an implementation that gives you 25x improvement here as well as
tips for getting even more out of it:
http://tolstoy.newcastle.edu.au/R/e17/help/12/01/2471.html
Michael
On Sat, Jan 28, 2012 at 12:28 PM, Kevin Ummel kevinum...@gmail.com wrote:
Thanks. I've played around with pure R
, at 9:14 AM, Peter Langfelder wrote:
ehm... this doesn't take very many ideas.
x = runif(n=10e6, min=0, max=1000)
x = round(x)
system.time( {
y = x[-1] - x[-length(x)]
})
I get about 0.5 seconds on my old laptop.
HTH
Peter
On Fri, Jan 27, 2012 at 4:15 PM, Kevin Ummel kevinum
Hi everyone,
Speed is the key here.
I need to find the difference between a vector and its one-period lag (i.e. the
difference between each value and the subsequent one in the vector). Let's say
the vector contains 10 million random integers between 0 and 1,000. The
solution vector will have
(temp1)])
temp2 = temp
temp2[chg=delta] = 0
y = y+temp1+temp2
chg = diff(c(y,y[1]))
}
#Solution vector:
y
Thank you,
Kevin
Kevin Ummel
CARMA Project Manager
Center for Global Development
[[alternative HTML
Take vector x and a subset y:
x=1:10
y=c(4,5,7,9)
For each value in 'x', I want to know how many elements in 'y' are less than
'x'.
An example would be:
sapply(x,FUN=function(i) {length(which(yi))})
[1] 0 0 0 0 1 2 2 3 3 4
But this solution is far too slow when x and y have lengths in the
expansion' approach was the fastest by about a factor of 8.
Cheers,
Kevin
On Jan 11, 2011, at 9:20 AM, Petr Savicky wrote:
Dear Kevin Ummel:
There were several suggestions on R-help concering your question below.
None of them suggests a base function. I would also expect that
there is a base
Hi everyone,
I'm looking for a way to 'explode' a matrix like this:
matrix(1:4,2,2)
[,1] [,2]
[1,]13
[2,]24
into a matrix like this:
matrix(c(1,1,2,2,1,1,2,2,3,3,4,4,3,3,4,4),4,4)
[,1] [,2] [,3] [,4]
[1,]1133
[2,]1133
[3,]22
(seq(ncol(m)), each = 2)]
On Wed, Jan 5, 2011 at 10:03 AM, Kevin Ummel kevinum...@gmail.com wrote:
Hi everyone,
I'm looking for a way to 'explode' a matrix like this:
matrix(1:4,2,2)
[,1] [,2]
[1,]13
[2,]24
into a matrix like this:
matrix(c
Two posts in one day is not a good day...and this question seems like it should
have an obvious answer:
I have a matrix where rows are unique combinations of 1's and 0's:
combs=as.matrix(expand.grid(c(0,1),c(0,1)))
combs
Var1 Var2
[1,]00
[2,]10
[3,]01
[4,]1
A bit embarrassed to post this seemingly trivial question, but I can't find
anything in the archive that's quite relevant:
a1=1
a2=2
obs=objects(pattern=glob2rx(a?))
I want to utilize 'obs' as a function argument to produce something like:
sum(a1,a2)
Obviously, sum(obs) doesn't work, but
' (potentially very long).
Cheers,
Kevin
On Nov 16, 2010, at 5:07 PM, David Winsemius wrote:
On Nov 16, 2010, at 10:34 AM, Kevin Ummel wrote:
A bit embarrassed to post this seemingly trivial question, but I can't find
anything in the archive that's quite relevant:
a1=1
a2=2
obs=objects
Thanks, David. That does, indeed, work. It didn't occur to me that a list would
do the job as an argument.
Thanks for the fix!
kevin
On Nov 16, 2010, at 6:58 PM, David Winsemius wrote:
On Nov 16, 2010, at 1:04 PM, Kevin Ummel wrote:
Sorry, I shouldn't have used 'sum' as an example; I
' issue for the R-Sig-Geo list...
On Nov 16, 2010, at 7:09 PM, Kevin Ummel wrote:
Thanks, David. That does, indeed, work. It didn't occur to me that a list
would do the job as an argument.
Thanks for the fix!
kevin
On Nov 16, 2010, at 6:58 PM, David Winsemius wrote:
On Nov 16, 2010
on top of nearby country/state borders, etc.
Many thanks,
Kevin Ummel
Central European University
Department of Environmental Science and Policy
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PLEASE do read the posting
.
Many thanks in advance,
Kevin Ummel
Central European University
Department of Environmental Science and Policy
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