Wonderful, David! Thank you much for your help on this.
AC
On Fri, Jul 9, 2010 at 6:49 PM, David Winsemius wrote:
>
> On Jul 9, 2010, at 7:19 PM, Erik Iverson wrote:
>
> AC Del Re wrote:
>>
>>> Hi David,
>>> I don't have the function in order yet but the names will be going into
>>> the
>>> for
Steve,
Couldn't he also just use the decision.value property to see the
equivilent of t(x) %*% b for each row?
-N
On 7/9/10 7:11 PM, Steve Lianoglou wrote:
Hi,
On Fri, Jul 9, 2010 at 12:15 PM, manuel.martin
wrote:
Dear all,
after having calibrated a svm model through the svm() comma
Hi,
The absolute function stopping criterion is not meant for any positive
objective function. It is meant for functions whose minimum is 0. Here is
what David Gay's documentation from PORT says:
"6 - absolute function convergence: |f (x)| < V(AFCTOL) = V(31). This test is
only of interest
Hi Matt,
This works great, thanks!
At first I got an error message saying BLOB is not implemented in RSQLite.
When I updated to the latest version it worked.
Is there any reason the string needs to be stored as type BLOB? It seems to
work the same when I swap "BLOB" with "TEXT" in the CREATE
Hi,
On Fri, Jul 9, 2010 at 12:15 PM, manuel.martin
wrote:
> Dear all,
>
> after having calibrated a svm model through the svm() command of the e1071
> package, is there a way to
> i) represent the modeled relationships between the y and X variables
> (response variable vs. predictors)?
Can you e
On 07/09/2010 05:33 PM, Saeed Abu Nimeh wrote:
Is there a function similar to combine.levels ( in the Hmisc package)
that combines the levels of factors, but not based on their frequency.
Alternatively, I am looking into using the significance of the dummy
variables of factors based on their Pr(>
Thanks so lot
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Thanks this works
On Fri, Jul 9, 2010 at 4:32 PM, Wu Gong [via R] <
ml-node+2284062-824667456-312...@n4.nabble.com
> wrote:
> Do you mean substring?
>
> sub(".txt","", "mytest.txt")
> A R learner.
>
>
> --
> View message @
> http://r.789695.n4.nabble.com/String-trunc
one string named as: mytest.txt
how can I remove the .txt and return the mytest only.
i tried split substr, and grep didn't work it out,
Thanks so lot
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And further to those two:
nlminb( obj = function(x) 2*(x+3), start=-2, lower=-Inf, upper=Inf )
On 9 July 2010 23:09, Matthew Killeya wrote:
> Yes clearly a bug... there are numerous variations ... problem seems to be
> for a linear function whenever the first function valuation is 1.
>
> e.g.
Hi there,
I tried the code that was sent. Thanks a lot. It doesn't seem to work.
Here is what I'm working with...
inc<-ts(incdat[,2], start=1968.000, freq=12)
par(mfrow=c(1,2))
plot(stl(log(inc), s.window="periodic"))
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View this message in context:
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Yes clearly a bug... there are numerous variations ... problem seems to be
for a linear function whenever the first function valuation is 1.
e.g. two more examples:
nlminb( obj = function(x) x+1, start=0, lower=-Inf, upper=Inf )
nlminb( obj = function(x) x+2, start=-1, lower=-Inf, upper=Inf )
(
Hi,
I'm using the sensitivity package to calculate the sobol indices. The case is
that I have 13 parameter and each one has 1000 samples and my model result is a
vector having 1000 values (based on the 1000 sample simulation).
I just wonder how to use the sobol function, the pdf looks straightfo
Jeremie Smaga 4ecap.com> writes:
>
> Good afternoon,
>
> I have been experiencing a lot of crashes working with large vectors in R.
>
> Specifically, I am using XTS of length of minimum 120k elements.
>
> My problem is that I cannot display the vector (otherwise R crashes), I
> cannot plot it
I should add that I'm using R 2.10.1 on a Windows 7 machine, thanks!
