On 10/28/19 2:17 PM, varin sacha via R-help wrote:
Dear R-experts,
My reproducible example here below is not working because of an error message :
Erreur : vecteurs de mémoire épuisés (limite atteinte ?)
My code perfectly works when n=3000 or n=5000 but as soon as n=1 my code
does not wo
You are describing an awkward way of doing this (and your example is unreadable
because you are not following posting instructions for the list) ... but the
following contains the essence of what I think is needed: parse() and eval().
v <- character()
v[1] <- sprintf("%d ^ %s", 2, "duck")
v[2]
I believe it is possible, but R is not really a full-fledged symbolic algebra
system so it wouldn't be an intuitive tool to use and what would you do with it
once you had it? It is much more useful in R to do something like
f1 <- function( M, A, B, C ) {
M %*% c( A, B, C )
}
m <- matrix( c(
Have you read any R tutorials?? This is about as basic as it gets.
?rep
-- Bert
P.S. Due to vectorized operations, there is probably no need to do this anyway!
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On T
Hi, Thanoon!
I want to simulate a vector of constants values with dimention = 200x2 and
all values of this vector are 1.
You mean "I want to _construct_ a _matrix_ of dimension 200 x 2 with all
entries equal to 1", don't you?
Homework? Take a look at
?matrix
Hth -- Gerrit
Any help
If you mean you need a matrix of that size,
matrix(1, nrow=200, ncol=2)
On Thu, Oct 29, 2015 at 1:07 PM, thanoon younis
wrote:
> Dear Members,
>
> I want to simulate a vector of constants values with dimention = 200x2 and
> all values of this vector are 1.
>
> Any help please.
>
>
> Regards
>
_
For the record, it seems that devEMF::emf() works fine on the Windows
Vista SP2 machine with R3.2.2 and Illustrator CS4: the text is
recognized as text and every point/line can be dissociated.
Ivan
--
Ivan Calandra, PhD
University of Reims Champagne-Ardenne
GEGENAA - EA 3795
CREA - 2 esplanade
Thanks Jean for the tip.
I'll try postscript() and devEMF::emf() and see if it works.
Still, it's very complicated to export vector graphics in R...
Ivan
--
Ivan Calandra, PhD
University of Reims Champagne-Ardenne
GEGENAA - EA 3795
CREA - 2 esplanade Roland Garros
51100 Reims, France
+33(0)3 26
Perhaps the discussion at this link will be helpful ...
http://stackoverflow.com/questions/9555889/producing-a-vector-graphics-image-i-e-metafile-in-r-suitable-for-printing-in
Jean
On Tue, Oct 6, 2015 at 9:42 AM, Ivan Calandra
wrote:
> Dear useRs,
>
> A colleague of mine is having a problem wit
And if we want to use the approach of William Dunlap for sequence.optimization
, then we can write:
rev( xr[ seq_len(sum(vec)) - rep.int(cumsum(c(0L, vec[-length(vec)])), vec)] -
rep.int( xr[ -1], vec))
Regards.
On 2015-09-22 23:43:10, Frank Schwidom wrote:
> Hi,
>
> xr <- rev( x)
> vec <- 1:
Hi,
xr <- rev( x)
vec <- 1:(length( x) - 1)
rev( xr[ sequence( vec)] - rep.int( xr[ -1], vec))
On 2015-09-21 14:17:40, Dan D wrote:
> I need an efficient way to build a new n x (n-1)/2 vector from an n-vector x
> as:
>
> c(x[-1]-x[1], x[-(1:2)]-x[2], ... , x[-(1:(n-1)] - x[n-1])
>
> x is incre
Use ?mappy and ?rep.int
> x[unlist(mapply(":",2:4,4))] - x[rep.int(1:3,3:1)]
[1] 3 7 20 4 17 13
Cheers,
Bert
Bert Gunter
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
-- Clifford Stoll
On Mon, Sep 21, 2015 at 2:17 PM, Dan D wrote:
> I
On 20/07/2014, 5:46 PM, David Winsemius wrote:
>
> On Jul 20, 2014, at 6:30 AM, Duncan Murdoch wrote:
>
>> On 17/07/2014, 10:00 PM, Dario Strbenac wrote:
>>> The example in the question was not inside a user function.
