On Fri, Oct 29, 2010 at 11:02 PM, Shi, Tao wrote:
> Hi Ben,
>
> That must be the case! In fact if I do:
>
>> difftime(strptime("24NOV2004", format="%d%b%Y"),
>> strptime("13MAY2004",format="%d%b%Y"), units="days", tz="GMT")
> Time difference of 195 days
>
>
> which supports your claim.
>
> Can s
Hi Ben,
That must be the case! In fact if I do:
> difftime(strptime("24NOV2004", format="%d%b%Y"),
> strptime("13MAY2004",format="%d%b%Y"), units="days", tz="GMT")
Time difference of 195 days
which supports your claim.
Can someone from the R development team confirm this?
Thanks!
...Tao
Hi Phil,
Thanks for the reply, but I don't think you have
explained where the decimal part is coming from
...Tao
- Original Message -
> From:Phil Spector
> To:"Shi, Tao"
> Cc:r-h...@stat.math.ethz.ch
> Sent:Friday, October 29, 2010 5:04:23 PM
> Subject:Re: [R] date calculation
>
I am trying to reproduce this setting, but it does not seem to work anymore.
I keep getting the error: execve failed: No such file or directory.
One of the reasons is that /usr/local/bin/Rserve is no longer a standalone
executable, and is now initiated using R CMD Rserve. However, after making
th
Hi,
I'm trying to install the gsl wrapper source code
(http://cran.r-project.org/src/contrib/gsl_1.9-8.tar.gz) on a Linux
system (OpenSuse 11.1), but encountering the following problem. I've
already installed 'gsl' version 1.14
(ftp://ftp.gnu.org/gnu/gsl/gsl-1.14.tar.gz) on the system. What's
miss
Hi:
x <- matrix(20:35, ncol = 1)
u <- c(1, 4, 5, 6, 11) # 'x values'
m <- c(1, 3, 1, 1, 0.5)
# Function to compute the inner product of the multipliers with the
extracted
# elements of x determined by u
f <- function(mat, inputs, mults) crossprod(mat[inputs], mults)
f(x, u, mults = c(1, 3, 1
Shi, Tao yahoo.com> writes:
> Could someone explain to me why the following result is not a integer?
>
> > difftime(strptime("24NOV2004", format="%d%b%Y"), strptime("13MAY2004",
> >format="%d%b%Y"), units="days")
> Time difference of 195.0417 days
Presumably because this goes across a daylig
Hello Jorge,
Thank you for the reply. I tried a few different things with if/else but
couldn't get them to go. I really appreciate your feedback. I learned
something new from this
Will
--
View this message in context:
http://r.789695.n4.nabble.com/grouping-question-tp3019922p3019952.html
Se
Hello Jim
Wow. I tried cut but i see you have an interim step with labels a,b,c but
levels night and day. i was really close to this. i have labels
night,day,night and it wouldn't let me duplicate labels. I am very greatful
for your input
Will
--
View this message in context:
http://r.7896
Hi Will,
One way would be:
> x
[1] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
24
> factor(ifelse(x>6 & x<18, 'day', 'night'))
[1] night night night night night night night day day day day day
day day day
[16] day day day night night night night nig
try this:
> x
[1] 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
> y <- cut(x, breaks=c(-Inf,6,18, Inf), labels=c('a','b','c'))
> levels(y) <- c('night','day','night')
> y
[1] night night night night night night night day day day day
day day day day day
Hello
I have what is probably a very simple grouping question however, given my
limited exposure to R, I have not found a solution yet despite my research
efforts and wild attempts at what I thought "might" produce some sort of
result.
I have a very simple list of integers that range between 1 a
Hi Mary,
I am not sure off hand why savePlot() is not doing what you expect,
but hopefully my inline comments answer your other questions and at
least get you the graphs you want.
Cheers,
Josh
On Fri, Oct 29, 2010 at 3:18 PM, wrote:
>
>
> I want to plot the unstadardized version of a normal p
Hi Duncan,
If vectors of unequal length is the problem, one way to go, using your
example, would be:
# your example
x <- list(c(9, 5, 7, 2, 14, 4, 4, 3), c(3, 6, 25, 2, 14, 3, 3, 4),
c(28, 4, 14, 3, 14, 2, 4, 5), 28)
x
# maximum number of components
k <- max(sapply(x, length))
k
# expanding
Hi.
