no further info on the reason for the crash - R just stops.
The for loop is there just to generate the error. I'm following a previous
example that others couldn't reproduce. I need Rprof to work on a different
set of functions that I haven't
Hi Jim,
Could you be a little more explicit of what the error message is. On
my Windows system 32-bit, I get the following:
Sorry - it just stops as per the message posted. The text is from eclipse.
Rprof()
z - 1
system.time(for (i in 1:1e8) z - z + 1/i)
Error: cannot allocate vector
Update: A bit more digging and I found duplicated() (
http://tolstoy.newcastle.edu.au/R/e4/help/08/06/13592.html). Sorry for the
premature request for help!
On Mon, Feb 4, 2013 at 6:29 PM, Dylan Arena dar...@stanford.edu wrote:
Hi there,
I have data frame with columns ID and Date. There
Hi,
Thanks. This extract every row that satisfy the condition, but I need look at
the last row (the maximum of cumulative sum) for each block (m1,n1). for
example, if I set the criteria
res2$cterm1_P1L0.6 res2$cterm1_P0H0.95, this should extract m1= 3, n1 = 2.
Hi,
I am not sure I
Hi,
I am not sure about what your end result should be:
From your code, it looks like you want to replace the rows of dat1 that
contain at least a `3` with the corresponding row of dat2. Or is it only the
cell with number `3` replaced by corresponding row in dat2.
dat1- read.table(text=
A1
Hi there,
I have data frame with columns ID and Date. There are multiple rows for
each ID, but I only want to keep the *first* such row--i.e., the row
corresponding to the earliest event. So if I had, say, 1000 rows of 100
IDs doing an average of ten events each, I'd run this trimming
Hello,
I have a dataframe with positive integers and
for every value 0
I would like to have 1 and the rest should be zero.
For instance
1 0 1
0 1 1
with df as reference.
df
V1 V2 V3
yes 23 0 5
no 0 5 7
dput (df)
structure(list(V1 = c(23, 0), V2 = c(0, 5), V3 = c(5, 7)), .Names =
Hi Sir,
I'm Roberto Palloni, student in economics. My question is about Multiple
Imputation with MICE.
I would like to know if skipping the imputation of a variable used as predictor
could affect the estimation of the dependent variable.
In other words, does the imputed values of a predictor are
Hi Bibek,
You can do this in different ways:
set.seed(15)
dat1-as.data.frame(matrix(sample(c(Yes,No),20,replace=TRUE),ncol=5),stringsAsFactors=FALSE)
dat1[dat1==Yes]-1
dat1[dat1==No]-0
dat1[]- sapply(dat1,as.numeric)
dat1
# V1 V2 V3 V4 V5
#1 0 1 0 0 1
#2 1 0 0 0 0
#3 0 0 1 0 1
I think that the excellent estimable function from gmodels should help you.
Cheers
Andrew
On Tuesday, February 5, 2013, chunlin liu wrote:
I am wondering how to obtain the model/equation at each level automatically
in a regression model with a few factors
without looking at summary of the
Hello,
I have a problem regarding calculation of Cumulative Incidence Function.
The event of interest is failure of bone-marrow transplantation, which may
occur due to relapse or death in remission. The data set that I have
consists of- lifetime variable, two indicator variables-one for relapse
Hi everyone,
I compared eigen() and irlba() (from the eponym package), for the eigen
decomposition of 1500x1500 and 3000x3000 dense matrices, retaining only the
two major eigenpairs.
I obtained the following results (VM running on two cores of a core
i7-2600) :
- 1500x1500 : 5.4s with eigen(),
On 02/05/2013 05:33 AM, Wim Kreinen wrote:
Hello,
I have a dataframe with positive integers and
for every value 0
I would like to have 1 and the rest should be zero.