On Fri, Jul 9, 2010 at 3:11 PM, Gene Leynes
> wrote:
> I thought the "apply" functions are faster than for loops, but my most
> recent test shows that apply actually takes a significantly longer than a
> for loop. Am I missing
On Jul 9, 2010, at 7:19 PM, Erik Iverson wrote:
AC Del Re wrote:
Hi David,
I don't have the function in order yet but the names will be going
into the
formula, e.g.,
in body of function:
{
...etc
out <- lm( y ~ TEMP) # where TEMP = mod1+mod2 (from previous post)
return(out)
}
So
AC Del Re wrote:
Hi David,
I don't have the function in order yet but the names will be going into the
formula, e.g.,
in body of function:
{
...etc
out <- lm( y ~ TEMP) # where TEMP = mod1+mod2 (from previous post)
return(out)
}
So why not just pass the formula to the function?
Hi David,
I don't have the function in order yet but the names will be going into the
formula, e.g.,
in body of function:
{
...etc
out <- lm( y ~ TEMP) # where TEMP = mod1+mod2 (from previous post)
return(out)
}
Thanks,
AC
On Fri, Jul 9, 2010 at 3:20 PM, David Winsemius wrote:
On 09/07/2010 6:09 PM, Matthew Killeya wrote:
Yes clearly a bug... there are numerous variations ... problem seems to be
for a linear function whenever the first function valuation is 1.
Not at all. You can get the same problem on a quadratic that happens to
have a zero at an inconvenient
Is there a function similar to combine.levels ( in the Hmisc package)
that combines the levels of factors, but not based on their frequency.
Alternatively, I am looking into using the significance of the dummy
variables of factors based on their Pr(>|t|) value using the linear
model, then deleting
Actually, it looks like any value other than 1.0
(and in (lower, upper)) for start will work.
-Peter Ehlers
On 2010-07-09 14:45, Ravi Varadhan wrote:
Setting abs.tol = 0 works! This turns-off the absolute function convergence
criterion.
nlminb( objective=function(x) x, start=1, lower=-2,
Let me clarify my statement in the previous email: by absolute criterion, I
meant |f(x_n)| < tol.
It is ok to use: |f(x_{n+1}) - f(x_n)| < abs.tol
Although I prefer the hybrid condition in my previous email.
Ravi.
-Original Message-
From: Ravi Varadhan [mailto:rvarad...@jhmi.edu]
S
An absolute criterion should NEVER be used. For the situation where f(x*) =
0, instead of the PORT criterion (6) we should use the following:
|f(x_{n+1}) - f(x_n)| < (rel.tol * |f(x_n)| + abs.tol)
Ravi.
-Original Message-
From: Ravi Varadhan [mailto:rvarad...@jhmi.edu]
Sent: Friday, J
Setting abs.tol = 0 works! This turns-off the absolute function convergence
criterion.
> nlminb( objective=function(x) x, start=1, lower=-2, upper=2,
control=list(abs.tol=0))
$par
[1] -2
$objective
[1] -2
$convergence
[1] 0
$message
[1] "both X-convergence and relative convergence (5)"
$iter
Duncan, `nlminb' is not intended for non-negative functions only. There is
indeed something strange happening in the algorithm!
start <- 1.0 # converges to wrong minimum
startp <- 1.0 + .Machine$double.eps # correct
startm <- 1.0 - .Machine$double.eps # correct
> nlminb( objective=obj, start
Do you mean substring?
sub(".txt","", "mytest.txt")
-
A R learner.
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On 2010-07-09 13:29, Duncan Murdoch wrote:
On 09/07/2010 2:36 PM, Jim Bouldin wrote:
> 1. The expression you gave us is clearly not the one that produced
the > error: it involved "ring.area" and "ba.beg".
> > 2. You don't tell us what x and y are, so we can't reproduce
anything.