On Fri, Oct 29, 2010 at 3:31 AM, Claudia Beleites wrote:
> Dear Henrik,
>
> sorry for bothering you with a report hastily pasted together and not
> particularly nice for you as I used my toy data flu from a non-standard
> package. I should have better used e.g. the iris.
>
> I'm aware that wr
Hi,
Ah! Thanks for your help! I have even seen Reduce before but for some reason
just didn't make the connection. This helps a ton! Thanks again.
Kind regards,
Greg
On Oct 29, 2010, at 6:53 PM, Jorge Ivan Velez wrote:
> Hi Greg,
>
> Here are two ways of doing it:
>
> > mylist <- list(x = rpoi
Tao -
The documentation for the difftime function says:
Function ‘difftime’ calculates a difference of two date/time
objects and returns an object of class ‘"difftime"’ with an
attribute indicating the units.
So that answers your question.
If you want it to be an integer, you're ce
Hi list,
Could someone explain to me why the following result is not a integer?
> difftime(strptime("24NOV2004", format="%d%b%Y"), strptime("13MAY2004",
>format="%d%b%Y"), units="days")
Time difference of 195.0417 days
I'm using R2.12.0 on WinXP.
Thanks!
...Tao
_
On Fri, Oct 29, 2010 at 6:34 PM, M.Ribeiro wrote:
>
> Hi R-helpers,
>
> I need to read some file with different lines (I don't know the number of
> lines to skip) and I would like to find a way to start reading the
> data.frame from the word "source".
>
> ex:
>
> djhsafk
> asdfhkjash
> shdfjkash
>
Aplogies to the original poster - I got names wrong in a cut and paste
Hi Jorge
I tried your methods for all (which work for complete rows) and then I
remove the first value of $y and repeated; both fail because of
the unequal numbers
The problem is when there are unequal numbers in the rows
Hi Jorge
I tried your methods for all (which work for complete rows) and then I
remove the first value of $y and repeated; both fail because of
the unequal numbers
The problem is when there are unequal numbers in the rows and trying to
make a matrix of them.
I was trying some things with Gr
Sorry, this isn't really an R solution, but here it goes anyway. You
can isolate the block from Source to the first following blank line by
the following unix/linux/cygwin command, assuming inFile is your input
file and outFile is the output file:
cat inFile | grep -A 100 Source | grep -m 1 -B 100
It might be easier to preprocess the file before passing it
to R, but you an do what you want using a connection:
f = file('s.dat','r')
while(1){
+ txt = readLines(f,1)
+ if(length(grep('Source',txt)))break;
+ }
while(1){
+ more = readLines(f,1)
+ if(nchar(more) == 0)break;
+ txt =
Hi Greg,
Here are two ways of doing it:
> mylist <- list(x = rpois(10, 10), y = rpois(10, 20), z = rpois(10, 5))
> mylist
$x
[1] 3 13 14 16 10 7 3 5 12 14
$y
[1] 17 16 26 13 23 24 16 28 23 12
$z
[1] 2 6 5 5 5 1 9 11 6 4
>
> colMeans(do.call(rbind, mylist), na.rm = TRUE)
[1] 7
So, I am having a tricky reference file to extract information from.
The format of the file is
x 1 + 4 * 3 + 5 + 6 + 11 * 0.5
So, the elements that are not being multiplied (1, 5 and 6) and the elements
before the multiplication sign (4 and 11) means actually the reference for
the row in a ma
Hi Everyone,
I have a list of vectors like this (in this case it's 3 vectors but assume the
vector count and the length of each vector is not known):
[[1]]
[1] 9 5 7 2 14 4 4 3
[[2]]
[1] 3 6 25 2 14 3 3 4
[[3]]
[1] 28 4 14 3 14 2 4 5
What I want to do is take the average ver
Sorry, the explanation wasn't very good...just to explain better.
I am writing a loop to read and process every time a different file in the
same script.
And what I want to load into a variable each time is a data.frame that is
bellow the word source in all of my files.
So I would like to recogn
Sorry, the explanation wasn't very good...just to explain better.
