On 30/07/2008, at 6:12 AM, dxc13 wrote:
useR's,
I am trying trying to find out if there is a faster way to do a
certain
computation. I have successfully used FOR loops and the apply
function to
do this, but it can take some time to fully compute, but I was
wondering if
anyone may know of a different function or way to do this:
x
[1] 1 2 3 4 5
xk
[1] 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
I want to do: abs(x-xk[i]) where i = 1 to length(xk)=13. It should
result
in a 13 by 5 matrix or data frame. Does anyone have a "quicker"
solution
than FOR loops or apply()?
outer(xk,x,function(a,b){abs(a-b)})
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