How to generate a matrix with certain conditions












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!










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










  • $begingroup$
    Same question posted here.
    $endgroup$
    – Rohit Namjoshi
    6 hours ago










  • $begingroup$
    tiffany, please go here to get your accounts merged, so you can easily access your question.
    $endgroup$
    – J. M. is computer-less
    2 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
    13 hours ago










  • $begingroup$
    Same question posted here.
    $endgroup$
    – Rohit Namjoshi
    6 hours ago










  • $begingroup$
    tiffany, please go here to get your accounts merged, so you can easily access your question.
    $endgroup$
    – J. M. is computer-less
    2 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 8 hours 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 13 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
    13 hours ago










  • $begingroup$
    Same question posted here.
    $endgroup$
    – Rohit Namjoshi
    6 hours ago










  • $begingroup$
    tiffany, please go here to get your accounts merged, so you can easily access your question.
    $endgroup$
    – J. M. is computer-less
    2 hours ago


















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










  • $begingroup$
    Same question posted here.
    $endgroup$
    – Rohit Namjoshi
    6 hours ago










  • $begingroup$
    tiffany, please go here to get your accounts merged, so you can easily access your question.
    $endgroup$
    – J. M. is computer-less
    2 hours ago
















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




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












$begingroup$
Same question posted here.
$endgroup$
– Rohit Namjoshi
6 hours ago




$begingroup$
Same question posted here.
$endgroup$
– Rohit Namjoshi
6 hours ago












$begingroup$
tiffany, please go here to get your accounts merged, so you can easily access your question.
$endgroup$
– J. M. is computer-less
2 hours ago




$begingroup$
tiffany, please go here to get your accounts merged, so you can easily access your question.
$endgroup$
– J. M. is computer-less
2 hours ago










4 Answers
4






active

oldest

votes


















5












$begingroup$

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


$$begin{pmatrix}
0&1&1&1&1\
1&0&1&1&1\
1&1&0&1&1\
1&0&1&0&1\
0&1&1&1&0end{pmatrix}$$



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



It's easy enough to make this into a function:



makeMat[n_, m_] := (mat = RandomVariate[BernoulliDistribution[m], {n, n}]; 
mat - DiagonalMatrix[Diagonal[mat]])


Then the above example is makeMat[5, 0.9]






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
    11 hours ago






  • 2




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



















4












$begingroup$

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
    8 hours 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
    12 hours ago










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






  • 2




    $begingroup$
    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
    8 hours ago



















0












$begingroup$

Since all the simple answers have been given, here is a SparseArray solution that may be useful if you want to generate large matrices without storing unneeded zero entries:



tiffany[n_Integer?Positive, m_] :=
SparseArray[{j_, k_} /; j != k && RandomReal < 1/m :> 1, {n, n}]


As an example:



BlockRandom[SeedRandom["tiffany"]; tiffany[7, 2.5] // Normal]
{{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1},
{0, 0, 0, 1, 0, 1, 1}, {0, 0, 0, 0, 1, 0, 0},
{0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 1, 0}}





share|improve this answer









$endgroup$













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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    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


    $$begin{pmatrix}
    0&1&1&1&1\
    1&0&1&1&1\
    1&1&0&1&1\
    1&0&1&0&1\
    0&1&1&1&0end{pmatrix}$$



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



    It's easy enough to make this into a function:



    makeMat[n_, m_] := (mat = RandomVariate[BernoulliDistribution[m], {n, n}]; 
    mat - DiagonalMatrix[Diagonal[mat]])


    Then the above example is makeMat[5, 0.9]






    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
      11 hours ago






    • 2




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
















    5












    $begingroup$

    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


    $$begin{pmatrix}
    0&1&1&1&1\
    1&0&1&1&1\
    1&1&0&1&1\
    1&0&1&0&1\
    0&1&1&1&0end{pmatrix}$$



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



    It's easy enough to make this into a function:



    makeMat[n_, m_] := (mat = RandomVariate[BernoulliDistribution[m], {n, n}]; 
    mat - DiagonalMatrix[Diagonal[mat]])


    Then the above example is makeMat[5, 0.9]






    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
      11 hours ago






    • 2




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














    5












    5








    5





    $begingroup$

    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


    $$begin{pmatrix}
    0&1&1&1&1\
    1&0&1&1&1\
    1&1&0&1&1\
    1&0&1&0&1\
    0&1&1&1&0end{pmatrix}$$



