I want to determine whether I have regularly sampled data or not, a là RegularlySampledQ. The data itself might be huge. I tried a bunch of different approaches and the best I could come up with was this:

regSampQ[d_, window_: 100] :=
  If[Length@d > window,
   Module[{eq = True},
     If[! Equal @@ Differences[d[[n - window ;; n]]], 
       eq = False; Break[]
      ], {n, window + 1, Length@d, window}];
   Equal @@ Differences[d]

It turns out this is a somewhat efficient way to do this, beating out my other best attempt, which was just

regSampNaive = Function[Null, Equal @@ Differences[#], HoldFirst];

The problem is that the data can be incredibly large. Here're some test cases I had:

test1 = RandomReal[{}, 50000];

regSampNaive@test1 // RepeatedTiming

{0.0037, False}

regSampQ@test1 // RepeatedTiming

{0.000097, False}

test2 = ConstantArray[0, 50000];

regSampNaive@test2 // RepeatedTiming

{0.0025, True}

regSampQ@test2 // RepeatedTiming

{0.0026, True}

test3 = Join[ConstantArray[0, 25000], ConstantArray[1, 25000]];

regSampNaive@test3 // RepeatedTiming

{0.0023, False}

regSampQ@test3 // RepeatedTiming

{0.0014, False}

But I feel like there has to be a really clean, fast way to do this... am I wrong?

  • 1
    $\begingroup$ On my machine (OS X V11.3) Length@DeleteDuplicates@Differences[d[[n - window ;; n]]] != 1 cuts the timings in half for True $\endgroup$ Oct 18, 2018 at 8:10
  • $\begingroup$ @MikeHoneychurch oh wow you're right. That's...embarrassing for Equal. But a very nice optimization. It cuts all my timings by a small factor at least. Actually, thinking about it the overhead might be in Apply instead. It might just be a data-copying issue. $\endgroup$
    – b3m2a1
    Oct 18, 2018 at 8:14
  • $\begingroup$ See if your method is faster if you widen the window to e.g. 300. I don't recall exactly whether Apply gets autocompiled but if it does it would be for 200-250+ list length. ps. DeleteDuplicates was a first "easy" thing that came to mind but there are probably other ways if you got creative $\endgroup$ Oct 18, 2018 at 23:13
  • $\begingroup$ @MikeHoneychurch compilation didn’t actually help when I used it procedurally. The data transfer step is the killer. $\endgroup$
    – b3m2a1
    Oct 18, 2018 at 23:35
  • $\begingroup$ I'm talking about autocompilation. mathematica.stackexchange.com/questions/311/… $\endgroup$ Oct 18, 2018 at 23:47


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