2
$\begingroup$

I want to generate triplets of random integers over where the integers are in the range -9 to 9 (inclusive) but excluding zero. How can I do that?

I have the following table.

Table[a -> RandomInteger[{-9, 9}, 3], {50}] // Column

{a -> {0, 9, 6}},
{a -> {7, -3, 1}},
{a -> {-8, -2, -6}},
{a -> {9, 0, 2}},
{a -> {-2, 5, 6}},
{a -> {0, 4, -2}},
{a -> {8, -5, -1}},
{a -> {3, -4, -9}},
{a -> {-8, 7, 6}},
{a -> {-9, 8, 5}},
{a -> {2, -1, 9}},
{a -> {7, 0, 2}},
{a -> {5, -2, 0}},
{a -> {1, -9, -6}},
{a -> {0, -8, 0}},
{a -> {-9, -9, -2}},
{a -> {9, -2, -3}},
{a -> {6, -1, -5}},
{a -> {-3, 2, 1}},
{a -> {-7, 3, 0}},
{a -> {8, 4, -5}},
{a -> {-9, -1, 0}},
{a -> {1, 0, -4}},
{a -> {0, -4, 3}},
{a -> {-1, -9, -6}},
{a -> {9, 8, -3}},
{a -> {6, -2, 4}},
{a -> {2, 3, 8}},
{a -> {-9, 1, -4}},
{a -> {8, -1, -9}},
{a -> {-7, 7, 9}},
{a -> {-8, -7, 2}},
{a -> {2, 3, -7}},
{a -> {-2, 9, 5}},
{a -> {9, 1, 0}},
{a -> {8, 0, -6}},
{a -> {-4, -8, 1}},
{a -> {8, -5, 6}},
{a -> {-9, -4, -9}},
{a -> {9, 7, -8}},
{a -> {9, 8, -5}},
{a -> {9, 5, 5}},
{a -> {-4, -5, -4}},
{a -> {-3, 2, 0}},
{a -> {-9, -9, 0}},
{a -> {-7, 1, -4}},
{a -> {7, 6, -5}},
{a -> {-4, 1, -6}},
{a -> {-5, -9, 0}},
{a -> {-6, 2, -6}}

I don't want the zeros that appear above. I have used DeleteCasesand Replace with no results,

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ How about Table[a -> Table[(-1)^RandomInteger[{1, 2}] RandomInteger[{1, 9}], {3}], {50}] // Column $\endgroup$
    – user46676
    May 27, 2017 at 22:15
  • $\begingroup$ @jose check the solution i posted below if you are interested in using DeleteCases and Replace $\endgroup$
    – Ali Hashmi
    May 28, 2017 at 12:31
  • $\begingroup$ How to remove repeated elements $\endgroup$
    – jose
    May 28, 2017 at 15:17

5 Answers 5

Reset to default
3
$\begingroup$

try

Table[a -> RandomChoice[Join[-Range[9], Range[9]], 3], 50] // Column
Improve this answer
$\endgroup$
5
  • $\begingroup$ RandomSample takes 3 distinct choices. It should be RandomChoice. $\endgroup$
    – Szabolcs
    May 27, 2017 at 22:36
  • $\begingroup$ Thanks, I'm left with this solution v / s the random $\endgroup$
    – jose
    May 27, 2017 at 22:54
  • $\begingroup$ How to remove repeated elements $\endgroup$
    – jose
    May 28, 2017 at 15:17
  • $\begingroup$ you mean DeleteDuplicates@ ? $\endgroup$
    – ZaMoC
    May 28, 2017 at 15:22
  • $\begingroup$ I tried with DeleteDuplicates but the list is reduced to 3 to 2 elements I need it to be 3 elements $\endgroup$
    – jose
    May 28, 2017 at 18:27
3
$\begingroup$

Replace 

RandomInteger[{-9, 9}, 3]

with

RandomChoice[Delete[Range[-9, 9], 10], 3]

Update

The OP asks in a comment below that the selected triples be constrained so that no element triples is duplicated. I would enforce that constraint with a recursive helper command.

With[{items = Delete[Range[-9, 9], 10]},
  helper :=
    Module[{triple},
      triple = DeleteDuplicates @ RandomSample[items, 3];
      If[Length[triple] == 3, triple, helper]]]

SeedRandom[42]; Column[rules = Table[a -> helper, 50]]

a -> {2, 9, -9}
a -> {5, -3, -6}
a -> {9, 7, 8}
...
a -> {4, -2, -6}
a -> {-1, -7, 9}
a -> {-7, 9, -5}

The following proves the above result contains no triple with duplicate elements.