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



    It's easy enough to make this into a function:



    makeMat[n_, m_] := (mat = RandomVariate[BernoulliDistribution[m], {n, n}]; 
    mat - DiagonalMatrix[Diagonal[mat]])


    Then the above example is makeMat[5, 0.9]






    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


    $$begin{pmatrix}
    0&1&1&1&1\
    1&0&1&1&1\
    1&1&0&1&1\
    1&0&1&0&1\
    0&1&1&1&0end{pmatrix}$$



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



    It's easy enough to make this into a function:



    makeMat[n_, m_] := (mat = RandomVariate[BernoulliDistribution[m], {n, n}]; 
    mat - DiagonalMatrix[Diagonal[mat]])


    Then the above example is makeMat[5, 0.9]







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 2 hours ago









    J. M. is computer-less

    97k10303463




    97k10303463










    answered 12 hours ago









    bill sbill s

    53.7k376153




    53.7k376153












    • $begingroup$
      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
      11 hours ago






    • 2




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






    • 2




      $begingroup$
      @tiffany, just do With[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
      $endgroup$
      – J. M. is computer-less
      8 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
    11 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
    11 hours ago




    2




    2




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




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











    4












    $begingroup$

    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
      8 hours ago
















    4












    $begingroup$

    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
      8 hours ago














    4












    4








    4





    $begingroup$

    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 11 hours ago









    Okkes DulgerciOkkes Dulgerci

    5,0991917




    5,0991917












    • $begingroup$
      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
      8 hours 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
      8 hours 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
    8 hours 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
    8 hours 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
      12 hours ago










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






    • 2




      $begingroup$
      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
      8 hours 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
      12 hours ago










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






    • 2




      $begingroup$
      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
      8 hours 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 13 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
      12 hours ago










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






    • 2




      $begingroup$
      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
      8 hours ago


















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










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






    • 2




      $begingroup$
      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
      8 hours ago
















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




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












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




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




    2




    2




    $begingroup$
    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
    8 hours 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
    8 hours ago











    0












    $begingroup$

    Since all the simple answers have been given, here is a SparseArray solution that may be useful if you want to generate large matrices without storing unneeded zero entries:



    tiffany[n_Integer?Positive, m_] :=
    SparseArray[{j_, k_} /; j != k && RandomReal < 1/m :> 1, {n, n}]


    As an example:



    BlockRandom[SeedRandom["tiffany"]; tiffany[7, 2.5] // Normal]
    {{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1},
    {0, 0, 0, 1, 0, 1, 1}, {0, 0, 0, 0, 1, 0, 0},
    {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 1, 0}}





    share|improve this answer









    $endgroup$


















      0












      $begingroup$

      Since all the simple answers have been given, here is a SparseArray solution that may be useful if you want to generate large matrices without storing unneeded zero entries:



      tiffany[n_Integer?Positive, m_] :=
      SparseArray[{j_, k_} /; j != k && RandomReal < 1/m :> 1, {n, n}]


      As an example:



      BlockRandom[SeedRandom["tiffany"]; tiffany[7, 2.5] // Normal]
      {{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1},
      {0, 0, 0, 1, 0, 1, 1}, {0, 0, 0, 0, 1, 0, 0},
      {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 1, 0}}





      share|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Since all the simple answers have been given, here is a SparseArray solution that may be useful if you want to generate large matrices without storing unneeded zero entries:



        tiffany[n_Integer?Positive, m_] :=
        SparseArray[{j_, k_} /; j != k && RandomReal < 1/m :> 1, {n, n}]


        As an example:



        BlockRandom[SeedRandom["tiffany"]; tiffany[7, 2.5] // Normal]
        {{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1},
        {0, 0, 0, 1, 0, 1, 1}, {0, 0, 0, 0, 1, 0, 0},
        {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 1, 0}}





        share|improve this answer









        $endgroup$



        Since all the simple answers have been given, here is a SparseArray solution that may be useful if you want to generate large matrices without storing unneeded zero entries:



        tiffany[n_Integer?Positive, m_] :=
        SparseArray[{j_, k_} /; j != k && RandomReal < 1/m :> 1, {n, n}]


        As an example:



        BlockRandom[SeedRandom["tiffany"]; tiffany[7, 2.5] // Normal]
        {{0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1},
        {0, 0, 0, 1, 0, 1, 1}, {0, 0, 0, 0, 1, 0, 0},
        {0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 0, 1, 0, 1}, {0, 0, 0, 0, 0, 1, 0}}






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 2 hours ago









        J. M. is computer-lessJ. M. is computer-less

        97k10303463




        97k10303463






















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










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