Dear Harrell,
Many thanks for your quick response!
However, after try and try, I still have difficulty to solve my questions.
I post my questions again. I hope someone can help me run the data and draw
the nomogram and calibration plot for me. I know that is not good but indeed
I have no way to
Fairly new at this.
Trying to create a conditional density plot.
cdplot(status~harvd.l,data=phy)
Error in cdplot.formula(status~harvd.l,data=phy):
dependent variable should be a factor
What does this error mean? Status is a binary response of infestation (0/1)
and harvd.l is the log of timber
What is teh reason some functions in the AICcmodavg package do not work with
'mer' class models?
One such example would be the 'importance' function.
Thanks
Ronny
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Sent
Hi:
Here's one way; using the mdply() function in the plyr package:
k - c(1,2,3,4,5)
i - c(0,1,3,2,1)
# Takes two scalars k and i as input, outputs a data frame
ff - function(k, i) data.frame(k = rep(k, i+1), i = seq(0, i, by = 1))
library('plyr')
mdply(data.frame(k, i), ff) # returns a data
On Mon, 01-Aug-2011 at 02:55PM -0400, Sushil Amirisetty wrote:
|
| Hi Everyone,
|
| When i try to install a package using
|
| install.packages(agricolae)
| --- Please select a CRAN mirror for use in this session ---
| |
|
|
| The cursor keeps blinking i dont get a popup menu to choose
Hi Jimmy
Years ago I think that Splus introduced an argument when dumping of
old.style = T or something similar to dump it into a form that could
be read into R.
This may only be for data.frames etc not things like random forest objects
Regards
Duncan
Duncan Mackay
Department of Agronomy
On 03/08/11 05:52, wildernessness wrote:
Fairly new at this.
Trying to create a conditional density plot.
cdplot(status~harvd.l,data=phy)
Error in cdplot.formula(status~harvd.l,data=phy):
dependent variable should be a factor
What does this error mean? Status is a binary response of
Hi
although the background is that it happened on an hpc cluster, this question
does *not* concern hpc computing with R.
I was using R on a cluster and had to install several packages in my home
directory. Now the head node was migrated to new hardware (new install as
well) and many dependencies
Does
xyplot(y ~ seq_along(y), xlab = Index)
do what you want?
Not exactly, because it does not work once multipanel conditioning comes
into play:
xyplot(y~seq_along(y)|factor(rep(1:2, each=5)), xlab = Index)
The points in the right panel are plotted from 6:10 while the points in
the
On 2011-08-02 21:52, wildernessness wrote:
Fairly new at this.
Trying to create a conditional density plot.
cdplot(status~harvd.l,data=phy)
Error in cdplot.formula(status~harvd.l,data=phy):
dependent variable should be a factor
What does this error mean? Status is a binary response of
Hello,
The idea is to grant access of remote users to R running on Linux.
Users must have ability to run their
R scripts but avoid corrupting the operating system.
How one can restrict/limit access of remote users to certain R
functions? For example, dealing with IO (file system), graphical
Dear all,
I would like to ask how one can access certain methods via fixInNamespace.
Is there some option / way for selecting a certain methods for a defined
signature.
Thank you for your answer and efforts in advance!
Best,
Michael
--
__
Hello,
I have a big data.frame, a piece of it as follows.
a b c d
1 58009 2010-11-02 0 NA
2 114761 NA 1 2008-11-05
3 184440 NA 1 2009-12-08
4
Hi
Hello,
I have a big data.frame, a piece of it as follows.
a b c d
1 58009 2010-11-02 0 NA
2 114761 NA 1 2008-11-05
3 184440 NA 1 2009-12-08
4 189372 NA 0 NA
5 105286
Please do your homework before asking the list:
An introduction to R, chapter 7
Am 03.08.2011 10:05, schrieb zcatav:
Hello,
I have a big data.frame, a piece of it as follows.
a b c d
1 58009 2010-11-02 0 NA
2
Wir sind bis am 20. August in den Ferien und werden keine e-mails beantworten.
