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How to generate random points without duplication?

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2

1

$begingroup$

In generating a list of random 2D integral coordinates e.g. u1,v1, u2,v2, u3,v3,..., how can one write a condition specifying that no point is repeated, i.e no ui,vi == uj,vj?

random

share|improve this question

edited 7 hours ago

user64494

3,90111323

asked 8 hours ago

spacedspaced

283

New contributor

spaced is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ui, vi are real numbers?
  $endgroup$
  – kglr
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Integers from 0 to 20, so it happens frequently.
  $endgroup$
  – spaced
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To add more detail to the question - the points are being generated by an optimization routine (NMaximize); I think the only way I can control for duplicates is by specifying a constraint on the ui, vi being generated.
  $endgroup$
  – spaced
  7 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

In generating a list of random 2D integral coordinates e.g. u1,v1, u2,v2, u3,v3,..., how can one write a condition specifying that no point is repeated, i.e no ui,vi == uj,vj?

random

share|improve this question

edited 7 hours ago

user64494

3,90111323

asked 8 hours ago

spacedspaced

283

New contributor

spaced is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ui, vi are real numbers?
  $endgroup$
  – kglr
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Integers from 0 to 20, so it happens frequently.
  $endgroup$
  – spaced
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  To add more detail to the question - the points are being generated by an optimization routine (NMaximize); I think the only way I can control for duplicates is by specifying a constraint on the ui, vi being generated.
  $endgroup$
  – spaced
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

In generating a list of random 2D integral coordinates e.g. u1,v1, u2,v2, u3,v3,..., how can one write a condition specifying that no point is repeated, i.e no ui,vi == uj,vj?

random

share|improve this question

edited 7 hours ago

user64494

3,90111323

asked 8 hours ago

spacedspaced

283

New contributor

spaced is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 7 hours ago

user64494

3,90111323

asked 8 hours ago

spacedspaced

283

New contributor

spaced is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 7 hours ago

user64494

3,90111323

asked 8 hours ago

spacedspaced

283

New contributor

spaced is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited 7 hours ago

user64494

3,90111323

edited 7 hours ago

user64494

3,90111323

asked 8 hours ago

spacedspaced

283

New contributor

spaced is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

spacedspaced

283

4 Answers
4

active

oldest

votes

4

$begingroup$

RandomSample[Tuples[Range[0, 20], 2], 10]

14, 13, 2, 17, 12, 7, 1, 11, 6, 6, 12, 4, 13, 14, 2,
16, 15, 6, 1, 0

Note: The length of the sample cannot exceed 21^2 (441).

share|improve this answer

answered 7 hours ago

kglrkglr

195k10216439

$endgroup$

add a comment | 

2

$begingroup$

Since you mentioned that your coordinates are drawn from integers in the regions 20 >= u >= 0 and 20 >= v >= 0, you can first generate a one-dimensional vector containing all lattice points:

V = Flatten[Table[x[i,j],i,0,20,j,0,20]];

Then you can select a random sample from this vector, e.g.:

RandomSample[V,5]/.x->List

16, 8, 10, 17, 8, 5, 0, 17, 16, 9

Since V has no duplicates, the random sample will also contain no duplicates.

share|improve this answer

answered 7 hours ago

KagaratschKagaratsch

5,16941351

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or even like this: RandomSample[Flatten[Table[x, y, x, 0, 20, y, 20], 1], 5].
  $endgroup$
  – kirma
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or... RandomSample[x, y /. Solve[(x | y) [Element] Integers, x, y [Element] Rectangle[0, 0, 20, 20]], 10]
  $endgroup$
  – kirma
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

For 10 points, oversample and select unique ones using RandomSample.

points = RandomInteger[20, 50, 2];
points = RandomSample[points, 10]

Actually, RandomInteger[20, 50, 2] appear to produce 50 unique 2D points.