For instance
1 0 1
0 1 1
with df as reference.
df
V1 V2 V3
yes 23 0 5
no 0 5 7
dput (df)
structure(list(V1 =
Dear Wim,
You could try
1*(df 0)
HTH,
Jorge.-
On Tue, Feb 5, 2013 at 5:33 AM, Wim Kreinen wrote:
Hello,
I have a dataframe with positive integers and
for every value 0
I would like to have 1 and the rest should be zero.
For instance
1 0 1
0 1 1
with df as reference.
df
I don't understand the problem: Do you mean that given either
a or b, solve for the other? Otherwise, I don't see an unknown in
the problem statement.
Further, do you mean that exp( ln(a) ) - exp( ln(0.1) + ln(b) )
= 0?
Or equivalently, exp( ln(a) ) = exp( ln(0.1) +
Hi Tasnuva,
You can read Scrucca, L.; Santucci, A. and Aversa, F. (2007) Competing risk
analysis using R: an easy guide for clinicians, Bone Marrow Transplantation,
40, 381-387, and also Brock, G.; Barnes, C.; Ramirez, J. and Myers, J. (2011)
R code for calculating the competing risks
Hello,
To average the values of z in case of duplicated x and y, you can use
s2 - aggregate(z ~ x + y, data = sorpe, FUN = mean)
Hope this helps,
Rui Barradas
Em 05-02-2013 07:06, Richard Müller escreveu:
Hello,
I have a long list of x-, y- and z-data and try to generate a heatmap.
Andre,
Does the batch script always fail, or is it sometimes successful? You've
probably tried running in batch and interactively one after the other, but
if not please do that to make sure it isn't a timing issue with the
Bloomberg connection.
Also, you can turn up the logging in Rbbg: conn -
Dear R-users,
a question concerning sparse matrices in package spam (spam_0.29-2).
On one hand I have a spam object (n X n) from which I cannot compute the
inverse. On the other hand, if I convert this object in a plain matrix, I can
find the inverse without any problem.
Specifically I get
Hola R, tengo las siguientes preguntas:
Pregunta 1:
Cargar las tablas de los datos de peliculas en R usando `ff`.
Cómo se construye una columna nueva que de, para cada cliente y cada
evaluación,
de el número de días que han pasado desde la primera evaluación del cliente?
Qué función se utiliza
Dear R users,
I need to perform a block bootstrap to a time series. The block size must be 3
and the blocks must NOT overlap.
I found a number of functions for bootstrap in R (tsboot, tsbootstrap in
tseries, and LocSig in SpatialVx), but neither of them seems to implement
non-overlapping
Hi all,
I realised that my last email question and code was probably going to be a bit
of an eyesore for some people and that perhaps the best thing for me to do is
to pose the question of what it is I want to achieve, rather than what I've
written, if it helps people:
I'm writing a
Hello,
I am trying to improve my skill on S4 classes. But there's something
strange (to me) happening with initialize methods. This is probably a
normal behaviour... but... something is unclear in my mind
I declare two classes, namely A, and B, which has a slot of type A. The
class A has an
HI,
If the `Date` column is not ordered:
Date1=as.Date(c(01/05/2012,01/07/2012,01/15/2012,01/09/2012,01/14/2012,01/25/2012,
01/08/2012,01/24/2012,01/03/2012),format=%m/%d/%Y)
dat1-data.frame(ID=rep(1:3,each=3),Date1)
aggregate(Date1~ID,data=dat1,function(x) min(x))
# ID Date1
#1 1
Dear R community,
is there an easy way to convert an adjacency list (or a data-frame) to a
non-symmetric matrix?
The adjacency list has the following form:
person group
1 Sam a
2 Sam b
3 Sam c
4 Greg a
5 Tom b
6 Tom c
7 Tom d
8 Mary b
9 Mary d
I need the data in a matrix with persons as rows
Good morning to you all,
Sorry for taking your time from your research and
teaching schedules.
