How about:
unlist(strsplit(x, split=" "))[c(4:5,10)]
That perl script looks like a good reason to avoid perl.
Simon.
On Fri, 2008-08-01 at 15:13 +0900, Edward Wijaya wrote:
> Hi,
>
> I have this string, in which I want to extract some of it's element:
>
> > x <- "Best-K Gene 11340 211952_at R
x[!(x %% 2 == 0)]
On Thu, Jul 31, 2008 at 10:01 PM, Gundala Viswanath <[EMAIL PROTECTED]>wrote:
> Dear all,
>
> How can I remove the even number from the following vector
>
> > x
> [1] 4 5 6 8 17 20 21 22 23 25 26 31 35 36 38 40 41 42
> 43
> [20] 44 50 74 75 82 84 89
On Thu, 31 Jul 2008, calundergrad wrote:
> i have a vector with values similar to the below text
> [1] 001-010-001-0
>
> I want to get rid of all leading zeroes.
> for example i want to change the values of the vector
> so that [1] 001-010-001-0 becomes [1] 1-010-001-0.
>
> Another example
> [1]08
Hi,
I have this string, in which I want to extract some of it's element:
> x <- "Best-K Gene 11340 211952_at RANBP5 Noc= 3 - 2 LL= -963.669 -965.35"
yielding this array
[1] "211952_at" "RANBP5" "2"
In Perl we would do it this way:
__BEGIN__
my @needed =();
my $str = "Best-K Gene 11340 21
On Fri, 1 Aug 2008, Gundala Viswanath wrote:
> How can I remove the even number from the following vector
>
> > x
> [1] 4 5 6 8 17 20 21 22 23 25 26 31 35 36 38 40 41 42
> 43
> [20] 44 50 74 75 82 84 89 90 91 95 96 97 100 101 102 118 119 121
> 122
> [39] 123 13
Yes, this is how it should be done!
--- On Fri, 1/8/08, Christos Hatzis <[EMAIL PROTECTED]> wrote:
> From: Christos Hatzis <[EMAIL PROTECTED]>
> Subject: Re: [R] cutting out numbers from vectors
> To: "'calundergrad'" <[EMAIL PROTECTED]>, r-help@r-project.org
> Received: Friday, 1 August, 2008,
Dear all,
How can I remove the even number from the following vector
> x
[1] 4 5 6 8 17 20 21 22 23 25 26 31 35 36 38 40 41 42 43
[20] 44 50 74 75 82 84 89 90 91 95 96 97 100 101 102 118 119 121 122
[39] 123 135 136 157 158
yielding
5, 17, 21, 23, 25, . (k
I install R in Ubuntu with apt-get,But I cannot see help.start() page.why
Thanks
[[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
y <- 2 - (x[,1] > x[,2])
you can also do
cbind(x,y)
if you wish.
--- On Fri, 1/8/08, Gundala Viswanath <[EMAIL PROTECTED]> wrote:
> From: Gundala Viswanath <[EMAIL PROTECTED]>
> Subject: [R] Grouping Index of Matrix Based on Certain Condition
> To: [EMAIL PROTECTED]
> Received: Friday, 1 Aug
This is something that is easier done in C than in R (to the best of my very
limited knowledge).