AllTrue[rules, Length[Union[#[[2]]]] === 3 &]

True

Improve this answer
$\endgroup$
6
  • 2
    $\begingroup$ RandomChoice[{-1, 1}, 3] RandomInteger[9, 3] should work, too. $\endgroup$
    – J. M.'s lack of A.I. ♦
    May 27, 2017 at 22:13
  • 5
    $\begingroup$ @J.M. That should be RandomInteger[{1,9}, 3] (I also thought it starts at 1, but it turns out it starts at 0.) $\endgroup$
    – Szabolcs
    May 27, 2017 at 22:35
  • $\begingroup$ How to remove repeated elements $\endgroup$
    – jose
    May 28, 2017 at 15:18
  • $\begingroup$ Instead of Delete referring to a position, I'd probably use either Cases[Range[-9, 9], Except[0]] or DeleteCases[Range[-9, 9], 0]. $\endgroup$
    – kirma
    Jun 3, 2017 at 14:48
  • $\begingroup$ @kirma. Seems obvious to me that in a sequence ranging from -n to n, the zero term is always the n + 1-th element. Why not use that fact? $\endgroup$
    – m_goldberg
    Jun 3, 2017 at 18:48
2
$\begingroup$

If you are interested in distinct triples, you might want to generate the entire list of possibilities and then use RandomSample:

samples = RandomSample[Tuples[DeleteCases[Range[-9,9],0], 3], 50];
Thread[a -> samples] //Short

{a->{-1,3,5},a->{5,-9,-7},a->{8,6,-6},<<44>>,a->{-9,-8,-1},a->{2,-1,9},a->{8,-9,-6}}

Improve this answer
$\endgroup$
0
$\begingroup$ 
randomSigns[] := RandomChoice[{1, -1}, {3}]
randomNumber[n_] := RandomInteger[{1, 9}, {n, 3}]
a -> randomSigns[] # & /@ randomNumber[50]
Improve this answer
$\endgroup$
0
$\begingroup$

This will delete all entries with zeros:

t = Table[a -> RandomInteger[{-9, 9}, 3], {50}];
DeleteCases[t, HoldPattern[a -> p : {__}] /; MemberQ[p, 0]]

also this will replace entries where zeros are present:

replacements = Range[-9, -1]~Join~Range[1, 9];
Replace[t, HoldPattern[a -> {OrderlessPatternSequence[(p : 0 ..), (t : Except[0] ..)]}] :> 
a -> {t}~Join~RandomChoice[replacements, Length@{p}], {1}]

Demo:

t // Short
(* {a -> {8, -2, -1}, a -> {8, 6, 0}, a -> {9, 5, -4}, a -> {9, 0, 0}, 
a -> {-7, 0, 4}, a -> {8, -5, -6}, a -> {4, 7, 2}, a -> {0, 3, 2}, <<35>>,
a -> {-5, -1, -6}, a -> {5, 7, -5}, a -> {-2, -3, 3}, a -> {-4, 0, 4},
a -> {1, 7, 3}, a -> {0, 9, -1}, a -> {-7, -4, -4}} *)

DeleteCases[t, HoldPattern[a -> p : {__}] /; MemberQ[p, 0]]//Short
(* {a -> {8, -2, -1}, a -> {9, 5, -4}, a -> {8, -5, -6}, a -> {4, 7, 2}, 
a -> {8, 5, -7}, a -> {5, 2, 3}, a -> {-5, 9, -5}, a -> {1, 5, -8}, <<22>>,
a -> {-9, -6, -7}, a -> {-4, 3, -9}, a -> {-5, -1, -6}, 
a -> {5, 7, -5}, a -> {-2, -3, 3}, a -> {1, 7, 3}, a -> {-7, -4, -4}} *)

(* if we need to keep the length of the list and replace entries where zero(s) exist *)

replacements = Range[-9, -1]~Join~Range[1, 9];
Replace[t, HoldPattern[a -> {OrderlessPatternSequence[(p : 0 ..),(t : Except[0] ..)]}] :>
a -> {t}~Join~ RandomChoice[replacements, Length@{p}], {1} ] // Short

(* {a -> {8, -2, -1}, a -> {6, 8, 2}, a -> {9, 5, -4}, a -> {9, -6, 3},
a -> {-7, 4, 8}, a -> {8, -5, -6}, a -> {4, 7, 2}, a -> {2, 3, -5},
<<35>>, a -> {-5, -1, -6}, a -> {5, 7, -5}, a -> {-2, -3, 3}, 
a -> {-4, 4, 5}, a -> {1, 7, 3}, a -> {-1, 9, -8}, a -> {-7, -4, -4}} *)
Improve this answer
$\endgroup$
3
  • $\begingroup$ How to remove repeated elements, thanks for help $\endgroup$
    – jose
    May 28, 2017 at 15:18
  • $\begingroup$ @jose sorry i am not able to understand. The code i posted in my solution either removes zeros or replaces them if found. What do you mean by repeated elements ?? $\endgroup$
    – Ali Hashmi
    May 28, 2017 at 15:53
  • $\begingroup$ @jose if what you mean is removing identical entries then you would need to apply DeleteDuplicates on the above results $\endgroup$
    – Ali Hashmi
    May 28, 2017 at 15:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.