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How to generate a matrix with certain conditions

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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!










share|improve this question









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
















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!










share|improve this question









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














1












1








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!










share|improve this question









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.







$endgroup$




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









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









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




share|improve this question








edited 1 hour ago









J. M. is computer-less

97k10303463




97k10303463






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




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.





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.






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












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


















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
















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




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










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


enter image description here



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









$endgroup$













  • $begingroup$
    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



















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









$endgroup$













  • $begingroup$
    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



















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









$endgroup$













  • $begingroup$
    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











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






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


enter image description here



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









$endgroup$













  • $begingroup$
    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
















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


enter image description here



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









$endgroup$













  • $begingroup$
    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














3












3








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


enter image description here



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









$endgroup$



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


enter image description here



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












share|improve this answer



share|improve this answer










answered 6 hours ago









bill sbill s

53.6k376153




53.6k376153












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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




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




$begingroup$
@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











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









$endgroup$













  • $begingroup$
    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
















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









$endgroup$













  • $begingroup$
    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














3












3








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









$endgroup$



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












share|improve this answer



share|improve this answer










answered 4 hours ago









Okkes DulgerciOkkes Dulgerci

5,0891917




5,0891917












  • $begingroup$
    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


















  • $begingroup$
    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
















$begingroup$
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




$begingroup$
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











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









$endgroup$













  • $begingroup$
    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
















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









$endgroup$













  • $begingroup$
    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














0












0








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









$endgroup$



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












share|improve this answer



share|improve this answer










answered 6 hours ago









Vsevolod A.Vsevolod A.

478211




478211












  • $begingroup$
    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


















  • $begingroup$
    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
















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




$begingroup$
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




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




1




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




$begingroup$
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










tiffany is a new contributor. Be nice, and check out our Code of Conduct.










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