If you have a non-stationary univariate time Series
data that has the transformation:
Say; l.dat-log (series)
d.ldat-diff (l.dat, differences=1)
and you fit say arima model.
predit.arima-predict
Hi,
You could try:
df[df0]-1
#or
df[]-as.integer(df!=0)
df
# V1 V2 V3
#yes 1 0 1
#no 0 1 1
A.K.
- Original Message -
From: Wim Kreinen wkrei...@gmail.com
To: r-help@r-project.org
Cc:
Sent: Monday, February 4, 2013 1:33 PM
Subject: [R] Transform every integer to 1 or 0
Hi,
I am trying to install the rJava package in a Linux environment (Platform:
x86_64-suse-linux-gnu (64-bit)), where I am facing an issue which I cannot
solve so far and where I could require some help:
The installation seems to go fine, however in the last step of the installation
where
Hi,
res3-with(res2,aggregate(cbind(cterm1_P1L,cterm1_P0H),by=list(m1,n1),max))
res3[res3[,3]0.6 res3[,4]0.95,] #this doesn't change the result as the
conditions are not met
# Group.1 Group.2 cterm1_P1L cterm1_P0H
#1 2 2 0.01440 0.00273750
#2 3 2 0.00032
R_BadLongVector is an entry point in 64-bit R-devel (pre-3.0.0). It
seems something got installed under the wrong version of R.
I'd also be concerned about your Java version. 1.6.0_18 is very old and
there are security alerts on old Java versions, even those from last month.
On 05/02/2013
Dear all
I have a model that looks like this:
m1 - lmer(Difference ~ 1+ (1|Examiner) + (1|Item), data=englisho.data)
I know it is not possible to estimate random effects but one can
obtain BLUPs of the conditional modes with
re1 - ranef(m1, postVar=T)
And then dotplot(re1) for the examiner
Hello,
I calculate the correlation between two matrices cor(x,y, method=spearman)
and I am wondering if it possible to see only the significant correlations. I
can do that for single OTUs with cor.test() command but I would like to have an
output for whole matrix. Besides,I would like to plot
For converting from first dataset to second,
library(reshape2)
#dat1 is data
dcast(dat1,person~group,value.var=group,length)
# person a b c d
#1 Greg 1 0 0 0
#2 Mary 0 1 0 1
#3 Sam 1 1 1 0
#4 Tom 0 1 1 1
A.K.
- Original Message -
From: Sebastian Haunss
Try table():
adj - table(d) # 'd' is your data.frame
adj
group
person a b c d
Greg 1 0 0 0
Mary 0 1 0 1
Sam 1 1 1 0
Tom 0 1 1 1
If you have duplicate rows in the d then it will
tally them up. You can can convert positives to
1's with adj[adj0] - 1.
Bill
On Mon, Feb 4, 2013 at 7:07 PM, Andrew Harley a.har...@bom.gov.au wrote:
Hello,
I'm trying to install ncdf4 on RHEL 5.8, R version 2.15.1.
Previously installed is netcdf 3.6.2 from Red Hat, so I've compiled and
installed netcdf 4.2.1.1 (with hdf5 and zlib as per install instructions, and
1) We don't have your previous email and I doubt anyone here committed your
code to memory
2) No offense but this post is still an eye sore. Actually I am guessing
even worse than the first because there is no working example. The idea is
to provide *minimal* code that reproduces the problem -
If the order of people in the original data frame is important, one more line
matches your example output:
d$person - factor(d$person, levels=unique(as.character(d$person)))
table(d)
group
person a b c d
Sam 1 1 1 0
Greg 1 0 0 0
Tom 0 1 1 1
Mary 0 1 0 1
Are there any tricks I can use to get a real result
for exp( ln(a) ) - exp( ln(0.1) + ln(b) ), either in logarithm or
exponential form?
log.a - 1347
log.b - 1351
f0 - function(log.a, log.b) exp(log.a) - exp(log(0.1) + log.b)
will not work because f0(1347,1351) is too big to represent as
On 02/05/2013 04:42 AM, Franck Doray wrote:
Hello,
I am trying to improve my skill on S4 classes. But there's something
strange (to me) happening with initialize methods. This is probably a
normal behaviour... but... something is unclear in my mind
I declare two classes, namely A, and B,
Daniel Caro dcarov at gmail.com writes:
Dear all
I have a model that looks like this:
m1 - lmer(Difference ~ 1+ (1|Examiner) + (1|Item), data=englisho.data)
I know it is not possible to estimate random effects but one can
obtain BLUPs of the conditional modes with
re1 - ranef(m1,
I stayed out of this one thinking it was probably a homework exercise.