>>
>> The explanations you were given were slightly inaccurate. The usual
>> r
On Jul 20, 2014, at 6:30 AM, Duncan Murdoch wrote:
> On 17/07/2014, 10:00 PM, Dario Strbenac wrote:
>> The example in the question was not inside a user function.
>
> The explanations you were given were slightly inaccurate. The usual
> rule is that results returned at the top level are printed
On 17/07/2014, 10:00 PM, Dario Strbenac wrote:
> The example in the question was not inside a user function.
The explanations you were given were slightly inaccurate. The usual
rule is that results returned at the top level are printed unless they
are marked as invisible. (There are a few cases
It's a plausible use-case. For example, in the example section of a help file.
if(require(aPackage))
{
# Do computations.
# Show beginning of first result vector.
# Show beginning of second result vector.
}
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R-help@r-project.org mailing list
http
On 18/07/14 15:00, Dario Strbenac wrote:
The example in the question was not inside a user function.
Don't be silly.
cheers,
Rolf Turner
--
Rolf Turner
Technical Editor ANZJS
__
R-help@r-project.org mailing list
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The example in the question was not inside a user function.
__
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.org/posting-guide.html
and provide commented, minimal, self
Rolf et.al
I have not followed this thread closely and so have nothing to say
about whose or what explanation is correct.
However, the following statement is misleading, if not wrong:
---
foo <- function(){
x <- 17
x
y <- 42
y
}
If you type foo() you get
[1] 42
which i
On 18/07/14 11:32, David Winsemius wrote:
On Jul 17, 2014, at 9:27 AM, Rui Barradas wrote:
Hello,
Also, unlike what the op says, if statements are functions, explaining the
behavior he got.
I'm not sure that is correct. The help page says if() is a control-construct. I think the
function
On Jul 17, 2014, at 9:27 AM, Rui Barradas wrote:
> Hello,
>
> Also, unlike what the op says, if statements are functions, explaining the
> behavior he got.
I'm not sure that is correct. The help page says if() is a control-construct. I
think the function is actually "{"
> {rnorm(10); rpois(1
This question is related to FAQ 7.16, which you should look up.
In general, whenever you just type a variable or function directly at the
command line, R prints the result for you. Inside code blocks of any kind,
it does not do that, so you need to use the print function yourself.
On Thu, 17
Hello,
Also, unlike what the op says, if statements are functions, explaining
the behavior he got.
Hope this helps,
Rui Barradas
Em 17-07-2014 16:53, Sarah Goslee escreveu:
Hi,
You can get the behaviour you want using the print() command:
if(1 < 2)
{
x<-rnorm(100)
y <- rpois(10, 5)
Hi,
You can get the behaviour you want using the print() command:
if(1 < 2)
{
x<-rnorm(100)
y <- rpois(10, 5)
print(head(x))
print(head(y))
}
Sarah
On Thu, Jul 17, 2014 at 2:00 AM, Dario Strbenac
wrote:
> Hello,
>
> I have a block of code that has two head calls at the end, but only t
Hi,
I have two integers a and b (with a
What did you try?Further, without knowing what your function f(x) is, we
can't tell you whether it accepts vector inputs.
--
View this message in context:
http://r.789695.n4.nabble.com/vector-where-elements-are-functions-evaluated-at-integers-bu
Hi,
It is not clear whether you read the file correctly or not. Please check
str(tbl_all)
The code you showed should work if the file was read correctly.
For example:
set.seed(468)
tbl_all<-
as.data.frame(matrix(sample(1:10,10*22,replace=TRUE),ncol=22,dimnames=list(NULL,c("A","V",LETTERS[3
Hi,
I prefer to build my packages on Linux!