E.g.

tries = 0;
points = RandomInteger[20, 50, 2];
While[(sample = Quiet[Check[RandomSample[points, 50], True]]),
 tries++;]
tries

0

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Chris DegnenChris Degnen

22.4k23787

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This doesn't guarantee success. A better approach is probably something like RandomSample[Range[0,20],10], right?
  $endgroup$
  – AccidentalFourierTransform
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @spaced I'm not sure I understand your comment, but I feel that it changes the question considerably. It would be better if you were more specific in what you are trying to accomplish (and watch out for XY problems). Also, please edit any extra information directly into the OP rather than in a comment on an answer. Thanks!
  $endgroup$
  – AccidentalFourierTransform
  7 hours ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

It kind of depends on what you want to do next. You only need to generate one such list, and only once? Then efficiency is not important, and so you can proceed very naively as follows: generate random lists until one works:

Module[l = RandomInteger[0, 20, 10, 2],
 While[Length[l] != Length[DeleteDuplicates[l]], 
 l = RandomInteger[0, 20, 10, 2]];
 l
]

On the other hand, if you need many of these lists, then a more efficient approach is as follows: first generate all possible pairs, and pick a random sample ever time you need one such list:

lists = Tuples[Range[0, 20], 2];
sample := RandomSample[lists,10];

Now every time you execute sample you will get one such list, picked at random.

share|improve this answer

answered 7 hours ago

AccidentalFourierTransformAccidentalFourierTransform

5,55511142

$endgroup$

add a comment | 

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4

$begingroup$

RandomSample[Tuples[Range[0, 20], 2], 10]

14, 13, 2, 17, 12, 7, 1, 11, 6, 6, 12, 4, 13, 14, 2,
16, 15, 6, 1, 0

Note: The length of the sample cannot exceed 21^2 (441).

share|improve this answer

answered 7 hours ago

kglrkglr

195k10216439

$endgroup$

add a comment | 

4

$begingroup$

RandomSample[Tuples[Range[0, 20], 2], 10]

14, 13, 2, 17, 12, 7, 1, 11, 6, 6, 12, 4, 13, 14, 2,
16, 15, 6, 1, 0

Note: The length of the sample cannot exceed 21^2 (441).

share|improve this answer

answered 7 hours ago

kglrkglr

195k10216439

$endgroup$

add a comment | 

$begingroup$

RandomSample[Tuples[Range[0, 20], 2], 10]

14, 13, 2, 17, 12, 7, 1, 11, 6, 6, 12, 4, 13, 14, 2,
16, 15, 6, 1, 0

Note: The length of the sample cannot exceed 21^2 (441).

share|improve this answer

answered 7 hours ago

kglrkglr

195k10216439

$endgroup$

RandomSample[Tuples[Range[0, 20], 2], 10]

share|improve this answer

answered 7 hours ago

kglrkglr

195k10216439

answered 7 hours ago

kglrkglr

195k10216439

answered 7 hours ago

kglrkglr

195k10216439

2

$begingroup$

Since you mentioned that your coordinates are drawn from integers in the regions 20 >= u >= 0 and 20 >= v >= 0, you can first generate a one-dimensional vector containing all lattice points:

V = Flatten[Table[x[i,j],i,0,20,j,0,20]];

Then you can select a random sample from this vector, e.g.:

RandomSample[V,5]/.x->List

16, 8, 10, 17, 8, 5, 0, 17, 16, 9

Since V has no duplicates, the random sample will also contain no duplicates.

share|improve this answer

answered 7 hours ago

KagaratschKagaratsch

5,16941351

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or even like this: RandomSample[Flatten[Table[x, y, x, 0, 20, y, 20], 1], 5].
  $endgroup$
  – kirma
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or... RandomSample[x, y /. Solve[(x | y) [Element] Integers, x, y [Element] Rectangle[0, 0, 20, 20]], 10]
  $endgroup$
  – kirma
  7 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Since you mentioned that your coordinates are drawn from integers in the regions 20 >= u >= 0 and 20 >= v >= 0, you can first generate a one-dimensional vector containing all lattice points:

