Say we have three elements, 0,1,2 and we want to generate 1000 random vectors of length 7 with these elements, i.e.
{2,1,0,0,0,0,2}
How can we do this without duplicates among these vectors?
$\begingroup$
$\endgroup$
3
2 Answers
$\begingroup$
$\endgroup$
3
First, there are $3^7=2187$ different 7-tuples, so there is indeed a considerable probability to encounter duplicates if we generate each 7-tuple independently. So, we use the following:
vectors=IntegerDigits[#, 3, 7] & /@ RandomSample[Range[3^7] - 1, 1000]
-
$\begingroup$ Can your answer be modified to work in case that the elements are not 3 successive numbers but a list to choose from? i.e. instead 0,1,2 to have {0,3,6} $\endgroup$John Matis– John Matis2021-01-14 10:50:14 +00:00Commented Jan 14, 2021 at 10:50
-
$\begingroup$ {0,3,6} is not so much difference, just multiply
vectorsby3. More general{a,b,c}would requirevectors/.{0->a,1->b,2->c}.$\endgroup$Roma Lee– Roma Lee2021-01-14 11:17:04 +00:00Commented Jan 14, 2021 at 11:17 -
$\begingroup$ If the sample universe is really large, it would be better to use Span instead of Range to avoid memory issues, e.g.,
RandomSample[Span[0, 3^7-1], 1000]instead. $\endgroup$Carl Woll– Carl Woll2021-01-14 16:32:08 +00:00Commented Jan 14, 2021 at 16:32
$\begingroup$
$\endgroup$
Here are two similar methods of generating a pseudorandom list of unique vectors. The first method uses Union. It is a little faster than the second, which uses DeleteDuplicates. The two methods are implemented as functions f and g
Clear[f]
f[nvect_, ndims_, list_] := Module[{a = {}, k},
If[TrueQ[nvect <= Length[list]^ndims],
While[(k = Length[a]) < nvect,
a = Union[a, RandomChoice[list, {nvect - k, ndims}]]
]];
RandomSample[a]]
Clear[g]
g[nvect_, ndims_, list_] := Module[{a = {}, k},
If[TrueQ[nvect <= Length[list]^ndims],
While[(k = Length[a]) < nvect,
a = DeleteDuplicates@Join[a, RandomChoice[list, {nvect - k, ndims}]]
]]; a]
Since Union returns an ordered array, function f uses RandomSample to return a pseudorandom permutation of the ordered array. Both functions return an empty set if you ask for an impossibly large number of unique vectors.
Timing tests on f and g give, for 1000 unique vectors,
tf = RepeatedTiming[f[1000, 7, {0, 1, 2}], 5];
tf // First
(* 0.002 *)
tg = RepeatedTiming[g[1000, 7, {0, 1, 2}], 5];
tg // First
(* 0.003 *)
TuplesandRandomSample. $\endgroup$RandomSample[Tuples[{0, 1, 2}, 7], 1000]$\endgroup$