Sorry, I guess
On Jul 9, 2010, at 3:51 PM, AC Del Re wrote:
Hi All,
I am interested in printing column names without quotes and am
struggling to
do it properly. The tough part is that I am interested in using
these column
names for a function within a function (e.g., lm() within a wrapper
function). The
Erik,
Can you store the data as a blob? For example:
> #create string, compress with gzip, convert to SQLite blob string
> string <- "gzip this string, store as blob in SQLite database"
> string.gz <- memCompress(string, type="gzip")
> string.sqlite <- paste("x'",paste(string.gz,collapse=""),"'"
On 09/07/2010 4:11 PM, Gene Leynes wrote:
I thought the "apply" functions are faster than for loops, but my most
recent test shows that apply actually takes a significantly longer than a
for loop. Am I missing something?
Probably not. apply() needs to figure out the shape of the results it
On Jul 9, 2010, at 3:37 PM, soeren.vo...@eawag.ch wrote:
Hello
Could you recommend a printed/electronic book that teaches time
series analysis (using R) for students? I am searching for something
similar to the MASS book, with a level lower but close, for TSA.
Easy examples would be fine
I thought the "apply" functions are faster than for loops, but my most
recent test shows that apply actually takes a significantly longer than a
for loop. Am I missing something?
It doesn't matter much if I do column wise calculations rather than row wise
## Example of how apply is SLOWER than f
Hi All,
I am interested in printing column names without quotes and am struggling to
do it properly. The tough part is that I am interested in using these column
names for a function within a function (e.g., lm() within a wrapper
function). Therefore, cat() doesnt seem appropriate and print() is n
On 09/07/2010 10:37 AM, Matthew Killeya wrote:
nlminb( obj = function(x) x, start=1, lower=-Inf, upper=Inf )
If you read the PORT documentation carefully, you'll see that their
convergence criteria are aimed at minimizing positive functions. (They
never state this explicitly, as far as I
Hello
Could you recommend a printed/electronic book that teaches time series analysis
(using R) for students? I am searching for something similar to the MASS book,
with a level lower but close, for TSA. Easy examples would be fine to
understand deeper statistical procedures. Functions, results
On 09/07/2010 2:36 PM, Jim Bouldin wrote:
> 1. The expression you gave us is clearly not the one that produced the
> error: it involved "ring.area" and "ba.beg".
>
> 2. You don't tell us what x and y are, so we can't reproduce anything.
Sorry, I guess that was unclear. I changed the respon
Try changing the plot call to:
> plot(x2, y2, type ="n")
look at:
?plot
-Roy
On Jul 9, 2010, at 12:20 PM, Marlin Keith Cox wrote:
> I need to hide scatter points and just leave the linear regression
> line. I have tried to color the points "white", but that does not
> work.
>
> A working e
plot(x2, y2, pch=16, type="n")
On Fri, Jul 9, 2010 at 3:20 PM, Marlin Keith Cox wrote:
> I need to hide scatter points and just leave the linear regression
> line. I have tried to color the points "white", but that does not
> work.
>
> A working example, I need the green plotted points to be tra
I need to hide scatter points and just leave the linear regression
line. I have tried to color the points "white", but that does not
work.
A working example, I need the green plotted points to be transparent or hidden.
x2 <- seq(1,200,.5)
y2 <- seq(5,204,.5)
plot(x2, y2, pch=16, col="green")
mod
depmixS4 can fit such models as a constrained latent markov mdoel,
best, Ingmar
On Fri, Jul 9, 2010 at 7:52 PM, Katherine Sanders <
sanders@buckeyemail.osu.edu> wrote:
> Does anyone know how to program mover stayer models in R? They are a type
> of Markov chain model where there is a group o
On Fri, 9 Jul 2010, steve_fried...@nps.gov wrote:
Hello,
I've been using ctree and have developed a 55 node - 28 terminal solution.
As can be imagined, the plot is difficult to travel down each of the major
branches.
I've read the help files for ctree I saw where terminal nodes can be color
c
On Fri, 9 Jul 2010, kateg wrote:
Hi all,
I'm working on decomposition and comparison of several time series. I'm
interested in extracting the trend components for each time series using the
stl function and overlaying them on one another.
I'm not sure how to plot the trend function alone and
Jonathan Flowers wrote:
Hi,
I would like to extract columns from a dataframe using a vector of desired
column names.