To do this in R you could do something like:
> x <- "082-232-232-1"
> y <-unlist(strsplit(x,""))
> i <- which(y != "0")[1]-1
> paste(y[-(1:i)],collapse="")
[1] "82-232-232-1"
--- On Fri, 1/8/08, c
on 07/31/2008 04:29 PM Alan Cox wrote:
Hello. I am hoping someone will be willing to help me understand
something about hazard plots created with muhaz(...). I have some
background in statistics (minor in grad school), but I haven't been
able to figure one thing about hazard plots. I am using
Have a look at gsub:
> x <- "001-010-001-0"
> gsub("^0+", "", x)
[1] "1-010-001-0"
-Christos
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of calundergrad
> Sent: Thursday, July 31, 2008 4:40 PM
> To: r-help@r-project.org
> Subject: [R] cutting ou
Dear Gundala,
Try this:
x=as.matrix(read.table(textConnection("
4.482909e-01 0.55170907
9.479594e-01 0.05204063
8.923553e-01 0.10764474
9.295003e-01 0.07049966
8.880434e-01 0.11195664
9.197367e-01 0.08026327
9.431232e-01 0.05687676
9.460356e-01 0.05396442
6.053829e-01 0.39461708
9.515173e-01 0.04
On 1/08/2008, at 2:56 PM, stephen sefick wrote:
z <- rnorm(5000)
f <- fft(z)
d <- fft(f, inverse=T)
plot(z, d)
z <- rnorm(5000)
z.ts <- ts(z)
f <- fft(z.ts)
d <- fft(f, inverse=T)
plot(z.ts, d)
temp <- matrix(c(1,4,2, 20), nrow=2)
d <- fft(temp)
f <- fft(d, inverse=T)
plot(temp, f)
this,
Marc,
Thank you so much! Your solution is perfect! I hadn't known about the "cut"
function. The graph is precisely what I needed. I have the ISwR book but not
anything more advanced. Need to get some more advanced books, maybe?
I have been stubbornly working on this for a few hours now, and getti
Hm didn't know about the abind package. That is pretty sweet.
I do often opt for the 'list of matrices' myself though because I'm more
familiar with the methods for lists...
To answer one of Gundala's questions directly (assuming list of matrices is
acceptable):
Since you can do this,
> x <-
Try this:
> x <- matrix(runif(20), ncol=2)
> x
[,1] [,2]
[1,] 0.33119833 0.4797847
[2,] 0.01339784 0.5218626
[3,] 0.78975940 0.8597246
[4,] 0.60849015 0.5217248
[5,] 0.91779777 0.9364047
[6,] 0.88302538 0.3467961
[7,] 0.87565986 0.4029147
[8,] 0.51594479 0.9885018
[9,] 0.
I am very much a beginner at Python and am trying to get RSPython to work
for me to solve a specific problem with geocoding addresses using the geopy
library for Python (which appears to be an easy way to geocode large numbers
of addresses).
While I can load up the RSPython library just fine, when
i have a vector with values similar to the below text
[1] 001-010-001-0
I want to get rid of all leading zeroes. for example i want to change the
values of the vector so that [1] 001-010-001-0 becomes [1] 1-010-001-0.
Another example
[1]082-232-232-1 becomes [1] 82-232-232-1
--
View this mes
You can use uniroot (see ?uniroot).
As an example, suppose you have a $100 bond which pays 3% every half year (6%
coupon) and lasts for 4 years. Suppose that it now sells for $95. In such a
case your time intervals are 0,0.5,1,...,4 and the payoffs are:
-95,3,3,...,3,103.
To find internal rate
Hi,
I have the following (M x N) matrix, where M = 10 and N =2
What I intend to do is to group index of (M) based on this condition
of "x_mn" , namely
For each M,
If x_m1 > x_m2, assign index of M to Group1
otherwise assign index of M into Group 2
> x
[,1] [,2]
[1,] 4.482
R also has arrays (?array). If all the matrices are the **same dimension**
and **same mode** (i.e. all numeric or all character), this is perhaps
easier, especially if you want to access various "pieces" of your matrices.
Given existing matrices, mat1, mat2,..., the slightly tricky way to combine
z <- rnorm(5000)
f <- fft(z)
d <- fft(f, inverse=T)
plot(z, d)
z <- rnorm(5000)
z.ts <- ts(z)
f <- fft(z.ts)
d <- fft(f, inverse=T)
plot(z.ts, d)
temp <- matrix(c(1,4,2, 20), nrow=2)
d <- fft(temp)
f <- fft(d, inverse=T)
plot(temp, f)
this, looks to me, to be the same. you have to take the
z <- rnorm(5000)
f <- fft(z)
d <- fft(f, inverse=T)
plot(z, d)
z <- rnorm(5000)
z.ts <- ts(z)
f <- fft(z.ts)
d <- fft(f, inverse=T)
plot(z.ts, d)
temp <- matrix(c(1,4,2, 20), nrow=2)
d <- fft(temp)
f <- fft(d, inverse=T)
plot(temp, f)
this, looks to me, to be the same. you have to take the
Gundala Viswanath wrote:
Thanks so much Erik,
But how do you include that in a loop.