I am writing a loop to read and process different files in the same script.
And what I want to load into a variable is a data.frame that is above the
word source in all of my files.
So I would like to recognize the word Source in
Hi R-helpers,
I need to read some file with different lines (I don't know the number of
lines to skip) and I would like to find a way to start reading the
data.frame from the word "source".
ex:
djhsafk
asdfhkjash
shdfjkash
asfhjkash #those lines contain numbers and words, I want to skip
I want to plot the unstadardized version of a normal plot. Can you explain
why that is not working?
Dev.set(1)
xcrit=-1.645
cord.x <- c(-3,seq(-3,xcrit,0.01),xcrit)
cord.y <- c(0,dnorm(seq(-3,xcrit,0.01)),0)Â Â Â Â Â Â Â Â Â Â Â # what does
final 0 do here?
curve(dnorm(x,
On 29/10/2010 5:19 PM, Andre Zege wrote:
Hi, everyone. I am using a fair amount of closures in my code. Problem i am
experiencing is i cannot figure out how to mtrace functions defined within a
function. There must be some way to name such function for mtrace to see it
and let me step into it. F
Actually, Berwin - you are right. I also did get 2 shortcuts. I just
thought that one of them was old - from my previous R installations.
And indeed the 4th line tells what version is running.
Great!
Thank you!
Dimitri
On Fri, Oct 29, 2010 at 5:02 PM, Berwin A Turlach
wrote:
> G'day Dimitri,
>
Hi, everyone. I am using a fair amount of closures in my code. Problem i am
experiencing is i cannot figure out how to mtrace functions defined within a
function. There must be some way to name such function for mtrace to see it
and let me step into it. For example, say i have code
mymodel<-func
Thank you for your suggestion, Juan. However, I have tried to specify
the innovations and this doesn't seem to solve the problem.
Are there really no opportunities to simulate from a (S)ARIMA model
using past observations in R, besides to make the whole script from
the ground?
Knut
Sit
On 29/10/2010 12:46 PM, DM2010 wrote:
Dear R Users
I have two questions about how R makes use of memory on a Windows computer.
On my machine certain R jobs seem to stop with messages such as...
"Error: cannot allocate vector of size 215.0 Mb"
...when, according to Windows Task Manager, there a
On 29/10/2010 4:45 PM, Dimitri Liakhovitski wrote:
Question: I installed R verison 2-12.0 on my Windows 7 (64 bit) PC.
When I was installing it, it did not ask me anything about 32 vs. 64 bit.
So, if I run R now - is it running as a 32-bit or a 64-bit?
If it didn't ask you about it, then it pro
***Summary:***
I'm setting up a cluster using netWorkSpace, and I'm having issues
with the sleigh initialization. My R function to initialize the sleigh
succeeds and the sleigh appears to be ready, but I get apparently
conflicting information from "status(s)", "rankCount(s)", and "s"; and
basic sl
Hi,
On Fri, Oct 29, 2010 at 4:24 PM, ANJAN PURKAYASTHA
wrote:
> Does any R package support one-class SVM?
> I'm trying to develop an application to detect anomalies in genome
> sequencing.
I'm pretty sure kernlab supports this:
http://cran.r-project.org/web/packages/kernlab/index.html
And:
http
G'day Dimitri,
On Fri, 29 Oct 2010 16:45:00 -0400
Dimitri Liakhovitski wrote:
> Question: I installed R verison 2-12.0 on my Windows 7 (64 bit) PC.
> When I was installing it, it did not ask me anything about 32 vs. 64
> bit. So, if I run R now - is it running as a 32-bit or a 64-bit?
Well, whe
Question: I installed R verison 2-12.0 on my Windows 7 (64 bit) PC.
When I was installing it, it did not ask me anything about 32 vs. 64 bit.
So, if I run R now - is it running as a 32-bit or a 64-bit?
thank you!
--
Dimitri Liakhovitski
Ninah Consulting
www.ninah.com
___
Hi Dimitri,
The help pages should be loaded from a local server (which is
corroborated by the 127...ip), so the internet connection should not
matter.
I would check whether IE is the default application for .html
extensions, and you could also try running R as an admin (though I do
not really thi
Problem solved. It turned out I had to reset my IE as my default
browser for all extentions (like .html). Now, R help is working!