V = Flatten[Table[x[i,j],i,0,20,j,0,20]];

Then you can select a random sample from this vector, e.g.:

RandomSample[V,5]/.x->List

16, 8, 10, 17, 8, 5, 0, 17, 16, 9

Since V has no duplicates, the random sample will also contain no duplicates.

share|improve this answer

answered 7 hours ago

KagaratschKagaratsch

5,16941351

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Or even like this: RandomSample[Flatten[Table[x, y, x, 0, 20, y, 20], 1], 5].
  $endgroup$
  – kirma
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or... RandomSample[x, y /. Solve[(x | y) [Element] Integers, x, y [Element] Rectangle[0, 0, 20, 20]], 10]
  $endgroup$
  – kirma
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Since you mentioned that your coordinates are drawn from integers in the regions 20 >= u >= 0 and 20 >= v >= 0, you can first generate a one-dimensional vector containing all lattice points:

V = Flatten[Table[x[i,j],i,0,20,j,0,20]];

Then you can select a random sample from this vector, e.g.:

RandomSample[V,5]/.x->List

16, 8, 10, 17, 8, 5, 0, 17, 16, 9

Since V has no duplicates, the random sample will also contain no duplicates.

share|improve this answer

answered 7 hours ago

KagaratschKagaratsch

5,16941351

$endgroup$

V = Flatten[Table[x[i,j],i,0,20,j,0,20]];

RandomSample[V,5]/.x->List

share|improve this answer

answered 7 hours ago

KagaratschKagaratsch

5,16941351

answered 7 hours ago

KagaratschKagaratsch

5,16941351

answered 7 hours ago

KagaratschKagaratsch

5,16941351

2

$begingroup$

For 10 points, oversample and select unique ones using RandomSample.

points = RandomInteger[20, 50, 2];
points = RandomSample[points, 10]

Actually, RandomInteger[20, 50, 2] appear to produce 50 unique 2D points.

E.g.

tries = 0;
points = RandomInteger[20, 50, 2];
While[(sample = Quiet[Check[RandomSample[points, 50], True]]),
 tries++;]
tries

0

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Chris DegnenChris Degnen

22.4k23787

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This doesn't guarantee success. A better approach is probably something like RandomSample[Range[0,20],10], right?
  $endgroup$
  – AccidentalFourierTransform
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @spaced I'm not sure I understand your comment, but I feel that it changes the question considerably. It would be better if you were more specific in what you are trying to accomplish (and watch out for XY problems). Also, please edit any extra information directly into the OP rather than in a comment on an answer. Thanks!
  $endgroup$
  – AccidentalFourierTransform
  7 hours ago
  
  

  
  
  
  

add a comment | 

2

$begingroup$

For 10 points, oversample and select unique ones using RandomSample.

points = RandomInteger[20, 50, 2];
points = RandomSample[points, 10]

Actually, RandomInteger[20, 50, 2] appear to produce 50 unique 2D points.

E.g.

tries = 0;
points = RandomInteger[20, 50, 2];
While[(sample = Quiet[Check[RandomSample[points, 50], True]]),
 tries++;]
tries

0

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Chris DegnenChris Degnen

22.4k23787

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This doesn't guarantee success. A better approach is probably something like RandomSample[Range[0,20],10], right?
  $endgroup$
  – AccidentalFourierTransform
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @spaced I'm not sure I understand your comment, but I feel that it changes the question considerably. It would be better if you were more specific in what you are trying to accomplish (and watch out for XY problems). Also, please edit any extra information directly into the OP rather than in a comment on an answer. Thanks!
  $endgroup$
  – AccidentalFourierTransform
  7 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

For 10 points, oversample and select unique ones using RandomSample.

points = RandomInteger[20, 50, 2];
points = RandomSample[points, 10]

Actually, RandomInteger[20, 50, 2] appear to produce 50 unique 2D points.