I tried this, doesn't seem to work. Please advice:
__BEGIN__
all_mat <- NULL
for (matno in 1:10) {
mat <- process_to_create_matrix(da[matno])
all_mat <- list(all_post, matno = mat)
}
I'm not exac
In the first case, 'crops' is an object that is 142MB in size, and
that alone will take up almost 50% of your memory. You are also
probably generating another object of like size in doing the
operation. Can you cut down the size of the objects that you need.
What operating system do you have and
Thanks so much Eric,
But how do you include that in a loop.
I tried this, doesn't seem to work. Please advice:
__BEGIN__
all_mat <- NULL
for (matno in 1:10) {
mat <- process_to_create_matrix(da[matno])
all_mat <- list(all_post, matno = mat)
}
print(all_mat) # it gives funny structure
I think a named list is probably the easiest way to start off, something
like:
all_mat <- list(mat1 = mat1, mat2 = mat2)
all_mat$mat2
Gundala Viswanath wrote:
Hi,
Suppose I have these two matrices (could be more).
What I need to do is to store these matrices into a hash.
So that I can cal
Hi,
Suppose I have these two matrices (could be more).
What I need to do is to store these matrices into a hash.
So that I can call back any of the matrix back later.
Is there a way to do it?
> mat_1
[,1][,2]
[1,] 9.327924e-01 0.067207616
[2,] 9.869321e-01 0.013067929
Hi,
say I have a loop that cannot complete. how do I abort within JGR?
I use unbuntu linux.
Thanks.
--
View this message in context:
http://www.nabble.com/How-to-abort-JGR-tp18766719p18766719.html
Sent from the R help mailing list archive at Nabble.com.
__
I came up with the same solution as Mark Leeds.
> a1roworders <- t(apply(a1, 1, order))
> a2ord <- t(sapply(seq(nrow(a1)), function(x) a2[x, a1roworders[x,]] ))
> a2ord
[,1] [,2] [,3]
[1,] 102 101 103
[2,] 102 101 103
[3,] 103 101 102
[4,] 101 102 103
Mark's question about the o
I am trying to write a function for P-I curve meansurament of Platt (1980). I
used the exemple of nonlinear least squares in R-intro.pdf but I can't finish
the analysis. Could you help me? Is there another way to do this analysis?
Thanks.
I<- c(0,0,100,100,200,200,500,500)
P<- c(1.35,-1.41,10.
Hi all,
I am doing some experiment studies...
It seems to me that with different combination of 5 parameters, the end
results ultimately converged to two scalars. That's to say, some
combinations of the 5 parameters lead to one end result and some other
combinations of the 5 parameters lead to t
I am getting an error "allocMatrix: too many elements specified" when
I am trying to create large matrix or vector (about 1 billion elements).
How can I find out limits on allocMatrix? Can I increase them?
?"Memory-limits", and you cannot increase them unless you have a
system which has lar
Dear all,
I was trying to understand how "multcomp" package works by running the
examples given in the documentation.
However I still don't understand when it comes to multiple comparison set by
user (please refer to "Ksub" in the code). Therefore I run 2 other cases
along with the original examp
below is another way ( maybe the same ? )but with an extra line to make
roworder. i'm also not clear on why I have to take the transpose of
the result that
comes back form the apply call. in ?apply, it says that the function
tries to convert the result back to a matrix. that's fine but why doe
On Thu, Jul 31, 2008 at 12:54 PM, GOUACHE David
<[EMAIL PROTECTED]> wrote:
> Hello R-helpers,
>
> I would like to produce a boxplot for dates, using lattice.
