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How to generate a matrix with certain conditions

Standardizing a coset table via matrix manipulationEigenvalues and Determinant of a large matrixHow to generate a random matrix with specific parameters?Generate Non-Singular Matrix of $ntimes n$ dimensionHow to create a matrix with some conditions?Problems with RandomChoicePicking first row of a matrix with a positive first elementSimplest way to construct a matrix its elements are defined by known functionsmatrix with chosen elements distributed in a random positionUsing Table to build a matrix

1

$begingroup$

I want to generate a $n times n$ matrix.

1. I want the diagonal entries to be all 0


2. I want a random choice of matrix elements with 0 or 1.


3. The probability of having a 1 as a matrix element is $1/m$ and the probability of having a 0 as a matrix element is $1-1/m$.

I used the following command but it is wrong.

A[n_, m_] :=Table[If[i == j, 0,RandomVariate[BernoulliDistribution[m],{n,n}]]]

And I tried to test this command with n=4, m=0.4 but it didn't work.

Could anyone kindly tell me how to do this please?
Thank you!

matrix probability-or-statistics random sampling

share|improve this question

edited 1 hour ago

J. M. is computer-less♦

97k10303463

asked 6 hours ago

tiffanytiffany

61

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For a start, your code seems to set the diagonal elements to 1 rather than zero.
  $endgroup$
  – MarcoB
  6 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

I want to generate a $n times n$ matrix.

1. I want the diagonal entries to be all 0


2. I want a random choice of matrix elements with 0 or 1.


3. The probability of having a 1 as a matrix element is $1/m$ and the probability of having a 0 as a matrix element is $1-1/m$.

I used the following command but it is wrong.

A[n_, m_] :=Table[If[i == j, 0,RandomVariate[BernoulliDistribution[m],{n,n}]]]

And I tried to test this command with n=4, m=0.4 but it didn't work.

Could anyone kindly tell me how to do this please?
Thank you!

matrix probability-or-statistics random sampling

share|improve this question

edited 1 hour ago

J. M. is computer-less♦

97k10303463

asked 6 hours ago

tiffanytiffany

61

New contributor

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

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For a start, your code seems to set the diagonal elements to 1 rather than zero.
  $endgroup$
  – MarcoB
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I want to generate a $n times n$ matrix.

1. I want the diagonal entries to be all 0


2. I want a random choice of matrix elements with 0 or 1.


3. The probability of having a 1 as a matrix element is $1/m$ and the probability of having a 0 as a matrix element is $1-1/m$.

I used the following command but it is wrong.

A[n_, m_] :=Table[If[i == j, 0,RandomVariate[BernoulliDistribution[m],{n,n}]]]

And I tried to test this command with n=4, m=0.4 but it didn't work.

Could anyone kindly tell me how to do this please?
Thank you!

matrix probability-or-statistics random sampling

share|improve this question

edited 1 hour ago

J. M. is computer-less♦

97k10303463

asked 6 hours ago

tiffanytiffany

61

New contributor

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

$endgroup$

A[n_, m_] :=Table[If[i == j, 0,RandomVariate[BernoulliDistribution[m],{n,n}]]]

share|improve this question

edited 1 hour ago

J. M. is computer-less♦

97k10303463

asked 6 hours ago

tiffanytiffany

61

New contributor

tiffany 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 1 hour ago

J. M. is computer-less♦

97k10303463

asked 6 hours ago

tiffanytiffany

61

New contributor

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

edited 1 hour ago

J. M. is computer-less♦

97k10303463

edited 1 hour ago

J. M. is computer-less♦

97k10303463

asked 6 hours ago

tiffanytiffany

61

New contributor

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

asked 6 hours ago

tiffanytiffany

61

3 Answers
3

active

oldest

votes

3

$begingroup$

Binary random variables are often modeled using the BernoulliDistribution. You can use the function RandomVariate to get a matrix of such variables.

mat = RandomVariate[BernoulliDistribution[0.9], {5, 5}]; 

mat - DiagonalMatrix[Diagonal[mat]]