Bei dringenden Fällen melden Sie sich bei Stefanie von Felten
steffi.vonfel...@oikostat.ch
We are on vacation until 20. August. In urgent cases, please contact Stefanie
von Felten steffi.vonfel...@oikostat.ch
Dear List,
i would like to mimic the behaviour or the following indexing with a do.call
construct to be able to supply the arguments to `[` as a list:
test = matrix[1:4,2]
result = test[2,]
My try, however, did not work:
result = do.call(`[`,list(test,2,NULL))
result =
On 2011-08-02 11:51, Sébastien Bihorel wrote:
Hi,
This might be a simple problem but I don't know how to calculate a random
variable density the way panel.histogram does it before it creates the
actual density rectangles. The documentation says that it uses the density
function but the actual
As others have noted, this is discussed in many free R tutorials, but if you
want to do it in one line I think this should do it:
X[is.NA(X[,b])(X[,c]==0),b]-2011-07-28 #where X is the name of the
data frame.
It's a somewhat convoluted line of code but if you read it inside out the logic
is
Petr Pikal wrote:
Hi
I believe there are better solutions but I would use two steps
select rows where c==0 (see also FAQ 7.31)
sel-which(big.data.frame$c==0)
change NA values in b column based on sel
big.data.frame$b[sel][is.na(big.data.frame$b[sel])]-20011-07-28
Beware of data
Your suggestion works perfect as i pointed previous message. Now have another
question about data editing. I try this code:
X[X[,c]==1,b]-X[,d]
and results with error: `[-.data.frame`(`*tmp*`, X[, c] == 1, b, value
= c(NA, :
replacement has 9 rows, data has 2
Logically i selected 2 rows with
Dear R Users,
Am using lm() with contrasts as below. If I skip the contrasts()
statement, I get the coefficient names to be
names(results$coef)
[1] (Intercept) VarAcat VarArat VarB
which are much more meaningful than ones based on integers.
Can anyone tell me how to get R to keep
As R notes, the problem is that you are trying to fit 9 rows into two:
specifically, note that on the left hand side you select only those rows
such that X[,c]==1 (which is 2 for your data) while on the right hand you
select all 9rows of column d so they simply don't fit. If you change the
code to
On Wed, Aug 3, 2011 at 8:09 AM, zcatav zca...@gmail.com wrote:
Your suggestion works perfect as i pointed previous message. Now have another
question about data editing. I try this code:
X[X[,c]==1,b]-X[,d]
and results with error: `[-.data.frame`(`*tmp*`, X[, c] == 1, b, value
= c(NA, :
Good question, Jannis.
I couldn't figure out how to specify the j argument in the [ function as
empty or missing either.
One work around is to specify ALL the columns as the j argument:
test - matrix(1:4, 2) # I think this is what you meant in your
original post, not matrix[1:4, 2]
The nomogram you included was produced by the Design package, the precursor
to the rms package. You will have to take the time to intensively read the
rms package documentation. Note that how you developed the model (e.g.,
allowing for non-linearity in log PSA, not using stepwise regression
Antonio Rodriges wrote:
The idea is to grant access of remote users to R running on Linux. Users
must have ability to run their
R scripts but avoid corrupting the operating system.
Check RStudio.org
Dieter
--
View this message in context:
Andrew Winterman wrote:
I'm trying to use the xlsx package to read a series of excel spreadsheets
into R, but my code is failing at the first step.
I setwd into my the directory with the spreadsheets, and, as a test ask
for
the first one:
read.xlsx(file = Argentina Final.xls,
So I take it 3D pie charts are out?
P.S. It is not about hiding anything. It is about consulting and being told
by your client to make 3D pie charts and change this font or that color to
make the graphs more apealing. Given that I am the one trying to open the
door to using R where I work it
On Aug 3, 2011, at 8:09 AM, zcatav wrote:
Your suggestion works perfect as i pointed previous message. Now
have another
question about data editing. I try this code:
X[X[,c]==1,b]-X[,d]
and results with error: `[-.data.frame`(`*tmp*`, X[, c] == 1,
b, value
= c(NA, :
replacement has 9
On Wed, Aug 3, 2011 at 9:04 AM, Antonio Rodriges antonio@gmail.com wrote:
Hello,
The idea is to grant access of remote users to R running on Linux.
