2
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What I want to do is be able to round all numbers below 0 to -1 and all other number to 1. So that no matter what the input is it becomes either 1 or -1.

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3
  • $\begingroup$ should an exact zero input yield 1? $\endgroup$
    – george2079
    Oct 21, 2016 at 1:17
  • $\begingroup$ It doesn't matter, I'm never rounding 0 $\endgroup$
    – John Smith
    Oct 21, 2016 at 1:29
  • $\begingroup$ "It doesn't matter, I'm never rounding 0" — Well, then you just want the Sign of the numbers, no? I didn't suggest it because Sign[0] == 0 in Mathematica. $\endgroup$
    – Szabolcs
    Oct 21, 2016 at 23:16

4 Answers 4

9
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Use

2 UnitStep[x] - 1

where x can be a number or any array of numbers.


If you never need to handle 0, use Sign.

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9
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Because the OP wrote "I'm never rounding 0", this will work fine:

Sign @ {-2, 0, 5}

{-1, 0, 1}


Issue with Sign[0] = 0:

sign[x_] := If[x == 0, 1, Sign[x]]

or

sign[x_] := Piecewise[{{Sign[x], x != 0}, {1, x == 0}}]

then

sign /@ {-2, 0, 5} (* not Listable though *)

{-1, 1, 1}


Usually not recommended:

Unprotect[Sign];
Sign[0] := 1
Protect[Sign];

Sign @ {-2, 0, 5}

{-1, 1, 1}

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3
  • $\begingroup$ (+1) But Sign is Listable... You can save a character! $\endgroup$
    – kale
    Oct 21, 2016 at 2:36
  • $\begingroup$ Sign[0] gives the wrong result, tho... $\endgroup$
    – Michael E2
    Oct 21, 2016 at 9:33
  • $\begingroup$ Ah, I hate it when someone can't write what they mean. After all, "So that no matter what the input is it becomes either 1 or -1" is pretty darn clear. $\endgroup$
    – Michael E2
    Oct 21, 2016 at 22:58
4
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Like this?

             MyRounding[x_] = If[x < 0, -1, 1];

Examples:

             MyRounding[5]
             MyRounding[-5]

give +1 and -1. Note that this rounds zero to 1.

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x/Abs[x]

Conveniently doesn't work for $x=0$.

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