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
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5
m + 1 - Unitize[m]
might be faster because it preserves packed arrays, but we'd need a real test.
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$\begingroup$ You likely need
Unitizein place ofUnitStep. $\endgroup$ Commented Oct 24, 2013 at 15:09 -
$\begingroup$ This is wrong if the matrix contains negative integers. Like @LeonidShifrin said,
Unitizeis the correct function here. $\endgroup$ Commented Oct 24, 2013 at 15:10 -
$\begingroup$ @Leonid Shifrin and RunnyKine - thanks. Indeed, I was taking the example quite literally --
Unitizeis more general and does handle negative numbers. $\endgroup$– bill sCommented Oct 24, 2013 at 15:13 -
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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.
answered Oct 24, 2013 at 15:33
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Unitizeis twice as fast on my PC for10^7element matrix. $\endgroup$ Commented 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),
Unitizeis about 10-20 % slower. Which version of Mathematica you used? $\endgroup$ Commented 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 $\endgroup$– RojoCommented Oct 24, 2013 at 15:46
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$\begingroup$ I'm using V9.0.1 Windows 8.1 64bit $\endgroup$ Commented Oct 24, 2013 at 15:46
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$\begingroup$ @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. $\endgroup$ Commented 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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$\begingroup$ @Timothy Wofford -- habit, I guess.
/.is shorter. Rojo -- habit again. I"m always nervous when a white space means something though, as in:/. 0->1is OK but/.0->1is not. $\endgroup$– bill sCommented Oct 24, 2013 at 14:58 -
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Replace[m, 0 -> 1, Infinity]
{{1, 1, 2}, {5, 1, 3}, {8, 1, 1}}
answered Oct 24, 2013 at 15:00
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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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$\begingroup$ Try with
RandomInteger[{0, 5}, {10^7, 3}]$\endgroup$ Commented 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. $\endgroup$ Commented Oct 24, 2013 at 17:28