>
> Here is a dummy example :
>
> dates<-as.Date(32768:32895,origin="1900-01-01")
> plouf<-data.frame(days=dates,group=factor(rep(1:2,times=1
On 31/07/2008 5:21 PM, Jacob Wegelin wrote:
I used read.dta() to read in a Stata 9 dataset to R. The "Sex01" variable
takes on two values in Stata: 0 and 1, and it is labeled "M" and "F"
respectively, analogous to an R factor. Thus, read.dta reads it in as a
factor.
Now, I wanted to see what thi
If you're not adverse to cbind-ing a1 and a2, you can use this:
a1a2 <- cbind(a1, a2)
a3 <- t(apply(a1a2, 1, function(x) x[order(x[1:ncol(a1)])+ncol(a1)]))
Eric
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Timothy W. Hilton
Sent: Thursday, July 31,
On 31/07/2008 4:07 PM, georgiana onicescu wrote:
Dear HelpeRs,
I have created a tar.gz package (the package was created on unix) and I would like to install it in R on a Windows operating system.
As far as I know, R in Windows accepts only packages in zip file for installation and I could not fi
See ?NumericConstants.
On Thu, 31 Jul 2008, Jacob Wegelin wrote:
I used read.dta() to read in a Stata 9 dataset to R. The "Sex01" variable
takes on two values in Stata: 0 and 1, and it is labeled "M" and "F"
respectively, analogous to an R factor. Thus, read.dta reads it in as a
factor.
Now, I
On Thu, 31 Jul 2008, Vadim Kutsyy wrote:
I am getting an error "allocMatrix: too many elements specified" when I am
trying to create large matrix or vector (about 1 billion elements).
How can I find out limits on allocMatrix? Can I increase them?
?"Memory-limits", and you cannot increase the
I clicked "Send" before making sure I thanked anyone who took the time to help
me out. Â Sorry about that. Â To all who read or respond: thanks!.
- Original Message -
From: "Alan Cox" <[EMAIL PROTECTED]>
To: r-help@r-project.org
Sent: Thursday, July 31, 2008 5:29:35 PM GMT -05:00 US/C
Dear HelpeRs,
I have created a tar.gz package (the package was created on unix)
and I would like to install it in R on a Windows operating system.
As far as I know, R in Windows accepts only packages in zip file for
installation and I could not find a way to install a tar.gz file.
I would like
I used read.dta() to read in a Stata 9 dataset to R. The "Sex01" variable
takes on two values in Stata: 0 and 1, and it is labeled "M" and "F"
respectively, analogous to an R factor. Thus, read.dta reads it in as a
factor.
Now, I wanted to see what this variable *really* is, in R. For instance,
so
rollapply over an index. Modify this appropriately:
> z <- zoo(101:109, 11:19)
> f <- function(ix) if (ix[1] < 5) min(z[ix]) else max(z[ix])
> rollapply(zoo(seq_along(z)), 3, by = 3, f)
2 5 8
101 104 109
On Thu, Jul 31, 2008 at 3:36 PM, rcoder <[EMAIL PROTECTED]> wrote:
>
> Hi everyone,
>
Hello. I am hoping someone will be willing to help me understand something
about hazard plots created with muhaz(...). I have some background in
statistics (minor in grad school), but I haven't been able to figure one thing
about hazard plots. I am using hazard plots to track customer cancell
Hello all,
I am trying to sort rows of one matrix by rows of another. Given a1 and a2:
--
> a1
[,1] [,2] [,3]
[1,]768
[2,]424
[3,]472
[4,]038
a1 <-
structure(c(7, 4, 4, 0, 6, 2, 7, 3, 8, 4, 2, 8), .Dim = c(4L, 3L))
> a2
[,1] [,2] [,
Hi All,
I have this problem. I dont understand the right code in R when I have an
anisotropy in the semivariogram model
plot(variogram(Z~1, subground, cutoff=1800, width=80, alpha=c(45, 135, 90,
135)))
I have a good model in 90° and eventually in 90° and 45°
v = variogram(Z~1, sub
I am getting an error "allocMatrix: too many elements specified" when I
am trying to create large matrix or vector (about 1 billion elements).