I changed the object trPaths as below,
After R CMD build TinnRcom on Linux and did the installation
on Windows from the source code: It worked very well!
I thank all who contributed to the solution and apologize for my lack
of knowledge in this area.
On 24/09/2013 03:51, Jose Claudio Faria wrote:
I think I found the main problem!
I am developing the package under Linux and after R CMD build,
manually compressing the folder TinnRcom inside of the folder
TinnRcom.RCheck and installing in Windows.
And how on earth did you expect the helpers h
Hi JCFaria,
You package is supposed to be used only under Windows, right? Then, use:
OS_type=windows
in the DESCRIPTION file… and, of course, use R CMD check/R CMD build/ R CMD
INSTALL under Windows only.
Best,
Philippe
On 23 Sep 2013, at 20:55, Jose Claudio Faria
wrote:
> I have been deve
I think I found the main problem!
I am developing the package under Linux and after R CMD build,
manually compressing the folder TinnRcom inside of the folder
TinnRcom.RCheck and installing in Windows.
As the APPDATA environment variable does not exist in Linux, the
function Sys.getenv ("APPDATA"
I appreciate your attention Gabor.
However, the result was the same. :(
Both only work when the trPath object is sent to a R session already running.
When inside the package the result was the same.
> remove.packages('TinnRcom')
> Install.packages('TinnRcom_1.0-09.zip', repos=NULL) # New versio
On Mon, Sep 23, 2013 at 2:55 PM, Jose Claudio Faria
wrote:
> trPaths <- paste(paste(Sys.getenv('APPDATA'),
>'\\Tinn-R\\tmp\\',
>sep=''),
> c('',
>'search.txt',
>'objects.txt',
>
to:r-help-boun...@r-project.org] On
> Behalf
> Of arun
> Sent: Monday, June 17, 2013 11:58 AM
> To: Andras Farkas
> Cc: R help
> Subject: Re: [R] vector question
>
> HI Andras,
> Sorry, I misunderstood your question:
>
> Try this:
> sapply(sapply(b,function(x) d
HI Andras,
Sorry, I misunderstood your question:
Try this:
sapply(sapply(b,function(x) d[x
To: arun
Cc:
Sent: Monday, June 17, 2013 2:48 PM
Subject: Re: [R] vector question
Arun,
thank you. Looking at it I am wondering if there is a way to get 36 in the
result instead of the 24. The 2nd
May be this helps:
d[cumsum(sapply(d,function(x) any(x>b)))>=1]
#[1] 12 24 36
A.K.
- Original Message -
From: Andras Farkas
To: r-help@r-project.org
Cc:
Sent: Monday, June 17, 2013 2:29 PM
Subject: [R] vector question
Dear All,
would you please help with the following:
let us say I h
On 08.05.2013 13:38, Ramon Hofer wrote:
Thanks for your answer Uwe
On Fri, 03 May 2013 23:36:24 +0200
Uwe Ligges wrote:
On 02.05.2013 14:37, Ramon Hofer wrote:
I'm trying to analyse the network speed and used iperf to create a
csv file containing the link test data. It's only about 6 MB
Thanks for your answer Uwe
On Fri, 03 May 2013 23:36:24 +0200
Uwe Ligges wrote:
> On 02.05.2013 14:37, Ramon Hofer wrote:
> >
> > I'm trying to analyse the network speed and used iperf to create a
> > csv file containing the link test data. It's only about 6 MB big but
> > contains about 40'00
On 02.05.2013 14:37, Ramon Hofer wrote:
Hi all
I'm trying to analyse the network speed and used iperf to create a csv
file containing the link test data. It's only about 6 MB big but
contains about 40'000 samples.