Dimitri
On Fri, Oct 29, 2010 at 4:20 PM, Dimitri Liakhovitski
wrote:
> I tried one more thing - I uninstalled R again, and this time
> installed it under:
> C:\Users\M
Does any R package support one-class SVM?
I'm trying to develop an application to detect anomalies in genome
sequencing.
Thanks in advance.
Anjan
--
===
anjan purkayastha, phd.
research associate
fas center for systems biology,
harvard university
52 oxford street
c
I tried one more thing - I uninstalled R again, and this time
installed it under:
C:\Users\Myname\R\R-2.12.0
And again - am gettting the same error:
Error in shell.exec(url) :
access to 'http://127.0.0.1:30111/library/stats/html/lm.html' denied
Any help is greatly appreciated!
Dimitri
On Fri, O
Dear R Users
I have two questions about how R makes use of memory on a Windows computer.
On my machine certain R jobs seem to stop with messages such as...
"Error: cannot allocate vector of size 215.0 Mb"
...when, according to Windows Task Manager, there are still hundreds of
megabytes of physic
Hello, dear R-community.
This is a question about TukeyHSD between factor combinations of a Three-Way
ANOVA, which is - since it is a multi measure ANOVA - not a simple ANOVA but
a Generalised Linear Mixed Model (GLMM), calculated with "lmer".
> growth <-
groupedData(length~meas|box_id,outer=~spe
the following code was used
library(akima)
library(clim.pact)
nc.1 <- "RF_80-05.nc"
nc.rf.in <- open.ncdf(nc.1)
x1 <- retrieve.nc(nc.1, v.nam="Rainfall",l.scale=FALSE, x.rng=c(70, 80),
y.rng=c(10, 13.5))
#dimension is checked for the subset. (lon, lat, time) is changed as (time,
lat,
On Fri, Oct 29, 2010 at 2:30 PM, Jason Kwok wrote:
> Thanks. 1 more question.
>
> When I use
>
> Plot(series1)
> lines(series2)
>
> The graph will use the y axis scaling for series 1 so some of series 2 is
> cut off. How do I control the y axis scaling?
>
Plot them together as a multivariate se
Thanks. 1 more question.
When I use
Plot(series1)
lines(series2)
The graph will use the y axis scaling for series 1 so some of series 2 is
cut off. How do I control the y axis scaling?
Thanks,
Jason
On Fri, Oct 29, 2010 at 2:10 PM, Gabor Grothendieck wrote:
> On Fri, Oct 29, 2010 at 1:41
On Fri, Oct 29, 2010 at 1:41 PM, Jason Kwok wrote:
> How do I plot two time series plots on the same chart?
>
Try this:
example(plot.ts)
example(ts.plot)
library(zoo)
example(plot.zoo)
library(lattice)
example(xyplot.zoo)
--
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel:
Thanks a lot for your answers, Dennis and Bernhard!
--
View this message in context:
http://r.789695.n4.nabble.com/Dickey-Fuller-Test-tp3018408p3019544.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing lis
How do I plot two time series plots on the same chart?
Thanks,
Jason
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.o
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of govin...@msu.edu
> Sent: Friday, October 29, 2010 8:33 AM
> To: r-help@r-project.org
> Subject: [R] doubt in climate variability analysis in R!
>
>
>
> Hello all,
>
> I am try
Dear Chris,
See the examples in ?Anova in the car package (but it works with a model fit
by lm).
I hope this helps,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
web
I had a look:
I fixed it 5 months ago and forgot to make a new release.
The version on R-forge contains the fix already.
New release on its way to CRAN.
Best wishes;
Uwe
On 28.10.2010 11:28, John Coulthard wrote:
Hi
I'm trying to scale the point symbols on a 3d plot so that the ones at the
Hello all,
Is there an r function that exists that will perform repeated measures MANOVAs?
For example, let's say I have 3 DVs, one between-subjects IV, and one
within-subjects IV. Based on the documentation for the manova command, a
function like that below is not appropriate because it cannot
On Oct 29, 2010, at 12:08 PM, David Winsemius wrote:
On Oct 29, 2010, at 11:37 AM, dpender wrote:
Apologies for being vague,
The structure of the output is as follows:
Still no code?
