Using reduce() with tuples created by zip()

[bearophile]

Craig Dillabaugh:

> 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!


import std.stdio, std.algorithm, std.range, std.math, 
std.functional;

alias sum(T) = curry!(reduce!q{a + b}, cast(T)0);

double euclidDistance(double[] xs, double[] ys)
in {
     assert(xs.length == ys.length);
} body {
     return zip(xs, ys)
            .map!(xy => (xy[0] - xy[1]) ^^ 2)
            .sum!double
            .sqrt;
}

void main() {
     auto xs = [0.0,  1.0, 0.1];
     auto ys = [1.0, -1.0, 0.0];
     euclidDistance(xs, ys).writeln;
}


Bye,
bearophile