%//MatrixForm

You can change the 0.9 to any value (this is your m). The second line sets all the diagonal elements to zero.

share|improve this answer

answered 6 hours ago

bill sbill s

53.6k376153

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you bill s How will the command be if I do not restrict what n and m be? For example, I would like to generate a set of commands which I can replace m and n easily by any number.
  $endgroup$
  – tiffany
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany, just do With[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

You can use weight option in RandomChoice

n = 5;

m = 2;

mat = RandomChoice[{1/m, 1 - 1/m} -> {1, 0}, {n, n}];

(mat - DiagonalMatrix[Diagonal@mat]) // MatrixForm

$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$

share|improve this answer

answered 4 hours ago

Okkes DulgerciOkkes Dulgerci

5,0891917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Altho using the BernoulliDistribution[] is best, this is likely to be more easy to read for someone who is not accustomed to discrete probability distributions.
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

0

$begingroup$

m = 2;

n = 5;

rnd[x_] := If[RandomReal[{0, 1}] < 1/m, 1, 0];

t = Table[rnd[x_]*(1 - KroneckerDelta[i, j]), {i, 1, n}, {j, 1, n}];

t // MatrixForm

share|improve this answer

answered 6 hours ago

Vsevolod A.Vsevolod A.

478211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you Vsevolod A. How will the command be if I do not restrict what n and m be?
  $endgroup$
  – tiffany
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany you set m and n in the first two lines...
  $endgroup$
  – Vsevolod A.
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just rnd := If[RandomReal[{0, 1}] < 1/m, 1, 0]; will do, since the function never uses its argument.
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

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3

$begingroup$

Binary random variables are often modeled using the BernoulliDistribution. You can use the function RandomVariate to get a matrix of such variables.

mat = RandomVariate[BernoulliDistribution[0.9], {5, 5}]; 

mat - DiagonalMatrix[Diagonal[mat]]

%//MatrixForm

You can change the 0.9 to any value (this is your m). The second line sets all the diagonal elements to zero.

share|improve this answer

answered 6 hours ago

bill sbill s

53.6k376153

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you bill s How will the command be if I do not restrict what n and m be? For example, I would like to generate a set of commands which I can replace m and n easily by any number.
  $endgroup$
  – tiffany
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany, just do With[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Binary random variables are often modeled using the BernoulliDistribution. You can use the function RandomVariate to get a matrix of such variables.

mat = RandomVariate[BernoulliDistribution[0.9], {5, 5}]; 

mat - DiagonalMatrix[Diagonal[mat]]

%//MatrixForm

You can change the 0.9 to any value (this is your m). The second line sets all the diagonal elements to zero.

share|improve this answer

answered 6 hours ago

bill sbill s

53.6k376153

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you bill s How will the command be if I do not restrict what n and m be? For example, I would like to generate a set of commands which I can replace m and n easily by any number.
  $endgroup$
  – tiffany
  5 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany, just do With[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

Binary random variables are often modeled using the BernoulliDistribution. You can use the function RandomVariate to get a matrix of such variables.

mat = RandomVariate[BernoulliDistribution[0.9], {5, 5}]; 

mat - DiagonalMatrix[Diagonal[mat]]

%//MatrixForm

You can change the 0.9 to any value (this is your m). The second line sets all the diagonal elements to zero.

share|improve this answer

answered 6 hours ago

bill sbill s

53.6k376153

$endgroup$

mat = RandomVariate[BernoulliDistribution[0.9], {5, 5}]; 

mat - DiagonalMatrix[Diagonal[mat]]

%//MatrixForm

share|improve this answer

answered 6 hours ago

bill sbill s

53.6k376153

answered 6 hours ago

bill sbill s

53.6k376153

answered 6 hours ago

bill sbill s

53.6k376153

3

$begingroup$

You can use weight option in RandomChoice

n = 5;

m = 2;

mat = RandomChoice[{1/m, 1 - 1/m} -> {1, 0}, {n, n}];

