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
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 increasing with x[1] = 0.
The following works but is not the greatest:
junk<-outer(x, x, '-')
junk[junk>0]
e.g.,
given
x<-c(0, 3, 7, 20)
junk<-outer
he room still a room when its empty? Does the room,
> the thing itself have purpose? Or do we, what's the word... imbue it."
> - Jubal Early, Firefly
>
> r-help-boun...@r-project.org wrote on 03/02/2011 10:42:12 AM:
>
> > [image removed]
> >
> > [
/02/2011 10:42:12 AM:
> [image removed]
>
> [R] Vector manipulations
>
> Benjamin Hartley
>
> to:
>
> r-help
>
> 03/02/2011 11:08 AM
>
> Sent by:
>
> r-help-boun...@r-project.org
>
> I have a question regarding the most efficient way to
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
I have a question regarding the most efficient way to select a substring of
a vector:
I have a vector of value v, and I want to select a subspace of this vector
called w such that:
w=v[1:n]
where
sum(w) = x
I am interested in what you thing would be the most efficient way to do this
- I would
ro.au/bill.venables/
>
> -Original Message-
>
> From: Duncan Murdoch [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, 5 March 2008 9:36 AM
> To: Venables, Bill (CMIS, Cleveland)
> Cc: r-help@r-project.org
> Subject: Re: [R] vector manipulations
>
> On 3/4/2008 5:41 PM,
://www.cmis.csiro.au/bill.venables/
-Original Message-
From: Duncan Murdoch [mailto:[EMAIL PROTECTED]
Sent: Wednesday, 5 March 2008 9:36 AM
To: Venables, Bill (CMIS, Cleveland)
Cc: r-help@r-project.org
Subject: Re: [R] vector manipulations
On 3/4/2008 5:41 PM, [EMAIL PROTECTED] wrote:
> Your problem
On 3/4/2008 5:41 PM, [EMAIL PROTECTED] wrote:
> Your problem is that your function log1( , ) is not vectorized with
> respect to its arguments. For a function to work in outer(...) it must
> accept vectors for its first two arguments and it must produce a
> parallel vector of responses.
>
> To
700
mailto:[EMAIL PROTECTED]
http://www.cmis.csiro.au/bill.venables/
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Pete Dorothy
Sent: Wednesday, 5 March 2008 3:38 AM
To: r-help@r-project.org
Subject: [R] vector manipulations
Hello,
I have simulated a set of
Thank you very much to both of you, and especially you Phil.
I will tell you if it works.
2008/3/4, Phil Spector <[EMAIL PROTECTED]>:
>
> Pete -
> As others have told you, outer only works with vectorized
> functions. An alternative is to use expand.grid to find all
> the combinations of bet
Did you read the help page for 'outer'?
?outer
It says
'FUN' is called with these two extended vectors as arguments.
Therefore, it must be a vectorized function (or the name of one),
expecting at least two arguments.
I don't think your logl function is vectorized according to
Hello,
I have simulated a set of data which i called "nir" (a vector).
I have created a function "logl" which calculates the log-likelihood.
logl is a function of 2 real parameters : "beta" and "zeta" (of length 1).
This function works perfectly well when I try for example "logl(0.1,0.2)"
Now
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