How can I find out limits on allocMatrix? Can I increase them?
Thanks,
Vadim.
PS: I am running R on SUSE10 on
___
This probably has to do with your ssh configuration and nothing to do
with R. How are you starting Rserve? Is it run with the same user
privileges as when you run R manually?
Best,
Jeff
Patil, Prasad wrote on 07/31/2008 02:51 PM:
Hello,
I've installed an Rserve instance on a remote serv
On Thu, Jul 31, 2008 at 1:30 PM, GOUACHE David
<[EMAIL PROTECTED]> wrote:
> R-helpers,
>
> I'm having difficulty with customizing strip names for a lattice graphic.
>
> Here is an example using the iris data set :
>
> xyplot(Sepal.Length+Sepal.Width~Petal.Length,groups=Species,data=iris)
>
> ## I'd
You are missing 'ranlib': take a closer look at the output from
configure (and the posting guide).
My guess is that either /usr/ccs/bin is not in your path or you don't have
the tools installed to build applications. This is not an R issue, and
you need to seek local Solaris help.
On Thu, 3 M
on 07/31/2008 03:07 PM georgiana onicescu wrote:
Dear HelpeRs, I have created a tar.gz package (the package was
created on unix) and I would like to install it in R on a Windows
operating system. As far as I know, R in Windows accepts only
packages in zip file for installation and I could not fin
Thanks for both replies.
Then I found the "ifft2" from Matlab gives different result from "fft( ,
inverse=T)" from R.
An example:
in R:
> temp <- matrix(c(1,4,2, 20), nrow=2)
> fft(temp)
[,1] [,2]
[1,] 27+0i -17+0i
[2,] -21+0i 15+0i
> fft(temp,inverse=T)
[,1] [,2]
[1,] 27+0i -
> It's not really equivalent, natural language has ambiguities and subtleties
> that computer languages, especially functional languages, intentionally
> don't have. By their nature, computer languages can be turned into parse
> trees unambiguously and then those trees can be manipulated.
But in
Please find below my command inputs, subsequent outputs and errors that I've
been receiving.
> crops <- read.table("crop2000AD.asc", colClasses = "numeric", na="-")
> str(crops[1:10])
'data.frame': 2160 obs. of 10 variables:
$ V1 : num NA NA NA NA NA NA NA NA NA NA ...
$ V2 : num NA N
On 5/2/07, Sundar Dorai-Raj <[EMAIL PROTECTED]> wrote:
>
>
> Iasonas Lamprianou said the following on 5/2/2007 8:25 AM:
> > Hi I am using R version 2.4.1. How can I upgrade to version 2.5 without
> > having to install all the packages again?
> > Thanks
> > Jason
> >
>
> You may find the following
on 07/31/2008 03:20 PM Economics Guy wrote:
I have a large data set where one of the columns needs be a unique
identifier (ID) for each row. However for a few of the rows they have
the same ID. What I need to do is randomly draw one of the rows and
keep it in the data frame and drop all the other
R-helpers,
I'm having difficulty with customizing strip names for a lattice graphic.
Here is an example using the iris data set :
xyplot(Sepal.Length+Sepal.Width~Petal.Length,groups=Species,data=iris)
## I'd like to change the 2 strip names to "Length" and "Width" for example,
this is what
Dear HelpeRs,
I have created a tar.gz package (the package was created on unix) and I would
like to install it in R on a Windows operating system.
As far as I know, R in Windows accepts only packages in zip file for
installation and I could not find a way to install a tar.gz file.
I would like
How about making your homeworks yourselfes?
lalitha viswanath wrote:
> Hi
> I have a dataframe which has 3 columns of numeric data
> A,B,C each of which has been obtained independent of
> the other.
