Optimise d3.mean.

Originally we were using Welford’s algorithm, but this is primarily
useful when computing the variance in a numerically stable manner, since
Welford’s approach requires an incremental mean.

I’ve removed a test for the mean of more than one instance of
Number.MAX_VALUE as this is unlikely to occur in practice; most likely
this was the reason I used Welford’s algorithm in the first place.

There’s a paper [1] comparing various algorithms for computing the mean,
and Welford’s is actually slightly less accurate than the naïve
approach.  There are some more accurate approaches but I think it’s
overkill for d3.mean.

[1] Youngs, Edward A., and Elliot M. Cramer. "Some results relevant to
choice of sum and sum-of-product algorithms." Technometrics 13.3 (1971):
657-665.

Related: d3#1842.

d3.js

@@ -91,13 +91,13 @@
     return x != null && !isNaN(x);
   }
   d3.mean = function(array, f) {
-    var n = array.length, a, m = 0, i = -1, j = 0;
+    var s = 0, n = array.length, a, i = -1, j = n;
     if (arguments.length === 1) {
-      while (++i < n) if (d3_number(a = array[i])) m += (a - m) / ++j;
+      while (++i < n) if (d3_number(a = array[i])) s += a; else --j;
     } else {
-      while (++i < n) if (d3_number(a = f.call(array, array[i], i))) m += (a - m) / ++j;
+      while (++i < n) if (d3_number(a = f.call(array, array[i], i))) s += a; else --j;
     }
-    return j ? m : undefined;
+    return j ? s / j : undefined;
   };
   d3.quantile = function(values, p) {
     var H = (values.length - 1) * p + 1, h = Math.floor(H), v = +values[h - 1], e = H - h;

src/arrays/mean.js

@@ -1,15 +1,15 @@
 import "../math/number";
 
 d3.mean = function(array, f) {
-  var n = array.length,
+  var s = 0,
+      n = array.length,
       a,
-      m = 0,
       i = -1,
-      j = 0;
+      j = n;
   if (arguments.length === 1) {
-    while (++i < n) if (d3_number(a = array[i])) m += (a - m) / ++j;
+    while (++i < n) if (d3_number(a = array[i])) s += a; else --j;
   } else {
-    while (++i < n) if (d3_number(a = f.call(array, array[i], i))) m += (a - m) / ++j;
+    while (++i < n) if (d3_number(a = f.call(array, array[i], i))) s += a; else --j;
   }
-  return j ? m : undefined;
+  return j ? s / j : undefined;
 };

test/arrays/mean-test.js

@@ -18,10 +18,6 @@ suite.addBatch({
       assert.equal(mean([1, 2, 3, 4, 5, NaN]), 3);
       assert.equal(mean([10, null, 3, undefined, 5, NaN]), 6);
     },
-    "can handle large numbers without overflowing": function(mean) {
-      assert.equal(mean([Number.MAX_VALUE, Number.MAX_VALUE]), Number.MAX_VALUE);
-      assert.equal(mean([-Number.MAX_VALUE, -Number.MAX_VALUE]), -Number.MAX_VALUE);
-    },
     "returns undefined for empty array": function(mean) {
       assert.isUndefined(mean([]));
       assert.isUndefined(mean([null]));