$ cluster1 : Named num [1:131] 3.05 2.71 3.26 2.91 2.88 3.11 3.21
-1 2.97
3.39 ...
..- attr(*, "nam
Hello, dear R-community.
This is a question about TukeyHSD between factor combinations of a Three-Way
ANOVA, which is - since it is a multi measure ANOVA - not a simple ANOVA but
a Generalised Linear Mixed Model (GLMM), calculated with "lmer".
> growth <-
groupedData(length~meas|box_id,outer=~spe
On Oct 29, 2010, at 11:37 AM, dpender wrote:
Apologies for being vague,
The structure of the output is as follows:
Still no code?
$ cluster1 : Named num [1:131] 3.05 2.71 3.26 2.91 2.88 3.11 3.21
-1 2.97
3.39 ...
..- attr(*, "names")= chr [1:131] "6667" "6668" "6669" "6670" ...
Wi
Dear John,
Thank you very much for your help! I appreciate it.
eshi
--
View this message in context:
http://r.789695.n4.nabble.com/draw-path-diagram-using-dot-tp3017987p3019359.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help
You could do a weighted cox ph model, with possible haplotype
configurations for each subject weighted by their posterior
probabilities given genotype data.
Are your markers SNPs? If so you can use a utility function from the
hapassoc package to get started. For example, if your data is in a
data
Hi dear Mi³ego dnia,
I am sorry I got lost here.
May be it it sound if I write you about what I am going to do.
"To test such null hypotheses we used the Monte Carlo test, i.e. we chose
randomly (5000 times) a value for the
current dependent characteristic (richness or actual species pool) in the
Apologies for being vague,
The structure of the output is as follows:
$ cluster1 : Named num [1:131] 3.05 2.71 3.26 2.91 2.88 3.11 3.21 -1 2.97
3.39 ...
..- attr(*, "names")= chr [1:131] "6667" "6668" "6669" "6670" ...
With 613 clusters. What I require is abstracting the first and last va
Thanks for your help.
May be I can not make clear by my writing. The statement which I am going to do
is like this
"To test such null hypotheses we used the Monte Carlo test, i.e. we chose
randomly (5000 times) a value for the
current dependent characteristic (richness or actual species pool) in
Hello all,
I am trying to use "clim.pact" package for my work, but since this is the
beginning for me to use gridded datasets in "R", I am having some trouble.
I want to do seasonal analyses like trends, anomalies, variograms, EOF and
probably kriging too to downscale my 1 degree gr
Job: Scientific programmer at Merck, Biostatistics, Rahway, NJ, USA
[Job Description]
This position works closely with statisticians to process and analyze
ultrasound, MRI, and radiotelemetry longitudinal studies using a series
of programs developed in R and Mathworks/Matlab. This position provid
On Fri, Oct 29, 2010 at 5:16 AM, Sarah Moens wrote:
> Hi all,
>
> I've been trying to find a solution for the problem of reading
> multiple files and storing them in a variable that contains the names
> by which I want to call the datasets later on.
>
> For example (5 filenames):
>
> - The filenam
I think that'll work... thanks!
matt
On Fri, Oct 29, 2010 at 9:45 AM, jim holtman wrote:
> Is this what you want:
>
>> df
> f1 f2
> 1 Maj I Minor A
> 2 Maj I Minor A
> 3 Maj I Minor A
> 4 Maj II Minor A
> 5 Maj II Minor B
> 6 Maj II Minor B
> 7 Maj III Minor B
> 8 Maj III Min
On Oct 29, 2010, at 6:55 AM, skan wrote:
Hello
Could anyone explain me the difference between strftime vs strptime,
please
?
I've read the help but it's a little bit cionfusing for me.
You should focus on the "Value" section of the help page. They return
vectors of different classes.
Is this what you want:
> df
f1 f2
1 Maj I Minor A
2 Maj I Minor A
3 Maj I Minor A
4 Maj II Minor A
5 Maj II Minor B
6 Maj II Minor B
7 Maj III Minor B
8 Maj III Minor C
9 Maj III Minor C
> df[!duplicated(df),]
f1 f2
1 Maj I Minor A
4 Maj II Minor A
5 Maj II Min
strptime() takes a character vector and makes a date-time object.