After others have responded, I'll go ahead with my gloss on
Bill's function...
The specific problem is really of the form
exp(a) - exp(a+eps) = exp(a)*(1-exp(eps))
So even though we can't compute exp(1347), we can
Functions log1p and expm1 may be of interest here. To understand
them, look at a 2- or 3-term Taylor expansion for log(x) [natural log,
of course] and exp(x). Spencer
On 2/5/2013 10:48 AM, Albyn Jones wrote:
I stayed out of this one thinking it was probably a homework exercise.
After others
Dear list members,
is there any maximum amount of lines of code that can be copy pasted
into the R console? Or some maximum amount of levels of nested
parantheses? Whenever I copy paste some big chunk of code into the R
console I get errors of missing parantheses etc. even though the
If you just want point forecasts, it's simple:
Let your original series be X_t, t=1, ..., N.
Let Y_t = log(X_t).
Let Z_t = Y_t - Y_{t-1}, t = 2, ..., N.
Fit your model and forecast, obtaining Z-hat__1, ..., Z-hat_10.
Then Y-hat_{N+1} = Y_N + Z-hat_1, Y-hat_{N+2} = Y-hat_{N+1} + Z-hat_2,
.,
On Mon, Feb 4, 2013 at 10:11 AM, francesca casalino
francy.casal...@gmail.com wrote:
Dear R experts,
I have the logarithms of 2 values:
log(a) = 1347
log(b) = 1351
And I am trying to solve this expression:
exp( ln(a) ) - exp( ln(0.1) + ln(b) )
But of course every time I try to
Presumably you wish to *calculate*
exp(1347) - exp(eps + 1351) where eps = ln(0.1)
(rather than *solve* anything).
This is equal to
exp(1347)*(1 - exp(eps + 4)) = - exp(1347 + log(exp(eps+4) - 1)
= -exp(1347 + 1.495107) = - exp(1348.495107)
This is, to all intents and purposes,
On Tue, Feb 5, 2013 at 3:35 PM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Mon, Feb 4, 2013 at 10:11 AM, francesca casalino
francy.casal...@gmail.com wrote:
Dear R experts,
I have the logarithms of 2 values:
log(a) = 1347
log(b) = 1351
And I am trying to solve this expression:
I am looking for a package that will allow me to choose R (a real number) that
minimizes the correlation of y and x^R, i.e.
find R such that corr(y,x^R) is minimized. Any suggestions for packages I might
look at would be helpful.
Thanks,
John
John David Sorkin M.D., Ph.D.
Chief,
The stats package has the optimize function that could be used for this.
On Tue, Feb 5, 2013 at 2:08 PM, John Sorkin jsor...@grecc.umaryland.eduwrote:
I am looking for a package that will allow me to choose R (a real number)
that minimizes the correlation of y and x^R, i.e.
find R such that
If you really mean minimize and not maximize, I doubt that any
such R exists.
-- Bert
On Tue, Feb 5, 2013 at 1:08 PM, John Sorkin jsor...@grecc.umaryland.edu wrote:
I am looking for a package that will allow me to choose R (a real number)
that minimizes the correlation of y and x^R, i.e.
I thought that I can use metacharacters such as \w to match word characters
with one backslash. But for some reason, I need to include two backslashes.
grepl(pattern='\w', x=what)
Error: '\w' is an unrecognized escape in character string starting \w
grepl(pattern='\\w', x=what)
[1] TRUE
I
Thanks very much for your help!