I can do boxplots (apart from printing the number of samples but I ask
separate
On Apr 22, 2013, at 6:28 PM, Marc Schwartz wrote:
> On Apr 22, 2013, at 6:21 PM, Ayyappa wrote:
>
>> Dear group,
>>
>> I want to generate a vector of 10 elements that always has 20% zeroes, but
>> with a random ordering of zeroes and ones. Can you please suggest a function
>> to do that in
On Apr 22, 2013, at 6:21 PM, Ayyappa wrote:
> Dear group,
>
> I want to generate a vector of 10 elements that always has 20% zeroes, but
> with a random ordering of zeroes and ones. Can you please suggest a function
> to do that in R? I tried 'sample' function but the 20% zeros was not always
On Mon, Apr 22, 2013 at 7:21 PM, Ayyappa wrote:
> Dear group,
>
> I want to generate a vector of 10 elements that always has 20% zeroes, but
> with a random ordering of zeroes and ones. Can you please suggest a function
> to do that in R? I tried 'sample' function but the 20% zeros was not alway
On Mar 21, 2013, at 4:40 PM, Waichler, Scott R wrote:
> I have a 3D field of a scalar variable (x, y, z, value). Is there a way to
> generate a vector field from this data--gradient at defined points? I found
> the rasterVis package for 2D data, but as yet nothing for 3D data.
Can't you just
On 12-10-23 5:39 AM, Rui Barradas wrote:
Hello,
Try the following.
y=c(2,3,5,2,4,6,8,3,6,2,5)
first <- function(x) min(which(x))
prefix <- function(x, v) x[seq_len(v)]
suffix <- function(x, v) x[-seq_len(v)]
first(y > 4)
prefix(y, first(y > 4))
suffix(y, first(y > 4))
Be careful with this:
Hello,
Try the following.
y=c(2,3,5,2,4,6,8,3,6,2,5)
first <- function(x) min(which(x))
prefix <- function(x, v) x[seq_len(v)]
suffix <- function(x, v) x[-seq_len(v)]
first(y > 4)
prefix(y, first(y > 4))
suffix(y, first(y > 4))
Hope this helps,
Rui Barradas
Em 23-10-2012 10:21, Al Ehan escr
Hi,
Is it what you're looking for?
which(y>4) ##all indexes for y>4
[1] 3 6 7 9 11
which(y>4)[1] ##the first index
[1] 3
HTH,
Ivan
--
Ivan CALANDRA
Université de Bourgogne
UMR CNRS/uB 6282 Biogéosciences
6 Boulevard Gabriel
21000 Dijon, FRANCE
+33(0)3.80.39.63.06
ivan.calan...@u-bourgogn
Thank you all for helping me out.
As Michael points out, I abused the rounding and formatting of print()
while debugging.
The default number of digits to print is 7 according to ?print.default,
which makes floating point numbers to be somewhat plausible for index vector
subsetting at first glance:
On Sat, Oct 6, 2012 at 3:23 PM, Bert Gunter wrote:
> But the OP should not be doing this **at all.** He apparently has not
> bothered to read the Intro to R tutorial as he appears not to know
> about vectorized calculations.
>
> -- Bert
>
I don't really think that's relevant or constructive here.
But the OP should not be doing this **at all.** He apparently has not
bothered to read the Intro to R tutorial as he appears not to know
about vectorized calculations.
-- Bert
On Sat, Oct 6, 2012 at 3:29 AM, R. Michael Weylandt
wrote:
> Forgot to cc the list.
>
> RMW
>
> On Sat, Oct 6, 2012 at 1
Hello,
This seems to be a case for FAQ 7.31 Why doesn't R think these numbers
are equal?
See this example:
3/5 - 1/5 - 2/5 # not zero
3/5 - (1/5 + 2/5) # not zero, different from above
In your case, try
for(idx in breaks){
print(idx / interval, digits = 16) # see problem indices
Forgot to cc the list.
RMW
On Sat, Oct 6, 2012 at 11:29 AM, R. Michael Weylandt
wrote:
> A case study of a good question! Would that all posters did such a good job.