Users must have ability to run their
R scripts but avoid corrupting the operating system.
Ordinary users can't corrupt the operating system
Hello David,
I encountered the same problem of yours.
What did you do to resolve it?
Thanks for your reply
Mohammad
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Sent from the R help mailing
Hello there,
Im computing the total value of an order from the price of the order items
using a for loop and the ifelse function. I do this on a large dataframe
(close to 1m lines). The computation of this function is painfully slow: in
1min only about 90 rows are calculated.
The
Dear R-users,
I am comparing differences in variance, skew, and kurtosis between two groups.
For variance the comparison is easy: just
var.test(group1, group2)
I am using agostino.test() for skew, and anscombe.test() for kurtosis.
However, I can't find an equivalent of the F.test or
Dear Caroline,
Here is a faster and more elegant solution.
n - 1
exampledata - data.frame(orderID = sample(floor(n / 5), n, replace = TRUE),
itemPrice = rpois(n, 10))
library(plyr)
system.time({
+ ddply(exampledata, .(orderID), function(x){
+ data.frame(itemPrice =
Hi,
I apologize for posting this here, I am also trying to post this on machine
learning emailing lists.
I have a set (18K) of sequences (22 nt long) and I have their counts at 4
different stages. The difference in counts from one stage to the next
represents how well the sequence performed in
Thank you very much Peter,
I'm going to dig deeper into the code of the functions you've listed.
On Wed, Aug 3, 2011 at 6:57 AM, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2011-08-02 11:51, Sébastien Bihorel wrote:
Hi,
This might be a simple problem but I don't know how to calculate a
Hi,
On Wed, Aug 3, 2011 at 10:06 AM, Vishal Thapar vishaltha...@gmail.com wrote:
Hi,
I apologize for posting this here, I am also trying to post this on machine
learning emailing lists.
I have a set (18K) of sequences (22 nt long) and I have their counts at 4
different stages. The
On 2011-08-03 00:24, Thaler,Thorn,LAUSANNE,Applied Mathematics wrote:
Does
xyplot(y ~ seq_along(y), xlab = Index)
do what you want?
Not exactly, because it does not work once multipanel conditioning comes
into play:
xyplot(y~seq_along(y)|factor(rep(1:2, each=5)), xlab = Index)
The
Hi there,
I had a problem when I hoped to get confidence intervals for the
parameters I got using mle() of stats4 package. This problem would not
appear if ``fixed'' option was not used. The following mini-example will
demo the problem:
x - c(100, 56, 32, 18, 10, 1)
r - c(18, 17, 10, 6, 4,
On Aug 3, 2011, at 12:30 , Jannis wrote:
Dear List,
i would like to mimic the behaviour or the following indexing with a do.call
construct to be able to supply the arguments to `[` as a list:
test = matrix[1:4,2]
result = test[2,]
My try, however, did not work:
result
On Aug 3, 2011, at 9:25 AM, Caroline Faisst wrote:
Hello there,
Im computing the total value of an order from the price of the
order items
using a for loop and the ifelse function.
Ouch. Schools really should stop teaching SAS and BASIC as a first
language.
I do this on a
On Wed, 3 Aug 2011, peter dalgaard wrote:
On Aug 3, 2011, at 12:30 , Jannis wrote:
Dear List,
i would like to mimic the behaviour or the following indexing with a do.call
construct to be able to supply the arguments to `[` as a list:
test = matrix[1:4,2]
result = test[2,]
My try,
Hello I am using the step function in order to do backward selection for a
linear model of more than 200 variables but it doesn't work correctly.
I think, there is a problem, if the matrix has same or more columns than
rows.