The following working example uses the select argument in the subset
function to accomplish what I am trying to do. Is there a better solution?
This is very much "Introducti
> 1. The expression you gave us is clearly not the one that produced the
> error: it involved "ring.area" and "ba.beg".
>
> 2. You don't tell us what x and y are, so we can't reproduce anything.
Sorry, I guess that was unclear. I changed the response and independent
variable names to y and
for (i in 0:3)
{
r = runif(1)
assign(paste('b',i,sep=''),r)
}
for (i in 1:4)
{
assign(paste('b',i,sep=''),eval(parse(text=paste('b',i-1,sep=''
}
First, b1 is set to b0
Then, b2 is set to b1 (which was just set to b0)
Then, b3 is set
How about
data[ , colnames]
Or
data[ , colnames[1]]
data[ , colnames[2]]
-Don
On 7/9/10 11:27 AM, "Jonathan Flowers" wrote:
> Hi
>
> I want to extract columns from a data frame using a vector with the desired
> column names.
>
> This short example uses the select argument in the subse
Does anyone know how to program mover stayer models in R? They are a type of
Markov chain model where there is a group of people who stays in the same
place, a group who always moves, and a group who does both.
Thanks.
Leanne Sanders
sanders@osu.edu
[[alternative HTML version dele
Hi all,
I'm working on decomposition and comparison of several time series. I'm
interested in extracting the trend components for each time series using the
stl function and overlaying them on one another.
I'm not sure how to plot the trend function alone and to do the overlay
using some kind
So I have an array A of length n with multiple attributes and for a
selected attribute y I need to list all valid permutations where a
valid permutation is of the form:
A[1,y] != A[n,y]
A[i,y] != A[i+1,y]
I've tried using the 'combinat' package, but with the vector lengths
I'm using the permn func
Hello,
I would like to compress a long string (character vector), store the compressed
string in the text field of a SQLite database (using RSQLite), and then load
the text back into memory and decompress it back into the the original string.
My character vector can be compressed considerably
I am using the R plug-in for spss 18, and would like to know if there is any R
code that will highlight and bold rows (particularly the "Totals" row) within
an spss table. The current option that spss has is to use a python plug-in
which doesn't seem to work on my windows 7 machine.
Chris Ande
BaselR meeting July 2010
Thank you to everyone who supported our inaugural BaselR meeting on April
28th; we were fortunate to have three excellent presentations which prompted
lively discussion - our thanks go to Andreas Krause, Yann Abraham and Charles
Roosen for their presentations, details o
deaR useRs,
I am trying to assign different values to different objects in a for loop.
The following is a toy example of the part that has been giving me a hard
time.
The first "for loop" generates four objects, b0, b1, b2, b3 with random
numbers.
And, the second "for loop" is equivalent to
b1 =
"n.via...@libero.it" writes:
Hi,
> Dear List I would like to ask you something concenting a better print of the
> R output:
> I have a bit data frame which has the following structure:
> CFISCALE RAGSOCBANNO VAR1VAR2.
> 9853312 a
nlminb( obj = function(x) x, start=1, lower=-Inf, upper=Inf )
$par
[1] 0
$objective
[1] 0
$convergence
[1] 0
$message
[1] "absolute function convergence (6)"
$iterations
[1] 1
$evaluations
function gradient
22
[[alternative HTML version deleted]]
_
How about
data[,colnames(data)%in%colnames]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Jonathan Flowers
Sent: Friday, July 09, 2010 11:27 AM
To: r-help@r-project.org
Subject: [R] select columns from vector of column names
H
Steve, I'm not sure if your task could be accomplished with a ready-made
function in party. But, if you could manage to convert your tree structure to a
dendrogram, then it's straightforward using dendrapply. In fact, there is an
example in dendrapply help page showing how leaves are colored.
Hi,
I would like to extract columns from a dataframe using a vector of desired
column names.
The following working example uses the select argument in the subset
function to accomplish what I am trying to do. Is there a better solution?
Thanks.
#my data
data <- data.frame("col1"=c(1,2,3),"col2
Dear Greg,
I keep on being amazed at the abundance of functions you have packed into
the TeachingDemos package - thank you!