>
> We are trying to find out, which of A or B cause C
> i.e. We are hypothesising that C is the ef
Tomas Mikoviny wrote:> Hi all R positive,>> does anyone know how to refer R in
article?>> thanks>> tomas> Like this (well, almost! This is from R 2.4.1 -
2007 on the most recentversion)
> citation()
To cite R in publications use:
R Development Core Team (2006). R: A language and environment
On 7/31/08 2:12 PM, "Duncan Murdoch" <[EMAIL PROTECTED]> wrote:
>
> expression() returns a list of language objects, and we only asked for
> one. We can look inside it:
Hey, cool. Now let me see if I can do anything useful with that. Thanks.
-Ken
--
Ken Williams
Research Scientist
The
Hi everyone,
I have a rollapply statement that applies a function, in steps, over a data
matrix as follows:
#Code start
testm<-rollapply(mat, 100, by=100, min, na.rm=F)
#Code end
This moves down matrix 'mat' and calculates the minimum value over a 100 row
range, every 100 rows (i.e. no overlaps
Iasonas Lamprianou said the following on 5/2/2007 8:25 AM:
> Hi I am using R version 2.4.1. How can I upgrade to version 2.5 without
> having to install all the packages again?
> Thanks
> Jason
>
You may find the following link relevant.
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/75359
I have a large data set where one of the columns needs be a unique
identifier (ID) for each row. However for a few of the rows they have
the same ID. What I need to do is randomly draw one of the rows and
keep it in the data frame and drop all the others which have the same
ID.
For example:
v1 <-
I am sorry if I misinterpreted this but this questions looks/looked very
like a very basic one about statistical inference... (have a look at the
Manuals/contributed documentation e.g. Faraway)
Lalitha Viswanath wrote:
> Hi
> This is not a homework assignment :)
> Me and my manager are trying to u
Hello all,
I have a question on bpplot (or more specifically panel.bpplot used with
bwplot) from package Hmisc.
I would like to produce a box-percentile plot but vertically. And I hqve not
been able to use the horizontal argument...
using one of the examples from ?panel.bpplot :
bwplot(g ~
Hello R-helpers,
I would like to produce a boxplot for dates, using lattice.
Here is a dummy example :
dates<-as.Date(32768:32895,origin="1900-01-01")
plouf<-data.frame(days=dates,group=factor(rep(1:2,times=128/2)))
bwplot(group~days,data=plouf)
# doesn't work, whereas :
bwplot(group~as.num
Hello,
I've installed an Rserve instance on a remote server. I have no problem
interfacing with it and running most commands. I have loaded some R
scripts on the remote server, and one of them contains a system command
to copy a file (created by the script) onto another server. When I run
the s
On 7/31/2008 2:08 PM, Ken Williams wrote:
On 7/31/08 11:01 AM, "hadley wickham" <[EMAIL PROTECTED]> wrote:
I think that would be a very hard task -
Well, at least medium-hard. But I think significant automatic steps could
be made, and then a human can take over for the last few steps. Tha
Hi,
Thank everyone for you help. I got my histogram written to a file. Thanks
once again.
Aiste
2008/7/31 Jian Zhang <[EMAIL PROTECTED]>
> Hi aiste,
> two easy example produce mypic.ps or mypic.pdf for you x<-rnorm(50)
> postscript("mypic.ps")
> hist(x)
> dev.off()
> pdf( mypic.pdf)
>
This is perfect, thanks! :)
Jorge Ivan Velez pretended :
Hi Max,
See ?write.table. Perhaps:
MAT <- matrix(runif(1 * 6), 1, 6)
# TXT format
write.table(MAT,
"C:/yourmatrix.txt",col.names=FALSE,row.names=FALSE,quote=FALSE)
# XLS format
write.table(MAT,
"C:/yourmatrix.xls",col.names=FA
?Devices for a start.
Aiste Aistike wrote:
Hello,
I would like to ask if anyone could help me. I want to save images I create
(e.g. histograms, boxplots, plots, etc.) to a file or files. Does anyone
know how to do this?
Thank you.
Aiste
[[alternative HTML version deleted]]
_
Thank you both very much for your assistance. Both the suggestions worked out
in the end, and I managed to achieve what I wanted.