And if the matrix has too much columns the step-function doesn't work
Gabor Grothendieck wrote:
On Wed, Aug 3, 2011 at 8:09 AM, zcatav lt;zca...@gmail.comgt; wrote:
Your suggestion works perfect as i pointed previous message. Now have
another
question about data editing. I try this code:
X[X[,c]==1,b]-X[,d]
and results with error: `[-.data.frame`(`*tmp*`,
David Winsemius wrote:
On Aug 3, 2011, at 8:09 AM, zcatav wrote:
You need to apply the same logical test/selection on the rows of the
RHS as you are doing on the LHS.
Possibly:
X[ X[,c]==1, b] - X[ X[,c]==1, d]
This solution was suggested by R. Michael
Hi Peter,
Thanks for these information.
I used a column concatenating the listBy data to do this aggregation : (I
don't know if it's the best solution, but it seems to work).
aggregateMultiBy - function(x, by, FUN){
tableBy = data.frame(by)
tableBy$byKey =
for(colBy in
Hi,
I have some difficulties to work with the function lmer from lme4. My
responses are binary form and i want to use forward selection to my 12
covariates but i dont know how can I choose them based on deviance. Can
someone pls give me a example so i can apply. For example my covariates are
Hi all,
I have been using WEKA to do some text classification work and I want to try
out R.
The problem is I cannot load the String to Vector ARFF files created by
WEKA's string parser into Rattle .
Looking at the logs I get something like:
/Error in scan(file, what, nmax, sep, dec, quote,
Hello All,
I was trying to generate a map of Europe with the following codes:
europe-map(database=world, fill=FALSE,
plot=TRUE,xlim=c(-25,70),ylim=c(35,71))
However, the world database is too old to have right European country
names. Could anyone help?
Thanks,
Tianchan
--
View this message
I thought Sarah's reply was great and, alas, should probably be
templated for this list.
Not sure it fits as a fortunes package entry, but I thought it at
least worthy of consideration.
Cheers,
Bert
...
I appreciate any suggestions for this problem.
Sarah Goslee replied:
Suggestions? Yes.
On Wed, 2011-08-03 at 11:04 +0300, Antonio Rodriges wrote:
Hello,
The idea is to grant access of remote users to R running on Linux.
Users must have ability to run their
R scripts but avoid corrupting the operating system.
How one can restrict/limit access of remote users to certain R
Dear R users,
I am trying to determine how many characters can be displayed within the
width of an open grid viewport. Unfortunately, the arithmetic operation that
seems obvious in this case is be permitted with unit objects (see example
below). Although it isa brut force way to get this number
On Aug 3, 2011, at 9:59 AM, ONKELINX, Thierry wrote:
Dear Caroline,
Here is a faster and more elegant solution.
n - 1
exampledata - data.frame(orderID = sample(floor(n / 5), n, replace
= TRUE), itemPrice = rpois(n, 10))
library(plyr)
system.time({
+ ddply(exampledata,
zcatav zcatav at gmail.com writes:
Your suggestion works perfect as i pointed previous message. Now have another
question about data editing. I try this code:
X[X[,c]==1,b]-X[,d]
and results with error: `[-.data.frame`(`*tmp*`, X[, c] == 1, b, value
= c(NA, :
replacement has 9 rows,
On 8/3/2011 6:07 AM, wwreith wrote:
So I take it 3D pie charts are out?
At least with ggplot, yes. 2D pie charts are somewhat tricky with
ggplot, even. They can be gone with stacked, normalized bar charts
projected into polar coordinates, if I recall properly.
Not limited to ggplot,
There was too many spelling mistakes in my original post so I have
decided to re-submit it. So here is it
Dear R users,
I am trying to determine how many characters can be displayed within
the width of an open grid viewport. Unfortunately, the arithmetic
operation that seems obvious in this case
1. you wrote to the mailing list rather than to the original poster.
2. you forgot to cite the original post, hence we do not know what you
are referring to.
PLease do read the posting guide to this list!
Uwe Ligges
On 03.08.2011 14:53, mohammad...@gmail.com wrote:
Hello David,
I
You can look at the code
coxme:::print.coxme
There you will see that the global test is a chisquare
chi1 - 2*diff(x$loglik[1:2])
with x$df[1] degrees of freedom.