That is input.
strftime() takes a date-time object and makes an character vector.
That is output, and it is normally called via format() or print().
Let's see what the help says (slightly edited to refer just to these
two):
Da
On Fri, 29 Oct 2010, Ron Michael wrote:
Hi all, I am wondering is there any way to check whether some
Directory exists or not, given the parent path of that directory?
After searching for a while I found that there is a function
dir.create() to create some directory. However I need to know
wh
On Oct 29, 2010, at 5:14 AM, dpender wrote:
That's helpful but the reason I'm using clusters in evd is that I
need to
specify a time condition to ensure independence.
I believe this is the first we heard about any particular function or
package.
I therefore have an output
We woul
Read them into a list; much easier to handle:
myList <- lapply(filenames, read.csv)
On Fri, Oct 29, 2010 at 5:16 AM, Sarah Moens wrote:
> Hi all,
>
> I've been trying to find a solution for the problem of reading
> multiple files and storing them in a variable that contains the names
> by which
Hi to all!
When I use the example from kernlab::ksvm this works fine.. Give me the
result…
> filter <-
> ksvm(type~.,data=spamtrain,kernel="rbfdot",kpar=list(sigma=0.05),C=5,cross=3)
But as soon as I change the type data as follows
> type_train<-spamtrain[,ncol(spamtrain)]
> filter <-
> ks
Hi all,
I've been trying to find a solution for the problem of reading
multiple files and storing them in a variable that contains the names
by which I want to call the datasets later on.
For example (5 filenames):
- The filenames are stored in one variable:
filenames = paste(paste('name', '_',
Perhaps, the following construct does what you need:
paste(c("a", "b", "c"), c(",", ":", ""), sep="",collapse="")
Regards,
Jan
On 29-10-2010 10:49, Ron Michael wrote:
Hi all, I want to club different objects (character type) to a single one and using
paste() function that can be done happil
Matt,
Below are three (of the probably many more) possible ways of doing this:
aggregate(1:nrow(df), df, length)
ftable(1 ~ f1 + f2, data=df)
library(plyr)
ddply(df, .(f1,f2), nrow)
Regards,
Jan
On 29-10-2010 15:53, Matthew Pettis wrote:
Hi,
I have a data frame with two factors (well, mo
Hi Sarah,
There is one thing you need to think about: how do you choose which
values should not be removed if you have more than 20 and which should
be if you have less than 20. In my code, I've just done it with
sample(), which might not be what you need.
Here is what I have:
if (length(whi
Hi,
I have a data frame with two factors (well, more, but 2 for simple
consideration), and I want to display the different combinations of
the them that actually occur in the data. In reality, there are too
many of them to do to do a 'table' call and have one col vertical and
one col horizontal (
I have just installed R 12.
I have Windows 7, 64-bit verison.
I currently have IE as my default browser. The internet connection is very good.
Whenever I try to run a help command (?lm, for example), I get this error:
Error in shell.exec(url) : access to
'http://127.0.0.1:20271/library/stats/html/
Hello,
I would like to run a script in which a loop is included. Since I'm new to
R, I cannot manage the following problem. I really hope someone could help
me out.
Data in the variable Y should be removed from the simulated data set with
probability 0.50 if the variable X has a value below zero
Hello,
I have a question about relsurv package particularly rsadd function:
Rsadd(Surv(time,cens)~sex+ratetable(age=age*365.24,sex=sex,year=year),data=,=ratetable=,int=5,method=”max.lik”).
In the tutorial, it is indicated that "the age and year must be given in the
date format, i.e. in number of d
Dear R users
I would like to group my barplot graph (see example on the R help link). The
proposed R code, adding individual bars to the plot, looks really overwhelming.
My specific dataset just consists of five groups and three different levels
within each groups (the individual bars). The .t
?memory.size
use before/after a sequence of commands to get an idea of the memory usage.
On Fri, Oct 29, 2010 at 5:21 AM, Joel wrote:
>
> Hi
>
> Is there any way to check an certain command or procedure's RAM usage?