There is something else I want to try, which is a slight deviation on the
theme. In the code I posted, I export Intercept and Slope regression
coefficients to an o/p m
On 7/31/08 11:01 AM, "hadley wickham" <[EMAIL PROTECTED]> wrote:
> I think that would be a very hard task -
Well, at least medium-hard. But I think significant automatic steps could
be made, and then a human can take over for the last few steps. That's why
I was enquiring about "tools" rathe
m <- runif(6)
mm <- matrix(m, ncol=6)
HTH
Dave
On Thu, Jul 31, 2008 at 6:24 PM, Max <[EMAIL PROTECTED]> wrote:
> Hi Everyone,
>
> I did a quick search of the list and it looks like this may not have been
> asked before... I'm trying to generate a matrix of random numbers between 0
> and 1,
Thanks so much for the link.
Scott
Gabor Grothendieck wrote:
>
> Vista works fine for developing packages. You might look at
>
> http://batchfiles.googlecode.com
>
> which has some batch files for use with Vista and R.
>
> On Thu, Jul 24, 2008 at 9:13 PM, FScottDahlgren <[EMAIL PROTECTED]>
Hello,
I would like to ask if anyone could help me. I want to save images I create
(e.g. histograms, boxplots, plots, etc.) to a file or files. Does anyone
know how to do this?
Thank you.
Aiste
[[alternative HTML version deleted]]
__
R-help@r
Hi All,
I wish to create a directional variograms at (0, 45, 90, 135) degrees from
north (y-axis).
v = variogram(Z~1, subground, cutoff=1800, width=80, alpha=c(0, 45, 90,
135))
90 and 45 deg. look good but I don't understand in the fit.variogram how to
create a ansi=c(X) with 90 an
?write.table
could help
PF
2008/7/31 Max <[EMAIL PROTECTED]>:
> Marc,
>
> this is very handy. My next question is, do you know a quick and easy way to
> transfer all of the output to a txt file? (or .xls)?
>
> Thanks,
>
> -Max
>
>
> Marc Schwartz explained on 07/31/2008 :
>>
>> on 07/31/2008 12:
Hi Max,
See ?write.table. Perhaps:
MAT <- matrix(runif(1 * 6), 1, 6)
# TXT format
write.table(MAT,
"C:/yourmatrix.txt",col.names=FALSE,row.names=FALSE,quote=FALSE)
# XLS format
write.table(MAT,
"C:/yourmatrix.xls",col.names=FALSE,row.names=FALSE,quote=FALSE,sep="\t")
HTH,
Jorge
On
I am trying to do this using the copula library and find a possible way out.
library(copula)
x=mvdc(claytonCopula(.75),c("exp","exp"),list(list(rate=1),list(rate=2)))
x.sample=rmvdc(x,100)
The above code gives a sample with two marginal exponential ditributions. But
what does the first argument
Marc,
this is very handy. My next question is, do you know a quick and easy
way to transfer all of the output to a txt file? (or .xls)?
Thanks,
-Max
Marc Schwartz explained on 07/31/2008 :
on 07/31/2008 12:24 PM Max wrote:
Hi Everyone,
I did a quick search of the list and it looks like t
Gustaf,
Summarizing things I don't understand:
- Honestly, I was thinking I can use bootstrap to obtain better
estimate of a mean - provided that I want it. So, I can't?
- If I can't obtain reliable estimates of CI and variance from a small
dataset, but I can do it with bootstrap - isn't it a "
on 07/31/2008 12:24 PM Max wrote:
Hi Everyone,
I did a quick search of the list and it looks like this may not have
been asked before... I'm trying to generate a matrix of random numbers
between 0 and 1, with 6 columns, 1 rows. About all I know is that
runif(1) gives me the random number
Hi Everyone,
I did a quick search of the list and it looks like this may not have
been asked before... I'm trying to generate a matrix of random numbers
between 0 and 1, with 6 columns, 1 rows. About all I know is that
runif(1) gives me the random number I'm looking for.