The fixed effects coefficients are found in x$coefficients$fixed, and
the variances are
This takes about 2 secs for 1M rows:
n - 100
exampledata - data.frame(orderID = sample(floor(n / 5), n, replace = TRUE),
itemPrice = rpois(n, 10))
require(data.table)
# convert to data.table
ed.dt - data.table(exampledata)
system.time(result - ed.dt[
+ ,
Hi All,
is there a way of using strsplit with a forward slash '/' as the splitting
point?
For data such as:
1 T/TC/C 16/33
2 T/TC/C 33/36
3 T/TC/C 16/34
4 T/TC/C 16/31
5 C/CC/C 28/29
6 T/TC/C 16/34
strsplit(my.data[1,1], /) # and any
Hello.
I am running the examples provided in the gstat help menus. When I try to
run the following in predict.gstat:
data(meuse)
coordinates(meuse)= ~x+y
v-variogram(log(zinc)~1, meuse)
I get the following error message:
Error in vector(double, length) : invalid 'length' argument
What's the
On 03/08/2011 12:37 PM, Federico Calboli wrote:
Hi All,
is there a way of using strsplit with a forward slash '/' as the splitting
point?
For data such as:
1 T/TC/C 16/33
2 T/TC/C 33/36
3 T/TC/C 16/34
4 T/TC/C 16/31
5 C/CC/C 28/29
6 T/T
Hi Federico,
A forward slash isn't a special character:
strsplit(T/T, /)
[[1]]
[1] T T
so there's some other problem.
Are you sure that your first column contains strings and not factors?
What does str(my.data) tell you?
Does
strsplit(as.character(my.data[1,1]), /)
work?
If you used
On 3 Aug 2011, at 17:41, Duncan Murdoch wrote:
It looks as though your my.data[1,1] value is a factor, not a character value.
strsplit(as.character(my.data[1,1]), /)
Thanks Duncan, this solved it.
Best
Federico
would work, or you could avoid getting factors in the first place,
On 3 Aug 2011, at 17:46, Sarah Goslee wrote:
Hi Federico,
A forward slash isn't a special character:
strsplit(T/T, /)
[[1]]
[1] T T
so there's some other problem.
Are you sure that your first column contains strings and not factors?
What does str(my.data) tell you?
Does
On 03/08/2011 12:47 PM, Baidya Nath Mandal wrote:
Dear Murdoch,
After setting CYGWIN=nodosfilewarning, i re-ran the R CMD check and got
following message:
* installing *source* package 'mypackage' ...
** libs
ERROR: compilation failed for package 'mypackage'
* removing
On Aug 3, 2011, at 12:20 PM, jim holtman wrote:
This takes about 2 secs for 1M rows:
n - 100
exampledata - data.frame(orderID = sample(floor(n / 5), n, replace
= TRUE), itemPrice = rpois(n, 10))
require(data.table)
# convert to data.table
ed.dt - data.table(exampledata)
On 2011-08-03 09:40, gbre...@ssc.wisc.edu wrote:
Hello.
I am running the examples provided in the gstat help menus. When I try to
run the following in predict.gstat:
data(meuse)
coordinates(meuse)= ~x+y
v-variogram(log(zinc)~1, meuse)
I get the following error message:
Error in
If you add column names to your contrast matrix (treat3) then those names will
be used in the coefficient names.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org
On Aug 3, 2011, at 2:01 PM, Ken wrote:
Hello,
Perhaps transpose the table attach(as.data.frame(t(data))) and use
ColSums() function with order id as header.
-Ken Hutchison
Got any code? The OP offered a reproducible example, after all.
--
David.
On Aug 3, 2554 BE, at 1:12
Here is my sessionInfo()
sessionInfo()
R version 2.12.2 (2011-02-25)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252 LC_CTYPE=English_United
States.1252
[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
I have a matrix that looks like this:
structure(c(0.0376673981759913, 0.111066500741386, 1, 1103,
18, OPEN, DEPR, 0.0404073656092023, 0.115186044704599,
1, 719, 18, OPEN, DEPR, 0.0665342096693433, 0.197570061769498,
1, 1103, 18, OPEN, DEPR, 0.119287147905722, 0.356427096010845,
1, 1103, 18,
Can I put limits on the lm() command? I only know that you can choose a
liniar model with or without an intercept, but can I put other limits on
the coefficients (for example- the intercept must be bigger than 1) ?