>
>
>
> On Sat, Oct 6, 2012 at 7:14 AM, 周果 wrote:
>> Hi there,
>> Here is a minimum working example:
>> -
On 06-10-2012, at 08:14, 周果 wrote:
> Hi there,
> Here is a minimum working example:
>
> lower = 0
> upper = 1
> n_bins = 50
> interval = (upper - lower) / n_bins
> bins = vector(mode="numeric", length=n_bins)
> breaks = seq(from=lo
On Aug 10, 2012, at 07:30 , Sean Ruddy wrote:
> Thanks for the help all! Good to know that there's an answer. Unfortunately,
> I don't have the rights to install programs so I wasn't able to try devel and
> I've never heard of R patched but I'm guessing I can't install that either.
> I'll see
Thanks for the help all! Good to know that there's an answer. Unfortunately, I
don't have the rights to install programs so I wasn't able to try devel and
I've never heard of R patched but I'm guessing I can't install that either.
I'll see if I can get someone to do that.
Much appreciated!
Sea
As the posting guide asked you to before posting, try R-patched. That
has the NEWS items
• duplicated(), unique() and similar now support vectors of
lengths above 2^29 on 64-bit platforms.
• unique() and similar would infinite-loop if called on a vector of
length > 2^29 (bu
On Aug 9, 2012, at 7:29 PM, Sean Ruddy wrote:
> Hi,
>
> First, thanks in advance. Some useful info:
>
>> version
> platform x86_64-unknown-linux-gnu
> arch x86_64
> os linux-gnu
> system x86_64, linux-gnu
> version.string R version 2.15.1 (2012-06-22)
>
>
On Aug 9, 2012, at 5:29 PM, Sean Ruddy wrote:
Hi,
First, thanks in advance. Some useful info:
version
platform x86_64-unknown-linux-gnu
arch x86_64
os linux-gnu
system x86_64, linux-gnu
version.string R version 2.15.1 (2012-06-22)
I'm trying to use the t
Hi Thomas,
This is non trivial to do, but if you will be working with this sort
of data and are inclined to do some programming, you might consider
creating a new class. S4 classes and methods are quite flexible, and
you can allow them to lean on or inherit from existing classes such as
matrices.
To second Bert Gunter: you may get better answers if you give us a complete
description.
On Thu, Jul 5, 2012 at 12:37 PM, Thomas C. wrote:
> i have number of triangles which i'd like to store in a list,
> matrix,...etc.
> i thought it could look sth. like that:
>
> trianglenode1
You have failed to provide a complete, coherent description of what you
wish to do. In the absence of such a description, all suggestions are just
guesses. You need to think carefully about what information you want to
associate with each triangle and the appropriate data structure to use to
do thi
Hi,
I think it is better to store vectors in to list when the vectors are of
different lengths.
#Consider these cases,
#vectors of equal length
set.seed(1)
list1<-list(a=rnorm(5,15),b=rnorm(5,25),c=1:5,d=runif(5,0.4),e=16:20,f=rep(1,5))
mat1<-cbind(rbind(list1[[1]],list1[[2]],list1[[3]]),(rbin
well thank you, i think we get closer to my problem. but i meant it in a
different way. maybe i describe what i intended to do:
i have number of triangles which i'd like to store in a list, matrix,...etc.
i thought it could look sth. like that:
trianglenode1node2no
On Jul 5, 2012, at 11:22 AM, R. Michael Weylandt wrote:
It is possible to put dimensionality on a list (i.e., a generic
vector), which might be what you're looking for.
x <- list(1:4, letters[1:4], function(x,y) x + y, rnorm(50))
dim(x) <- c(2,2)
x[[1,2]]
x[[2,2]]
x[[3,2]] # Error
That co
It is possible to put dimensionality on a list (i.e., a generic
vector), which might be what you're looking for.
x <- list(1:4, letters[1:4], function(x,y) x + y, rnorm(50))
dim(x) <- c(2,2)
x[[1,2]]
x[[2,2]]
x[[3,2]] # Error
Best,
Michael
On Thu, Jul 5, 2012 at 8:19 AM, Thomas C. wrote:
> h
On 15.05.2012 23:47, rl269 wrote:
Hello,
I am having trouble asking R to read individual numeric vectors for a box
plot of the residuals of a linear regression. It is performing arithmetic
addition on the 16 individual variables that I want individual box plots
for.