Any help would b
Hi Michal,
>> It is that you just don't estimate mean, or CI, or variance on PK profile
>> data!
>> It is as if you were trying to estimate mean, CI and variance of a
>> "Toccata_&_Fugue_in_D_minor.wav" file. What for? The point is in the
>> music!
>> Would the mean or CI or variance tell you
A more general solution:
strip.fun <- function(x, split=".") {
xx <- strsplit(x, split, fixed=TRUE)
txx <- table(unlist(xx))
nxx <- names(txx)[txx > 1]
setdiff(unlist(xx), nxx)
}
> x <- c("dog.is.an.animal", "cat.is.an.animal", "rat.is.an.animal")
> strip.fun(x)
[1
There MUST be a better way but this will work.
x <- c("dog.is.an.animal", "cat.is.an.animal", "rat.is.an.animal")
bb <- strsplit(x, "\\.")
myfun <- function(m) m[1]
animals <- unlist(lapply(bb, myfun))
animals
--- On Thu, 7/31/08, Daren Tan <[EMAIL PROTECTED]> wrote:
> From: Daren Tan <[EMA
Alessandro unifi.it> writes:
>
> Hi All,
>
> I have this question. I wish to create a kriging map with R but I haven't a
> auxiliary map. I have only one txt file with X, Y and Z records.
>
> When I use this code:
>
...
> subground.ok <- krige (Z~sqrt(X+Y), subground, subground.ovgm, nmax=4
Hello R-User!
I have a data.frame with 82 variables (columns) and 290 rows.
The variables are set to classes factor, ordered factor and numeric.
I used the following code
Matrix.My.data<-as.matrix(Df.My.Data[2:82])
Matrix.My.data.rcorr<-rcorr(Matrix.My.data, type="spearman")
and got the foll
Hi all.
I am an R newbie and trying to grasp how the simple optimization routines in
R work. Specifically, I would like some guidance on how to set up a code to
calculate the internal rate of return on an investment project
(http://en.wikipedia.org/wiki/Internal_rate_of_return).
My main problem
on 07/31/2008 08:35 AM Ken Williams wrote:
On 7/30/08 1:59 PM, "Marc Schwartz" <[EMAIL PROTECTED]> wrote:
I (and many others) use ESS (Emacs Speaks Statistics), in which case, I
have an R source buffer in the upper frame and an R session in the lower
frame.
I also use ESS to edit my R code
on 07/31/2008 07:40 AM Wim Bertels wrote:
On Wed, 2008-07-30 at 12:13 -0500, Marc Schwartz wrote:
on 07/30/2008 08:48 AM Wim Bertels wrote:
Hi,
does anyone know of a function to calculate odds ratios in multiway
tables (stratified) (+ the other usual statistics involved)
i mean:
say we have a
On Thu, Jul 31, 2008 at 4:30 PM, Michal Figurski
<[EMAIL PROTECTED]> wrote:
> Frank and all,
>
> The point you were looking for was in a page that was linked from the
> referenced page - I apologize for confusion. Please take a look at the two
> last paragraphs here:
> http://people.revoledu.com/ka
dear all,
I apologize for a second post on the same subject. I still have the
problem. I'm going to describe the problem in a different setting:
I have 3 matrix A,B,W such that
dim(A) = c(n,M)
dim(B) = c(n,M)
dim(W) = c(M,M)
what I'm searching for is an efficient computation of vector R,
length(R
Amazing responses to ,y question, and very fast, thanks all of you!
Dr. Iasonas Lamprianou
Department of Education
The University of Manchester
Oxford Road, Manchester M13 9PL, UK
Tel. 0044 161 275 3485
[EMAIL PROTECTED]
- Original Message
From: Marc Schwartz <[EMAIL PROTECTED]>
To: [E
Dear Pf. Ripley,
I use the nls() function coupled with the confint() to obtain the confidence
interval of assessed parameters of fitting model.
I don't understand why the confidence intervals estimated with such a method
aren't consistent?
Regards
B. Boulinguiez
Ph.D. in Chemistry
Ecole Nation
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