_
Hello,
We are trying to use R to simulate a model based on parameters 'a' and 'b'.
This involves the following integration:
model-function(s,x,a,b)(exp(-s*x*10^-5.5)*(s^(a-1)*(1-s)^(b-1)))
g- function(x,a,b){
out-c()
for (i in 1:length(x)){
out[i]-1-
Dear Murdoch,
After setting CYGWIN=nodosfilewarning, i re-ran the R CMD check and got
following message:
* installing *source* package 'mypackage' ...
** libs
ERROR: compilation failed for package 'mypackage'
* removing 'C:/Rpackages/mypackage.Rcheck/mypackage'
The log file contained following.
Hey,
Is there any function plotting several implicit functions (F(x,y)=0) on
the same fig. Is there anyone who has an example code of how to do this?
The contour3d function in the misc3d package only work with the functions
with three dimensions.
Thanks a lot.
Many thanks
Many many thanks, working now.
Best,
BN Mandal
On Wed, Aug 3, 2011 at 10:34 PM, Duncan Murdoch murdoch.dun...@gmail.comwrote:
On 03/08/2011 12:47 PM, Baidya Nath Mandal wrote:
Dear Murdoch,
After setting CYGWIN=nodosfilewarning, i re-ran the R CMD check and got
following message:
*
Since we got this the x-th time now:
Dear Fränzi Korner,
please please please never ever add auto-replies to your account that
also reply to mailing list messages! Thousands of readers of R-help get
your auto reply everey day now!
Best,
Uwe Ligges
On 03.08.2011 12:11,
Hello R experts,
I am trying to fit an lme model within a function, using a formula that I
passed into the function, and then predict that model from a different
function. Could you please advise me on how to do this? The following code
illustrates the essence of what I'm trying to do. The
I see a 'variogram' function in both spatial and gstat when I use ??
variogram on my machine that probably does not have even all of those
packages installed. Are you sure they are the same (I looked they
are not) or failing that that the one you expect is being chosen? And
are you
Hello R Help,
I am attempting to install/build a 64-bit version of R to hopefully resolve
some memory.limit problems for a user who is running a simulation. The
'configure' runs fine and the compilation (make) runs fine until the very last
part (see below). I have libiconv in /usr/local/lib
Dear List,
I have some difficulties to work with the function lmer from lme4. My
responses are binary form and i want to use forward selection to my 12
covariates but i dont know how can I choose them based on deviance. Can
someone pls give me a example so i can apply. For example my covariates
Sorry about the lack of code, but using Davids example, would:
tapply(itemPrice, INDEX=orderID, FUN=sum)
work?
-Ken Hutchison
On Aug 3, 2554 BE, at 2:09 PM, David Winsemius dwinsem...@comcast.net wrote:
On Aug 3, 2011, at 2:01 PM, Ken wrote:
Hello,
Perhaps transpose the table
Have tried to install R.app several times (6, in fact: versions 2.12, 13 14,
both 32 and 64 bit versions), using packages freshly downloaded from the
official project page, and failed every time, given exception reports such as
the following (appended below, the 2 reports arising out of my 1st
On 03/08/2011 3:04 PM, Jeffrey Joh wrote:
I have a matrix that looks like this:
structure(c(0.0376673981759913, 0.111066500741386, 1, 1103,
18, OPEN, DEPR, 0.0404073656092023, 0.115186044704599,
1, 719, 18, OPEN, DEPR, 0.0665342096693433, 0.197570061769498,
1, 1103, 18, OPEN, DEPR,
Hello,
Perhaps transpose the table attach(as.data.frame(t(data))) and use ColSums()
function with order id as header.