>
> Im after something similar to system.time(bla) that gives me the time the
>
On Fri, Oct 29, 2010 at 4:49 AM, Ron Michael wrote:
> Hi all, I want to club different objects (character type) to a single one and
> using paste() function that can be done happily. However the problem with
> paste function is the separator field is unique for all underlying objects.
> If I pu
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
On 29/10/10 13:08, Ron Michael wrote:
> Hi all, I am wondering is there any way to check whether some Directory
> exists or not, given the parent path of that directory? After searching for a
> while I found that there is a function dir.create() to c
Try this:
if(!file.info('myfolder')$isdir) dir.create('myfolder')
On Fri, Oct 29, 2010 at 9:08 AM, Ron Michael wrote:
> Hi all, I am wondering is there any way to check whether some Directory
> exists or not, given the parent path of that directory? After searching for
> a while I found that th
Hi, something like this perhaps,
if("myfolder"%in%dir()==FALSE) dir.create("myfolder")
Command dir.create() is not overwriting an existing folder.
Best regards,
Andris
On Fri, Oct 29, 2010 at 12:08 PM, Ron Michael wrote:
> Hi all, I am wondering is there any way to check whether some Director
Perhaps the following construct does what you need:
paste(c("a", "b", "c"), c(",", ":", ""), sep="",collapse="")
Regards,
Jan
On 29-10-2010 10:49, Ron Michael wrote:
> Hi all, I want to club different objects (character type) to a single one and
> using paste() function that can be done happ
Hi
I am fitting an arima model to some time series X. When I was comparing
the fitted values of the model to the true time series I realized that
the true time series lags one time step behind the fitted values of the
arima model. And this is the case for any model. When I did a simple
linear
Hi all, I am wondering is there any way to check whether some Directory exists
or not, given the parent path of that directory? After searching for a while I
found that there is a function dir.create() to create some directory. However I
need to know whether such directory already exists or not,
Hi all
I have 7 responses that I want to fit with MANOVA. I have accumulated the
responses into a matrix Y and each response has a weight vector associated with
it, so I have accumulated the weights into a matrix WT of the same dimensions
as Y. When I try
fit <- manova( Y ~ X1 + X2
Hello
Could anyone explain me the difference between strftime vs strptime, please
?
I've read the help but it's a little bit cionfusing for me.
cheers
--
View this message in context:
http://r.789695.n4.nabble.com/strftime-vs-strptime-tp3018865p3018865.html
Sent from the R help mailing list
G'day Jose,
On Fri, 29 Oct 2010 11:25:05 +0200
José Manuel Gavilán Ruiz wrote:
> Hello, I want to maximize a likelihood function expressed as an
> integral that can not be symbolically evaluated. I expose my problem
> in a reduced form.
>
> g<- function(x){
> integrand<-function(y) {ex
Dear Henrik,
sorry for bothering you with a report hastily pasted together and not
particularly nice for you as I used my toy data flu from a non-standard package.
I should have better used e.g. the iris.
I'm aware that writeMat doesn't deal with S4 objects. In fact, if I'd overlook
the erro
G'day Jose,
On Fri, 29 Oct 2010 11:25:05 +0200
José Manuel Gavilán Ruiz wrote:
> Hello, I want to maximize a likelihood function expressed as an
> integral that can not be symbolically evaluated. I expose my problem
> in a reduced form.
>
> g<- function(x){
> integrand<-function(y) {ex
Jan Theodore Galkowski acm.org> writes:
>
> I'm interested in the Rsolnp package. For their primary function
> "solnp", one example is given, and there is a reference to "unit
> tests". Anyone know where these can be found? Also, Rsolnp is
> used in a few other packages (e.g., depmixS4), but I
Hello, I want to maximize a likelihood function expressed as an
integral that can not be symbolically evaluated. I expose my problem
in a reduced form.
g<- function(x){
integrand<-function(y) {exp(-x^2)*y}
g<-integrate(integrand,0,1)
}
h<-function(x) log((g(x)))
Hi
Is there any way to check an certain command or procedure's RAM usage?
Im after something similar to system.time(bla) that gives me the time the
command took to preform but for RAM usage.
Hope you understand what i mean.
Best regards
Joel
--
View this message in context:
http://r.789695.n
1 - 100 of 106 matches
Mail list logo