I have 16 race*treatment
You're thinking about it wrong. This is an arithmetic sequence:
seq(from = 1000, by = -30, length.out = 15)
Michael
On Fri, Apr 20, 2012 at 5:14 AM, uday wrote:
> I would like to calculate vector from existing value
> e.g
> v <- 1000
> s <- 30
> d1 <- v-s
> d
Hi Petr,
Thank you so much for your help.
You're right, the issue was that I needed to center my predictors by adding
the code 'na.rm=T':
verbal.ability_C <- verbal.ability - mean(verbal.ability, na.rm=T)
That removed the NA values from being included in my centering and allowed
me to run the r
Hi
>
> Here is my code:
>
>
> ##Centering predictors###
> verbal.ability_C <- verbal.ability - mean(verbal.ability)
> children_C <- children - mean(children)
> age_C <- age - mean(age)
> education_C <- education - mean(education)
> work.from.home.frequency_C <- work.from.home.frequency
Here is my code:
##Centering predictors###
verbal.ability_C <- verbal.ability - mean(verbal.ability)
children_C <- children - mean(children)
age_C <- age - mean(age)
education_C <- education - mean(education)
work.from.home.frequency_C <- work.from.home.frequency -
mean(work.from.home.fre
Hi
>
> Hi Petr!
>
> Thank you for responding to my post.
>
> I checked out all my variables in the way you suggested and they are all
in
> integer form, but here are many missing values in some of my vectors,
> denoted with NA.
>
> So, they are in the correct form, I am just wondering if there
Hi Petr!
Thank you for responding to my post.
I checked out all my variables in the way you suggested and they are all in
integer form, but here are many missing values in some of my vectors,
denoted with NA.
So, they are in the correct form, I am just wondering if there is something
else I need
Hi
>
> Hi,
>
> I am trying to run two Non-Gaussian regressions: logistic and probit. I
am
> receiving two different errors when I try to run these regressions and I
am
> not sure what they mean or how to fix my syntax.
>
> Here is the logistic regression error:
>
> Error in family$linkfun(mu
Great Dimitris. It helps indeed! Thanks a lot!
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View this message in context:
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Hi mentor_,
Try
c(sapply(vec[-1], function(x) c(vec[1], x)))
# [1] 2 4 2 6 2 9 2 10
HTH,
Jorge.-
On Sun, Feb 12, 2012 at 12:54 PM, syrvn <> wrote:
> Hello,
>
>
> I am stuck with the following problem. Consider the vector:
>
> vec <- c(2,4,6,9,10)
>
> I now want to use R to manipulate t
One way is:
vec <- c(2,4,6,9,10)
c(rbind(vec[1], vec[-1]))
I hope it helps.
Best,
Dimitris
On 2/12/2012 6:54 PM, syrvn wrote:
Hello,
I am stuck with the following problem. Consider the vector:
vec<- c(2,4,6,9,10)
I now want to use R to manipulate the vector as follows:
[1] 2, 4, 2, 6,
On Thu, Dec 1, 2011 at 3:11 PM, " Majid " wrote:
> Hi,
> yes, It is homework,
Then ask your TA/instructor for help.
>
> These are 2 command:
> first for generating data:
> (1:10)
> that output is 1 2 310
> ok ?
> second is :
> a1<-c( 1:10)
> what is the output ?I didnot see any thing.
Exact
Hi,
On Thu, Dec 1, 2011 at 2:43 PM, " Majid " wrote:
> Dear Sarah.
> Thanks so much,Really I am new in this software,I am wrking to learn the
> software.