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com
Hi
I want to extract columns from a data frame using a vector with the desired
column names.
This short example uses the select argument in the subset function to
accomplish what I am trying to do. Is there a better solution?
#names of desired columns
colnames <- c("col1","col3")
#my data
data
On 09/07/2010 1:51 PM, Jim Bouldin wrote:
I am trying to perform an nls for a valid negative exponential function:
zz=nls(y~constant+a.est*2.7183^(b.est*x),start=list(constant=4.0,a.est=-4,b.est
= -.005),trace=T)
and am getting a number of different error messages, the most problematic
of whic
Hi Jeremie,
Maybe you can take a look at the bigmemory package.
If you have multi core or have access to clusters, you may want to use any
parallel computing strategy.
For plotting of large data, if you are using basic R graphics, first try to
use "line" instead of using 'point',
if this still
I am trying to perform an nls for a valid negative exponential function:
zz=nls(y~constant+a.est*2.7183^(b.est*x),start=list(constant=4.0,a.est=-4,b.est
= -.005),trace=T)
and am getting a number of different error messages, the most problematic
of which is "Error in nls(ring.area ~ constant + a
Don't do this. The overlapping will confuse.
Plot them in a lattice display with one group above the other on the same
horizontal scale. See ?histogram.
Bert Gunter
Genentech Nonclinical Biostatistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-proj
Others pointed you to the text function for the base graphics system. But if
what you want to do is use text, but have a simple way of specifying the center
of the plot without computing the user coordinates by hand of the center, then
look at the grconvertX and grconvertY functions.
--
Grego
Hello,
I've been using ctree and have developed a 55 node - 28 terminal solution.
As can be imagined, the plot is difficult to travel down each of the major
branches.
I've read the help files for ctree I saw where terminal nodes can be color
coded.
plot(airct, type = "simple")
> plot(airct, te
There is a basic interface between R and gnuplot in the TeachingDemos package,
see ?gp.open
Not much interest has been shown in this, so it is still pretty alpha level,
but you can send your R data to gnuplot and have it create a basic plot.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Cen
Hi Ted,
Well since you mentioned data.table (!) ...
If risk_input is a data.table consisting of 3 columns (m_id, sale_date,
return_date) where the dates
are of class IDate (recently added to data.table by Tom) then try :
risk_input[, fitdistr(return_date-sale_date,"normal"), by=list(m_id,
y
On Fri, Jul 9, 2010 at 2:29 PM, Mao Jianfeng wrote:
> Dear R-help listers,
>
> I am new. I just want to get helps on how to plot two histograms
> overlapped in the same plane coordinate. What I did is very ugly.
> Could you please help me to improve it? I want to got a plot with semi-
> transparen
Empirical CDFs are much better for this purpose, and allow
superpositioning (see e.g. the Ecdf function in the Hmisc package).
Otherwise look at histbackback in Hmisc.
Frank
On 07/09/2010 11:40 AM, Andrew Miles wrote:
I'm not sure what you are trying to do. Do you want one histogram for
mal
I'm not sure what you are trying to do. Do you want one histogram for
males and one for females on the same graph? If so, the simplest way
to put two histograms together is to simply use the add parameter:
age.males=age[which(sex=="M")]
age.females=age[which(sex=="F")]
hist(age.males, co
No - devAskNewPage is not what the OP asked for.
1. The following are, but probably only work when R is in interactive mode
(?interactive), e.g. in the GUI:
2. ?winDialog ?file.choose ?choose.files ?select.list ?readline
and friends
and for keyboard: ?getGraphicsEvent
Bert Gunter
Genente
Dear all,
after having calibrated a svm model through the svm() command of the
e1071 package, is there a way to
i) represent the modeled relationships between the y and X variables
(response variable vs. predictors)?
ii) rank the influence of the predictors used in the model?
Right now I am m
Hi Nathaniel ,
Could you give us a simple example of your data using the
?dput
Function?
Basically you might want to draw the axis yourself, and connect the lines is
possible through using points(..., type = "l")
But I'd rather try and answer this with simple example data to be sure I
understand
Possible fortune.
