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In the following matrix m every 0 should be replaced with a 1:

m = {{0,1,2},{5,0,3},{8,0,0}}

Desired result:

m' = {{1,1,2},{5,1,3},{8,1,1}}

What is the fastest way to do this for a matrix with 200-1000 elements?

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5 Answers 5

13
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m + 1 - Unitize[m]

might be faster because it preserves packed arrays, but we'd need a real test.

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5
  • $\begingroup$ You likely need Unitize in place of UnitStep. $\endgroup$ Oct 24, 2013 at 15:09
  • $\begingroup$ This is wrong if the matrix contains negative integers. Like @LeonidShifrin said, Unitize is the correct function here. $\endgroup$
    – RunnyKine
    Oct 24, 2013 at 15:10
  • $\begingroup$ @Leonid Shifrin and RunnyKine - thanks. Indeed, I was taking the example quite literally -- Unitize is more general and does handle negative numbers. $\endgroup$
    – bill s
    Oct 24, 2013 at 15:13
  • $\begingroup$ It's definitely fast so +1. $\endgroup$
    – RunnyKine
    Oct 24, 2013 at 15:16
  • $\begingroup$ For me this is the fastest $\endgroup$
    – Danvil
    Oct 26, 2013 at 15:42
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Just for fun: the following hack is even slightly faster than the solution using Unitize and vectorization, on really large matrices:

replaceZeros[m_?MatrixQ] := 
   Normal[
      SparseArray[m] /. HoldPattern[SparseArray[s___]] :>
          Module[{parts = {s}},
            parts[[3]] = 1;
            SparseArray @@ parts
          ]
   ];

but, in this forms at least, it explicitly uses the fact that elements being replaced are zeros. This is just for fun, in any case.

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  • $\begingroup$ Unitize is twice as fast on my PC for 10^7 element matrix. $\endgroup$
    – RunnyKine
    Oct 24, 2013 at 15:36
  • $\begingroup$ @RunnyKine I used also 10^7 elements, and on my machine (Mac OS X 10.7.5 64 bit), Unitize is about 10-20 % slower. Which version of Mathematica you used? $\endgroup$ Oct 24, 2013 at 15:40
  • $\begingroup$ Perhaps you 2 used different proportion of zeros? If lots of zeros probably Leonid's wins, and viceversa $\endgroup$
    – Rojo
    Oct 24, 2013 at 15:46
  • $\begingroup$ I'm using V9.0.1 Windows 8.1 64bit $\endgroup$
    – RunnyKine
    Oct 24, 2013 at 15:46
  • $\begingroup$ @Rojo Yes, I already noticed that too. But even if the zeros are very scarce, on my machine my code then is just the same speed as the one with Unitize. $\endgroup$ Oct 24, 2013 at 15:47
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m = {{0, 1, 2}, {5, 0, 3}, {8, 0, 0}} /. 0 -> 1
{{1, 1, 2}, {5, 1, 3}, {8, 1, 1}}

It's certainly the fastest to write down.

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3
  • $\begingroup$ Why //. instead of /.? $\endgroup$ Oct 24, 2013 at 14:56
  • $\begingroup$ @Timothy Wofford -- habit, I guess. /. is shorter. Rojo -- habit again. I"m always nervous when a white space means something though, as in: /. 0->1 is OK but /.0->1 is not. $\endgroup$
    – bill s
    Oct 24, 2013 at 14:58
  • $\begingroup$ For me /. is 25% faster than //. $\endgroup$
    – Danvil
    Oct 24, 2013 at 15:02
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Replace[m, 0 -> 1, Infinity]

{{1, 1, 2}, {5, 1, 3}, {8, 1, 1}}

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0
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{{0, 1, 2}, {5, 0, 3}, {8, 0, 0}} //. {a___, 0, b___} -> {a, 1, b}

{{1, 1, 2}, {5, 1, 3}, {8, 1, 1}}

For,

RandomInteger[0, {100000, 3}] //. {a___, 0, b___} -> {a, 1, b};

2.62 Sec

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2
  • $\begingroup$ Try with RandomInteger[{0, 5}, {10^7, 3}] $\endgroup$
    – RunnyKine
    Oct 24, 2013 at 17:13
  • $\begingroup$ @RunnyKine : ya you seem to be right, upto 10^6 its taking 16 sec, after that sort of bottleneck. $\endgroup$ Oct 24, 2013 at 17:28

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