(mat - DiagonalMatrix[Diagonal@mat]) // MatrixForm

$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$

share|improve this answer

answered 4 hours ago

Okkes DulgerciOkkes Dulgerci

5,0891917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Altho using the BernoulliDistribution[] is best, this is likely to be more easy to read for someone who is not accustomed to discrete probability distributions.
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

You can use weight option in RandomChoice

n = 5;

m = 2;

mat = RandomChoice[{1/m, 1 - 1/m} -> {1, 0}, {n, n}];

(mat - DiagonalMatrix[Diagonal@mat]) // MatrixForm

$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$

share|improve this answer

answered 4 hours ago

Okkes DulgerciOkkes Dulgerci

5,0891917

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Altho using the BernoulliDistribution[] is best, this is likely to be more easy to read for someone who is not accustomed to discrete probability distributions.
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

You can use weight option in RandomChoice

n = 5;

m = 2;

mat = RandomChoice[{1/m, 1 - 1/m} -> {1, 0}, {n, n}];

(mat - DiagonalMatrix[Diagonal@mat]) // MatrixForm

$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$

share|improve this answer

answered 4 hours ago

Okkes DulgerciOkkes Dulgerci

5,0891917

$endgroup$

n = 5;

m = 2;

mat = RandomChoice[{1/m, 1 - 1/m} -> {1, 0}, {n, n}];

(mat - DiagonalMatrix[Diagonal@mat]) // MatrixForm

share|improve this answer

answered 4 hours ago

Okkes DulgerciOkkes Dulgerci

5,0891917

answered 4 hours ago

Okkes DulgerciOkkes Dulgerci

5,0891917

answered 4 hours ago

Okkes DulgerciOkkes Dulgerci

5,0891917

0

$begingroup$

m = 2;

n = 5;

rnd[x_] := If[RandomReal[{0, 1}] < 1/m, 1, 0];

t = Table[rnd[x_]*(1 - KroneckerDelta[i, j]), {i, 1, n}, {j, 1, n}];

t // MatrixForm

share|improve this answer

answered 6 hours ago

Vsevolod A.Vsevolod A.

478211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you Vsevolod A. How will the command be if I do not restrict what n and m be?
  $endgroup$
  – tiffany
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany you set m and n in the first two lines...
  $endgroup$
  – Vsevolod A.
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just rnd := If[RandomReal[{0, 1}] < 1/m, 1, 0]; will do, since the function never uses its argument.
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

0

$begingroup$

m = 2;

n = 5;

rnd[x_] := If[RandomReal[{0, 1}] < 1/m, 1, 0];

t = Table[rnd[x_]*(1 - KroneckerDelta[i, j]), {i, 1, n}, {j, 1, n}];

t // MatrixForm

share|improve this answer

answered 6 hours ago

Vsevolod A.Vsevolod A.

478211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you Vsevolod A. How will the command be if I do not restrict what n and m be?
  $endgroup$
  – tiffany
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @tiffany you set m and n in the first two lines...
  $endgroup$
  – Vsevolod A.
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Just rnd := If[RandomReal[{0, 1}] < 1/m, 1, 0]; will do, since the function never uses its argument.
  $endgroup$
  – J. M. is computer-less♦
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

m = 2;

n = 5;

rnd[x_] := If[RandomReal[{0, 1}] < 1/m, 1, 0];

t = Table[rnd[x_]*(1 - KroneckerDelta[i, j]), {i, 1, n}, {j, 1, n}];

t // MatrixForm

share|improve this answer

answered 6 hours ago

Vsevolod A.Vsevolod A.

478211

$endgroup$

m = 2;

n = 5;

rnd[x_] := If[RandomReal[{0, 1}] < 1/m, 1, 0];

t = Table[rnd[x_]*(1 - KroneckerDelta[i, j]), {i, 1, n}, {j, 1, n}];

t // MatrixForm

share|improve this answer

answered 6 hours ago

Vsevolod A.Vsevolod A.

478211

answered 6 hours ago

Vsevolod A.Vsevolod A.

478211

answered 6 hours ago

Vsevolod A.Vsevolod A.

478211