E.g.

tries = 0;
points = RandomInteger[20, 50, 2];
While[(sample = Quiet[Check[RandomSample[points, 50], True]]),
 tries++;]
tries

0

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Chris DegnenChris Degnen

22.4k23787

$endgroup$

points = RandomInteger[20, 50, 2];
points = RandomSample[points, 10]

tries = 0;
points = RandomInteger[20, 50, 2];
While[(sample = Quiet[Check[RandomSample[points, 50], True]]),
 tries++;]
tries

share|improve this answer

edited 7 hours ago

answered 8 hours ago

Chris DegnenChris Degnen

22.4k23787

answered 8 hours ago

Chris DegnenChris Degnen

22.4k23787

answered 8 hours ago

Chris DegnenChris Degnen

22.4k23787

1

$begingroup$

It kind of depends on what you want to do next. You only need to generate one such list, and only once? Then efficiency is not important, and so you can proceed very naively as follows: generate random lists until one works:

Module[l = RandomInteger[0, 20, 10, 2],
 While[Length[l] != Length[DeleteDuplicates[l]], 
 l = RandomInteger[0, 20, 10, 2]];
 l
]

On the other hand, if you need many of these lists, then a more efficient approach is as follows: first generate all possible pairs, and pick a random sample ever time you need one such list:

lists = Tuples[Range[0, 20], 2];
sample := RandomSample[lists,10];

Now every time you execute sample you will get one such list, picked at random.

share|improve this answer

answered 7 hours ago

AccidentalFourierTransformAccidentalFourierTransform

5,55511142

$endgroup$

add a comment | 

1

$begingroup$

It kind of depends on what you want to do next. You only need to generate one such list, and only once? Then efficiency is not important, and so you can proceed very naively as follows: generate random lists until one works:

Module[l = RandomInteger[0, 20, 10, 2],
 While[Length[l] != Length[DeleteDuplicates[l]], 
 l = RandomInteger[0, 20, 10, 2]];
 l
]

On the other hand, if you need many of these lists, then a more efficient approach is as follows: first generate all possible pairs, and pick a random sample ever time you need one such list:

lists = Tuples[Range[0, 20], 2];
sample := RandomSample[lists,10];

Now every time you execute sample you will get one such list, picked at random.

share|improve this answer

answered 7 hours ago

AccidentalFourierTransformAccidentalFourierTransform

5,55511142

$endgroup$

add a comment | 

$begingroup$

It kind of depends on what you want to do next. You only need to generate one such list, and only once? Then efficiency is not important, and so you can proceed very naively as follows: generate random lists until one works:

Module[l = RandomInteger[0, 20, 10, 2],
 While[Length[l] != Length[DeleteDuplicates[l]], 
 l = RandomInteger[0, 20, 10, 2]];
 l
]

On the other hand, if you need many of these lists, then a more efficient approach is as follows: first generate all possible pairs, and pick a random sample ever time you need one such list:

lists = Tuples[Range[0, 20], 2];
sample := RandomSample[lists,10];

Now every time you execute sample you will get one such list, picked at random.

share|improve this answer

answered 7 hours ago

AccidentalFourierTransformAccidentalFourierTransform

5,55511142

$endgroup$

Module[l = RandomInteger[0, 20, 10, 2],
 While[Length[l] != Length[DeleteDuplicates[l]], 
 l = RandomInteger[0, 20, 10, 2]];
 l
]

lists = Tuples[Range[0, 20], 2];
sample := RandomSample[lists,10];

share|improve this answer

answered 7 hours ago

AccidentalFourierTransformAccidentalFourierTransform

5,55511142

answered 7 hours ago

AccidentalFourierTransformAccidentalFourierTransform

5,55511142

answered 7 hours ago

AccidentalFourierTransformAccidentalFourierTransform

5,55511142