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 <-
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
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
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)
4 matches
Mail list logo