First, you should always send your replies to the list. That way
information can help others,
and more people are available to provide advice.
Hi,
On Thu, Dec 1, 2011 at 10:35 AM, " Majid " wrote:
> Hi.
> Can you please answer to my questions about R ?
> 1.how can I write command for vector ?
>
> for exaple in this sample :
> I have this :
> a1 <- c (1:10)
> now how can I put in the vector ?
I'm afraid I don't understand your question.
Thank you very much. Learned something again!
Joh
William Dunlap wrote:
> You can use [1] on the output of FUN to ensure that
> exactly one value (perhaps NA from numeric(0)[1]) is
> returned. E.g.
>
> > index <- 1
> > sapply(list(c(1,2,3),c(1,2),c(1)),function(x){x[max(length(x)-
index,0)
You can use [1] on the output of FUN to ensure that
exactly one value (perhaps NA from numeric(0)[1]) is
returned. E.g.
> index <- 1
> sapply(list(c(1,2,3),c(1,2),c(1)),function(x){x[max(length(x)-index,0)][1]})
[1] 2 1 NA
I'll also put in a plug for vapply, which throws an
error if FUN
Hi:
Here's one approach:
X1 <- sample(1:4, 10, replace = TRUE, prob = c(0.4, 0.2, 0.2, 0.2))
foo <- function(x) {
m <- matrix(NA, nrow = length(x), ncol = length(x))
m[, 1] <- x
idx <- seq_len(length(x))
for(j in idx[-1]) {
k <- sample(idx, 2)
x <- replace(x, k, 5)
You are storing the results in to a vector that is converted to
numeric; that is why you see the numbers. Try this:
> dataval=as.Date("2011/07/01")
> date_val=seq(dataval,length=260,by="-7 day")
> date_inizio=c()
> date_condizione=c()
> for (k in 1:length(date_val)){
+ date_inizio[
Hi,
To amplify on Josh's cogent remarks, the reason you can't create a
vector of functions is because vectors are composed of atomic objects
and functions are not atomic objects:
> is.atomic(f.3)
[1] FALSE
> is.atomic(1)
[1] TRUE
> is.atomic(TRUE)
[1] TRUE
> is.atomic('a')
[1] TRUE
'Vectors' of
Hi Thomas,
On Sat, Jul 2, 2011 at 9:36 AM, Thomas Stibor wrote:
> Hi there,
>
> I have a vector of some functions e.g.
> #-#
> f.1 <- function(a) {
> return( a );
> }
> f.2 <- function(a) {
> return( 1 - (tanh(a))^2 );
> }
> f.3 <- function(a) {
> return( 1 / (1+exp
On Jun 25, 2011, at 4:01 PM, Håvard Wahl Kongsgård wrote:
glm( V0 ~ "HARRY" + "KLINE" + "Brown" + "Larry")
No. You are mistaken. Assuming that V0 is defined and has the same
length as
there are rows in that data.frame, then V1 and V2 would be the
arguments in
the formula and they would
>> glm( V0 ~ "HARRY" + "KLINE" + "Brown" + "Larry")
>
> No. You are mistaken. Assuming that V0 is defined and has the same length as
> there are rows in that data.frame, then V1 and V2 would be the arguments in
> the formula and they would not be quoted.
>
> ?glm # and work through the examples
On Jun 25, 2011, at 2:13 PM, Håvard Wahl Kongsgård wrote:
Hi, sorry my question was not really clear
|Are you very early in efforts at learning R?
No, have been a long term user of R, but only use R for the
statistical stuff.
The heart of the issue is that I have list of keywords that I w
Hi, sorry my question was not really clear
|Are you very early in efforts at learning R?
No, have been a long term user of R, but only use R for the statistical stuff.
The heart of the issue is that I have list of keywords that I want to
analyse with a machine learning algorithm (20 000 keywords
On Jun 25, 2011, at 1:33 PM, Håvard Wahl Kongsgård wrote:
Well, when a think of it using an array would be a simple solution.