:)
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
---
Sounds like you want devAskNewPage(TRUE) or the related
options("device.ask.default"). See help("devAskNewPage",
package="grDevices").
Hope this helps.
Allan
On 09/07/10 13:54, vijaysheegi wrote:
Hi R Experts,
I have certain code ,i want to achive interactive execution .
For ex:
1. as part
Really? I don't usually think of Vectorize as a performance
enhancement, probably because my use of with a complex function then
gets applied to 4.5 million records. I need to go out, get a cup of
coffee, and leave it alone for about half an hour. I tried recently
to figure out how I can d
Original poster wanted a simple way to do it, but when R has three
graphics systems, four OO systems, and a zillion helpful people
there's never a simple way :)
-- Rather, I'd say it has a zillion simple ways. :)
Bert
Barry
--
blog: http://geospaced.blogspot.com/
web: http://www.maths.la
Dear list,
someone knows why the print.xtable doesnt print row.names? I dident do
anything with the options.may depends on the size of my table???
This is my code:
\documentclass[a4paper]{article}
\title{SCHEMA DI BILANCIO PER SINGOLE AZIENDE}
\begin{document}
\maketitle
\hline
<>=
library(xt
just to satisfy my curiousity,
aggregate(bla, list(bla$cat), max)
works for me and resulted in
Group.1 x catv1v2v3v4
1cat1 5 cat1 0.6337076 0.2887081 0.3629962 0.5328683
2cat2 10 cat2 0.5519426 0.6076447 0.4593770 0.9632341
3cat3 11 cat3 0.6094089
Could you elaborate?
Both
x <- 1:4
set <- matrix(nrow = 50, ncol = 11)
for(i in c(1:11)){
set[,i] <-sample(x,50)
print(c(i,"->", set), quote = FALSE)
}
and
x <- 1:4
set <- matrix(nrow = 50, ncol = 11)
for(i in c(1:50)){
set[i,] <-sample(x,11)
print(c(i,"->", set)
Hi,
Thanks a lot.
The Vectorize method worked and its much faster than looping through the
data frame.
Regards,
Harsh Yadav
On Thu, Jul 8, 2010 at 11:06 PM, David Winsemius wrote:
>
> On Jul 8, 2010, at 10:33 PM, Erik Iverson wrote:
>
>
>> I have a data frame:
>>> id url
>>> urlType
>>
Wow, Gabor - that's amazing - thank you so much!
Dimitri
On Fri, Jul 9, 2010 at 10:22 AM, Gabor Grothendieck
wrote:
> On Fri, Jul 9, 2010 at 9:35 AM, Dimitri Liakhovitski
> wrote:
>> Hello!
>>
>> Any hint would be greatly appreciated.
>> I have a data frame that contains (a) monthly dates and (b
I am attempting to plot a trellis object on a grid.
vplayout = viewport(layout.pos.row=x, layout.pos.col=y)
grid.newpage()
pushViewport(viewport(layout=grid.layout(2,2)))
g1 = ggplot() ...
g2 = ggplot() ...
g3 = ggplot() ...
p = xyplot() ...
# works as expected
print(g1, vp=vplayout(1,1))
print
On Jul 9, 2010, at 10:26 AM, Eik Vettorazzi wrote:
you are right. But maybe "aggregate" is close to the desired result?
aggregate(bla, list(bla$cat), max)
Right. I couldn't get it to work until I removed the first two columns:
aggregate(bla[,-(1:2)], list(bla$cat), max)
Then I got pretty m
On 09/07/2010 10:27 AM, Trafim Vanishek wrote:
Thanks everybody for referring me to FAQ 7.31 but I don't see how to solve
it.
I am giving a concrete number and I need to get 58 not 57. Seems, there is
no way?
You aren't giving a "concrete number". You're giving a string of three
characters,
On 2010-07-09 8:27, Trafim Vanishek wrote:
Thanks everybody for referring me to FAQ 7.31 but I don't see how to solve
it.
