Optimizing Chi-square distance between two distribution

I want to program the chi-square between two distribution Xu and Vu with length N. This function is defined as below: Xu and Vu are real and positive.

chiDistance[U_, V_] := Module[{vecs, vselect}, (
(*delete the {0,0} occurrence to avoid the division by zero*)
vecs = Join[Transpose[{U}], Transpose[{V}], 2];
vselect = Select[vecs, # != {0., 0.} &];
(1/2) Total[(vselect[[All, 1]] -
vselect[[All, 2]])^2/(vselect[[All, 1]] +
vselect[[All, 2]])]

)]


This function take

0.047625 when N=40000

When I replaced Select[] by DeleteCases[], I got the same complexity. I am not sure whether the intention is vectors with all positive entries. If not then potential pairs (a,-a) will also be division by zero.

Here is another implementation of formula, removing zero denominators:

cd[u_, v_] := Module[{pos, us, vs},
pos = Position[u + v, _?(# != 0 &)];
us = Extract[u, pos];
vs = Extract[v, pos];
Total[(us - vs)^2/(us + vs)]/2]


or exploiting SparseArray properties:

cdsa[u_, v_] := Module[{pos, us, vs},
pos = SparseArray[u + v]["NonzeroPositions"];
us = Extract[u, pos];
vs = Extract[v, pos];
Total[(us - vs)^2/(us + vs)]/2]


Appears to confer only small advantage:

Needs["GeneralUtilities"]
BenchmarkPlot[{cdsa @@ # &,
chiDistance @@ # &}, {RandomInteger[{0, 10}, #],
RandomInteger[{0, 10}, #]} &, {100, 1000, 10000, 40000},
"IncludeFits" -> True] • using Total[N@(us - vs)^2/(us + vs)]/2 in the last line of cdsa` seems to make it much faster (+1). – kglr Nov 8 '14 at 9:17
• @kguler thank you, I should have tested on reals but done in haste... – ubpdqn Nov 8 '14 at 9:22
• @ubpdqn, I edited my question. the two vectors are real and positive. So, I think that the use of Total[N@(us - vs)^2/(us + vs)]/2 does not change the complexity. – BetterEnglish Nov 8 '14 at 15:34