-Ken Hutchison
On Aug 3, 2554 BE, at 1:12 PM, David Winsemius dwinsem...@comcast.net wrote:
On Aug 3, 2011, at 12:20 PM, jim holtman wrote:
This takes
How about
Matrix[1:5,]=as.numeric(Matrix[1:5,])
-Ken Hutchison
On Aug 3, 2554 BE, at 3:04 PM, Jeffrey Joh johjeff...@hotmail.com wrote:
I have a matrix that looks like this:
structure(c(0.0376673981759913, 0.111066500741386, 1, 1103,
18, OPEN, DEPR, 0.0404073656092023,
On 03/08/2011 11:21 AM, KnifeBoot wrote:
Hey,
Is there any function plotting several implicit functions (F(x,y)=0) on
the same fig. Is there anyone who has an example code of how to do this?
The contour3d function in the misc3d package only work with the functions
with three
Please use R's search capabilities before posting.
RSiteSearch(Linear Model with Constraints)
appears to give you what you're looking for. Incidentally, with
constraints, the model is no longer linear, I believe.
-- Bert
2011/8/3 ראובן אברמוביץ gantk...@walla.com:
Can I put limits on the
Hi Jeffrey,
On Wed, Aug 3, 2011 at 3:04 PM, Jeffrey Joh johjeff...@hotmail.com wrote:
I have a matrix that looks like this:
structure(c(0.0376673981759913, 0.111066500741386, 1, 1103,
18, OPEN, DEPR, 0.0404073656092023, 0.115186044704599,
1, 719, 18, OPEN, DEPR, 0.0665342096693433,
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Ken
Sent: Wednesday, August 03, 2011 12:13 PM
To: Jeffrey Joh
Cc: r-help@r-project.org
Subject: Re: [R] Convert matrix to numeric
How about
On Aug 3, 2011, at 3:05 PM, Ken wrote:
Sorry about the lack of code, but using Davids example, would:
tapply(itemPrice, INDEX=orderID, FUN=sum)
work?
Doesn't do the cumulative sums or the assignment into column of the
same data.frame. That's why I used ave, because it keeps the sequence
Here is one more question,
How could I input an asymmetry in volatility speci cation in the BEKK
function?
As far as I know, the BEKK estimation function is
mvBEKK.est(eps, order = c(1,1), params = NULL, fixed = NULL, method =
BFGS, verbose = F)
I totally have no idea to exert an asymmetry
Dear ALL,
I use BEKK package to estimate Bivariate GARCH model. But when the results
come out, there's no t-stat or p-value of the estimated coeffients. Does
anyone know how to get the significance?
Followings are the codes I input,
P1=data.frame(x,y)
y1=mvBEKK.est(P1)
mvBEKK.diag(y1)
Anyhelp
To add to David's comments (nice catch, BTW), I found three
variogram() functions as a result of ??variogram. The one that gets
used is from the package that is highest in the search path (notice
that gstat is 55th (!!)) - that would be the one from the spatial
package. [The other is in the
Here's what you _should_ do
1) transpose
2a) as.data.frame
3a) fix the stupid default stringsAsFactor behavior
4a) convert the first 5 columns to numeric
dfrm - as.data.frame( t( structure(.) ) )
dfrm[, 1:5] -lapply(dfrm[, 1:5], as.character)
dfrm[, 1:5] -lapply(dfrm[, 1:5], as.numeric)
Or:
It is hard to prove a negative, but to the best of my knowledge lm will not do
what you want. This does not mean there is not a function that will perform
your analyses; the sort of thing you want to do is often accomplished using
non-linear methods.
John
ראובן אברמוביץgantk...@walla.com
On Aug 3, 2011, at 18:35 , Walter Ludwick wrote:
Have tried to install R.app several times (6, in fact: versions 2.12, 13
14, both 32 and 64 bit versions), using packages freshly downloaded from the
official project page, and failed every time, given exception reports such as
the
Did you install R first? R.app is just a GUI around the actual R code
that could run without any assistance in a terminal session. Generally
one installs both R and R.app from the super-bundle. Since you
provided no details of which .pkg files were chosen we are left
guessing.
(And
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