Could someone confirm for me whether DistanceMatrix is behaving differently in V11?

In V10 I could return the signed differenced between two vectors using


However, in V11 the returned values are all positive. Using an undefined function f for DistanceFunction reveals that DistanceMatrix is adding the Abs internally in V11 thus prohibiting signed returns:

DistanceMatrix[Range[2], Range[3], DistanceFunction -> f]

V11 output

{{Abs[f[1, 1]], Abs[f[1, 2]], Abs[f[1, 3]]}, {Abs[f[2, 1]], Abs[f[2, 2]], Abs[f[2, 3]]}}

V10.3 output

{{f[{1}, {1}], f[{1}, {2}], f[{1}, {3}], f[{2}, {1}], f[{2}, {2}], f[{2}, {3}]}}

I feel like including Abs by default isn't very helpful as I could always add it in if I wanted it!

Thus I have two questions:

  1. Is there any way to remove it? I
  2. Is there a smarter way to get the signed differences of two lists - I know I can use Outer but have been using DistanceMatrix following the discussion here

2 Answers 2


This is a bit of a hack, and I am not sure how it will impact performance, but it goes around the "improvement" put in place in v.11:

 DistanceMatrix[Range[2], Range[3], DistanceFunction -> HoldForm[Subtract]] /. Abs[a_] :> a

(* Out: {{0, -1, -2}, {1, 0, -1}} *)

Compare to the built-in in v.11:

DistanceMatrix[Range[2], Range[3], DistanceFunction -> Subtract]

(* Out: {{0, 1, 2}, {1, 0, 1}} *)
  • $\begingroup$ A similar hack: Block[{Abs = Identity}, DistanceMatrix[Range[2], Range[3], DistanceFunction -> Subtract] ] $\endgroup$
    – Jason B.
    Nov 15, 2016 at 22:37
  • $\begingroup$ @JasonB Your suggestion is arguably more elegant as well. $\endgroup$
    – MarcoB
    Nov 16, 2016 at 4:57
  • $\begingroup$ ReleaseHold and Block both great ways to get around the issue, thanks. ReleaseHold suffers a performance hit whereas Block does not. $\endgroup$ Nov 16, 2016 at 12:58

Outer is pretty fast when used with Plus so I daresay Outer[Plus, u, -v] will be competitive:

u = Range[1000];
v = Range[2000];

 a = DistanceMatrix[u, v, DistanceFunction -> Subtract];]
(* {0.499, Null} *)

 b = Outer[Plus, u, -v];]
(* {0.00918, Null} *)

Abs[b] == a
(* True *)
  • $\begingroup$ +1 for pointing out that Outer[Plus, u, -v] is orders of magnitude faster than Outer[Subtract, u, v]. I like the methods of circumventing the Abs issue in MarcoB's answer but the Outer Plus route is by far the fastest! And what I'll use form now on! $\endgroup$ Nov 16, 2016 at 13:00

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