I am giving a concrete number and I need to get 58 not 57. Seems, there is
no way?
Sure there is: round(100*0.58).
-Peter Ehlers
On Fri, Jul 9, 2010 at 4:05 PM, David
On 09/07/10 12:18, Ralf B wrote:
I am using RGUI, the command line or the StatET Eclipse environment.
Should this not all be the same?
No, there is no particular reason why they should.
Allan
Ralf
On Fri, Jul 9, 2010 at 7:11 AM, Allan Engelhardt wrote:
I'm assuming you are using
On Jul 9, 2010, at 10:27 AM, Trafim Vanishek wrote:
Thanks everybody for referring me to FAQ 7.31 but I don't see how to
solve
it.
I am giving a concrete number and I need to get 58 not 57. Seems,
there is
no way?
Building on an example on the help page for as.integer, this seems to
be
On 09/07/2010 7:37 AM, p...@orbit.umbr.cas.cz wrote:
Hi,
Is it possible to install packages without the testing if installed
package can be loaded?
I need to install bunch of packages on multiple computers over ssh. Some
packages witch interact with X11 display cannot be installed in this way.
fo
I am attempting to plot a trellis object on a grid.
vplayout = viewport(layout.pos.row=x, layout.pos.col=y)
grid.newpage()
pushViewport(viewport(layout=grid.layout(2,2)))
g1 = ggplot() ...
g2 = ggplot() ...
g3 = ggplot() ...
p = xyplot() ...
# works as expected
print(g1, vp=vplayout(1,1))
print
Thanks everybody for referring me to FAQ 7.31 but I don't see how to solve
it.
I am giving a concrete number and I need to get 58 not 57. Seems, there is
no way?
On Fri, Jul 9, 2010 at 4:05 PM, David Winsemius wrote:
>
> On Jul 9, 2010, at 9:46 AM, Trafim Vanishek wrote:
>
> Dear all,
>>
>> migh
you are right. But maybe "aggregate" is close to the desired result?
aggregate(bla, list(bla$cat), max)
Am 09.07.2010 16:01, schrieb David Winsemius:
>
> On Jul 9, 2010, at 9:46 AM, Eik Vettorazzi wrote:
>
>> Hi Nils,
>> have a look at
>> ?tapply
>> hth.
>
> Perhaps this will be part way there (I
Hi R Experts,
I have certain code ,i want to achive interactive execution .
For ex:
1. as part of input ,it should ask file name or table name as input.
2.in script so many graphs i need to draw,it should wait till certain key is
pressed .
3:i am using windows R,rscript is not working.
Please
cr<-cReturns(spData,ttr="MACD")
Error in ind[t - k] <- pos[t - k + 1] - pos[t - k] :
replacement has length zero
> cr<-cReturns(spData,ttr="macd4")
Why is the above error coming? macd4 works alright ( which is a custom
function built by the ttrTests author) while MACD doesnt work.
Thanks
--
'
On Fri, Jul 9, 2010 at 9:35 AM, Dimitri Liakhovitski
wrote:
> Hello!
>
> Any hint would be greatly appreciated.
> I have a data frame that contains (a) monthly dates and (b) a value
> that corresponds to each month - see the data frame "monthly" below:
>
> monthly<-data.frame(month=c(20100301,2010
Hi,
Is it possible to install packages without the testing if installed
package can be loaded?
I need to install bunch of packages on multiple computers over ssh. Some
packages witch interact with X11 display cannot be installed in this way.
for example after:
> install.packages('cairoDevice',dep=T
Hi Tal, Thanks for your help.
I've had a look at the site, and what i wanted to do was to plot X and Y
where X is a characters and Y is numeric. The problem I'm having now is that
the X axis isn't characters but just numbers from 1 onwards and when i plot
it, the data i have is in descending ord
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One solution is to put these unwanted entries to ""
repor$9853312 [1:2,2:3] <- ""
Cheers
Joris
On Fri, Jul 9, 2010 at 12:18 PM, n.via...@libero.it wrote:
>
> Dear List I would like to ask you something concenting a better print of the
> R output:
> I have a bit data frame which has the follo
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