I want to add some more columns based on the results. Is the following code
good way to create such a data frame and How to see the column m and n in
the updated data?
d2-reres3[res3[,3]0.01 res3[,4]0.01,]
colnames(d2)[1:2]- c(m1,n1);
d2
d3-data.frame(d2)
for
I've been away from this list but reentered recently. Now when I send a
question I get an automated answer saying:
---
Your mail to 'R-help' with the subject
..
Is being held until the list moderator can review it for approval.
The reason it is being held:
Post to
When I run the following function
HQ2 - function(n) {
nv - 6 * sqrt(n)
fcn - function(z) {
pchisq(z^2 / 36, n - 1) * dnorm(nv - z)
}
## I want the integral from 0 to infinity:
f.Inf - integrate(fcn, 0, Inf)
## Doc: Don't do this:
f.100 - integrate(fcn,
Both Greg and David's methods are working fine for me under R version 2.15.2
without warnings.
setClass(Currency)
setAs(character, Currency,
function(from) as.numeric(gsub([$,],,sub($,,from, fixed=TRUE
# method 2: ,|\\$, working as desired
setClass(Currency2)
setAs(character, Currency2,
Thanks (acutally same as Richard's) this was very helpful
2013/2/5 arun smartpink...@yahoo.com
Hi,
You could try:
df[df0]-1
#or
df[]-as.integer(df!=0)
df
#V1 V2 V3
#yes 1 0 1
#no 0 1 1
A.K.
- Original Message -
From: Wim Kreinen wkrei...@gmail.com
To:
Hi,
You can reduce the steps to reach d2:
res3- with(res2,aggregate(cbind(cterm1_P1L,cterm1_P0H),by=list(m1,n1),max))
#Change it to:
res3new- aggregate(.~m1+n1,data=res2[,c(1:2,12:13)],max)
res3new
m1 n1 cterm1_P1L cterm1_P0H
1 2 2 0.01440 0.00273750
2 3 2 0.00032 0.0025
3 2 3
On 05/02/2013 12:49 PM, Seth Dickey wrote:
I thought that I can use metacharacters such as \w to match word characters
with one backslash. But for some reason, I need to include two backslashes.
grepl(pattern='\w', x=what)
Error: '\w' is an unrecognized escape in character string starting \w
I really mean minimize.
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913
The moderation has changed. Postings from new subscribers are held until
reviewed and approved as passing weakly structured Turing test. (I just removed
your moderation flag and your postings will not be further reviewed after
whatever day long review cycle completes.)
On Feb 5, 2013, at
On Feb 5, 2013, at 9:49 AM, Seth Dickey wrote:
I thought that I can use metacharacters such as \w to match word characters
with one backslash. But for some reason, I need to include two backslashes.
grepl(pattern='\w', x=what)
Error: '\w' is an unrecognized escape in character string
On 2/4/2013 3:55 PM, Simon Urbanek wrote:
On Feb 4, 2013, at 7:14 PM, Spencer Graves wrote:
On 2/4/2013 7:03 AM, Simon Urbanek wrote:
On Feb 4, 2013, at 10:27 AM, Spencer Graves wrote:
On 2/4/2013 5:22 AM, Milan Bouchet-Valat wrote:
Le lundi 04 février 2013 à 08:19 -0400, Simon Urbanek a
On Feb 5, 2013, at 5:51 PM, Spencer Graves wrote:
On 2/4/2013 3:55 PM, Simon Urbanek wrote:
On Feb 4, 2013, at 7:14 PM, Spencer Graves wrote:
On 2/4/2013 7:03 AM, Simon Urbanek wrote:
On Feb 4, 2013, at 10:27 AM, Spencer Graves wrote:
On 2/4/2013 5:22 AM, Milan Bouchet-Valat wrote:
Le
Hi,
I would like to calculate the area under segmented regression lines (single
breakpoints). I had thought that I could do this using the predict() function
to ascertain some key x y values in order to determine the dimensions for two
trapezoids (under each segment of my regression). However,
Hi,
I'm trying to calculate some functions loop, unfortunately my data are
decimal numbers that not allow me to get an accurate estimate. I do not want
to use neither round nor signif but I would prefer to have all the
information available, (the function loop is generic so you should adapt to
John, thanks for your reply. Batch job fails always. I tried running
interactively at the same time as batch tries and at different times.
