I think reduce takes two arguments: the growing seed and the current front.
You're trying to get (a[0]-b[0])^^2 + (a[1]-b[1])^^2 + ..., right?
Try this:
module euclid;
import std.stdio;
double euclid_dist( double[] pt1, double[] pt2 ) {
assert( pt1.length == pt2.length );
import std.algorithm;
import std.math;
import std.range;
return sqrt(0.0.reduce!( (sum,pair) => sum + (pair[0]-pair[1])^^2
)(zip(pt1, pt2)));
}
void main()
{
double[] arr1 = [0.0, 1.0, 0.1];
double[] arr2 = [1.0, -1.0, 0.0];
writeln(euclid_dist(arr1,arr2));
}
On Thu, Oct 31, 2013 at 8:12 PM, Craig Dillabaugh <
cdill...@cg.scs.carleton.ca> wrote:
> I am trying to calculate the Euclidean Distance between two
> points. I currently have the following function (that doesn't
> compile):
>
> double euclid_dist( double[] pt1, double[] pt2 ) {
> assert( pt1.length == pt2.length );
>
> return sqrt(
> zip(pt1, pt2).reduce!(function(e) { return
> (e[1]-e[0])*(e[1]-e[0]); })(0.0)
> );
> }
>
> Hopefully it is clear what I am trying to do. I want zip to
> create a range that is a tuple of the point's coordinate pairs,
> and then use reduce to sum the squared differences. I get the
> following error but can't figure out how to fix my syntax:
>
> euclid.d(13): Error: template
> euclid.euclid_dist.reduce!(__**funcliteral2).reduce does not match
> any function template declaration. Candidates are:
> /usr/include/dmd/phobos/std/**algorithm.d(688):
> euclid.euclid_dist.reduce!(__**funcliteral2).reduce(Args...)(**Args
> args) if (Args.length > 0 && Args.length <= 2 &&
> isIterable!(Args[__dollar - 1]))
> euclid.d(13): Error: template
> euclid.euclid_dist.reduce!(__**funcliteral2).reduce(Args...)(**Args
> args) if (Args.length > 0 && Args.length <= 2 &&
> isIterable!(Args[__dollar - 1])) cannot deduce template function
> from argument types !()(Zip!(double[], double[]), double)
> Failed: 'dmd' '-v' '-o-' 'euclid.d' '-I.'
>
>
> I know I could use a simple foreach loop with my zip command, or
> even std.numeric.euclidean( ... ), but I am trying to be cute
> here!
>
> Cheers,
> Craig
>