But to what extent are arrays supported by other R functions?
Arrays are well supported by arithmetic, logical and accessor
functions. Your questions seem very unfoc
On Jun 25, 2011, at 12:49 PM, Håvard Wahl Kongsgård wrote:
Hi, this seems like a strange question, but in R is there a function
that
can handle vectors containing factors inside lists/tuples?
Yes, A very strange question, indeed ... since dataframes are lists
that commonly contain vectors
Well, when a think of it using an array would be a simple solution.
But to what extent are arrays supported by other R functions?
-Håvard
2011/6/25 Håvard Wahl Kongsgård
> Hi, this seems like a strange question, but in R is there a function that
> can handle vectors containing factors inside li
On 02.05.2011 09:55, andre bedon wrote:
Hi,
I'm quite new to R so this question will sound quite fundamental. I need to
create a vector of length 160. The first element should be (1+r)^159 and each
element thereafter should decrease by a factor of (1+r) until the 160th element
that should b
On 02-May-11 07:55:27, andre bedon wrote:
> Hi,
> I'm quite new to R so this question will sound quite fundamental.
> I need to create a vector of length 160. The first element should
> be (1+r)^159 and each element thereafter should decrease by a
> factor of (1+r) until the 160th element that shou
On Sun, May 1, 2011 at 8:08 AM, azam jaafari wrote:
> Dear All
>
> I want to import the vector file ( .shp) to R. I could import the file by
> rgdal package before, by following:
>
> geology<-readOGR('C:/geology//saga/geo.geom','finalgeology')
>
> but now there is an error:
>
> Error in ogrInfo
On Wed, Mar 9, 2011 at 4:13 PM, Dimitri Liakhovitski
wrote:
> I realize I can do something like this:
> date<-20081229
> mydates<-as.Date(as.character(date),"%Y%m%d")
> for(i in 1:52){ #
> mydates<-c(mydates,mydates[length(mydates)]+7)
> }
>
> I just thought maybe there is a more elegant way...
On Wed, Mar 9, 2011 at 3:04 PM, Dimitri Liakhovitski
wrote:
> Hello!
>
> I have a date (a Monday):
>
> date<-20081229
> mydates<-as.Date(as.character(date),"%Y%m%d")
>
> What package would allow me to create a vector that starts with that
> date (mydates) and contains dates for 51 Mondays that fol
I realize I can do something like this:
date<-20081229
mydates<-as.Date(as.character(date),"%Y%m%d")
for(i in 1:52){ #
mydates<-c(mydates,mydates[length(mydates)]+7)
}
I just thought maybe there is a more elegant way...
Thank you!
Dimitri
On Wed, Mar 9, 2011 at 4:04 PM, Dimitri Liakhovitski
Please ask the author of parts() in the partitions package or the author
of the function that calls the former:
your function calls generate a call
parts(J) where J is 1272.
Internally, a J*P(J) (1272 * 1.514126e+19) vector is generated (and
that one is too large for R).
Uwe Ligges
On 0
Thanks Jonathan and Ista, that's really helpful. Got it running much better
now.
Ben
On 2 March 2011 17:33, Jonathan P Daily wrote:
> Is this what you want? I don't know what your v looks like, but this won't
> work if there are cases in which v won't sum to exactly x.
>
> x <- 20
> v <- sample
Is this what you want? I don't know what your v looks like, but this won't
work if there are cases in which v won't sum to exactly x.
x <- 20
v <- sample(0:1, 100, T)
w <- v[1:which(cumsum(v)==x)]
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
1
Hi Benjamin,
There may be faster ways, but
v <- 1:100
x <- 10
n <- which(cumsum(v) == x)
w <- v[1:n]
seems pretty straightforward.
Best,
Ista
On Wed, Mar 2, 2011 at 10:42 AM, Benjamin Hartley
wrote:
> I have a question regarding the most efficient way to select a substring of
> a vector:
>
> I
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