Interactively, job always works. I set log level to finest, but it didn't
seem to add new anything to logs. I included new log messages below just in
case.
Dear R experts---
after many years, I am planning to give in and write my first R
package. I want to combine my collection of collected useful utility
routines.
as my guide, I am planning to use Friedrich Leisch's Creating R
Packages: A Tutorial from Sep 2009. Is there a newer or better
-Original Message-
From: davidwilliampie...@gmail.com
[mailto:davidwilliampie...@gmail.com] On Behalf Of David W. Pierce
Sent: Wednesday, 6 February 2013 3:59 AM
To: Andrew Harley
Cc: r-help@R-project.org
Hi Andrew,
I think it's reasonably likely that some components of the old
On 02/06/2013 10:40 AM, mary wrote:
Hi,
I'm trying to calculate some functions loop, unfortunately my data are
decimal numbers that not allow me to get an accurate estimate. I do not want
to use neither round nor signif but I would prefer to have all the
information available, (the function
My short answer is to watch this video by Jeffrey Horner
http://youtu.be/ScV7XXlBZww and learn roxygen2.
And the long answer is to read the manual which has everything you
need: http://cran.r-project.org/doc/manuals/r-release/R-exts.html
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone:
Dear ALL,
I need a function that takes character string as argument value and replaces
corresponding argument( which is behind dollar sign $) in function body with
its value.
For example:
function(dataframe, argument=ANYTHING I WANT)
{
return(dataframe$argument)
}
thanks for the responses. this is what I have learned so far:
primarily, I need to learn roxygen2 and devtools, because they do what
I was planning to reinvent.
roxygen2 allows R users to write R code that embeds its documentation.
the .R user source file has a family semblance with POD (perl's
actually, I may have found what I was looking for in
https://github.com/hadley/devtools/wiki/Philosophy
---
Ivo Welch (ivo.we...@gmail.com)
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting
Hi everyone,
I'd like to be able to apply lda to each 2D matrix slice of a 3D array, and
then use the scalings to obtain the corresponding lda scores.
I can use 'apply' to get a list of the lda output for each 2D slice, and
can create a list of the resulting scalings, but I'm not sure how to
I have a minimal package here based on roxygen2:
https://github.com/yihui/rmini
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Tue, Feb 5, 2013 at 11:49 PM, ivo welch
On Feb 5, 2013, at 8:07 PM, 谢一鸣 wrote:
Dear ALL,
I need a function that takes character string as argument value and
replaces corresponding argument( which is behind dollar sign $) in
function body with its value.
For example:
function(dataframe, argument=ANYTHING I WANT)
{
Thanks David, your solution perfectly solve my problem. But Do you know what
exactly the mechanism of Computing on the language in pdf file: R language
definition, Can that THING do the same job?
Thank you very much!
Xie YM
2013-02-06 14:22:57，David Winsemius dwinsem...@comcast.net 写道：
The $
Try:
n - length(mlist)
result - lapply(1:n,function(i,m,v){m[[i]]%*%v[[i]]},m=mlist,v=vlist)
These days the use of lapply is unlikely to be much, if at all, faster than
the use of a for loop.
cheers,
Rolf Turner
On 02/06/2013 06:50 PM, David Romano wrote:
Hi everyone,
I'd like
Hi everyone,
Do we have GARCH-in-mean VAR in R?
Is there any GARCH package in R?
Thanks,
Miao
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