How to generate a matrix with certain conditions
1
$begingroup$
I want to generate a $n times n$ matrix.
- I want the diagonal entries to be all 0
- I want a random choice of matrix elements with 0 or 1.
- 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
edited 9 hours ago
J. M. is computer-less♦
97k10303463
asked 14 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.
$endgroup$
add a comment |
1
$begingroup$
I want to generate a $n times n$ matrix.
- I want the diagonal entries to be all 0
- I want a random choice of matrix elements with 0 or 1.
- 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
edited 9 hours ago
J. M. is computer-less♦
97k10303463
asked 14 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
14 hours ago
$begingroup$
Same question posted here.
$endgroup$
– Rohit Namjoshi
7 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
add a comment |
1
1
1
$begingroup$
I want to generate a $n times n$ matrix.
- I want the diagonal entries to be all 0
- I want a random choice of matrix elements with 0 or 1.
- 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
edited 9 hours ago
J. M. is computer-less♦
97k10303463
asked 14 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.
$endgroup$
I want to generate a $n times n$ matrix.
- I want the diagonal entries to be all 0
- I want a random choice of matrix elements with 0 or 1.
- 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
matrix probability-or-statistics random sampling
edited 9 hours ago
J. M. is computer-less♦
97k10303463
asked 14 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 9 hours ago
J. M. is computer-less♦
97k10303463
asked 14 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 9 hours ago
J. M. is computer-less♦
97k10303463
edited 9 hours ago
J. M. is computer-less♦
97k10303463
edited 9 hours ago
J. M. is computer-less♦
97k10303463
97k10303463
asked 14 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 14 hours ago
tiffanytiffany
61
asked 14 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
14 hours ago
$begingroup$
Same question posted here.
$endgroup$
– Rohit Namjoshi
7 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
add a comment |
$begingroup$
For a start, your code seems to set the diagonal elements to 1 rather than zero.
$endgroup$
– MarcoB
14 hours ago
$begingroup$
Same question posted here.
$endgroup$
– Rohit Namjoshi
7 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
14 hours ago
$begingroup$
For a start, your code seems to set the diagonal elements to 1 rather than zero.
$endgroup$
– MarcoB
14 hours ago
$begingroup$
Same question posted here.
$endgroup$
– Rohit Namjoshi
7 hours ago
$begingroup$
Same question posted here.
$endgroup$
– Rohit Namjoshi
7 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
add a comment |
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]
edited 3 hours ago
J. M. is computer-less♦
97k10303463
answered 13 hours ago
bill sbill s
53.7k376153
$endgroup$
$begingroup$
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
12 hours ago
2
$begingroup$
@tiffany, just doWith[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
$endgroup$
– J. M. is computer-less♦
9 hours ago
add a comment |
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
$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$
answered 12 hours ago
Okkes DulgerciOkkes Dulgerci
5,0991917
$endgroup$
$begingroup$
Altho using theBernoulliDistributionis 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♦
9 hours 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
answered 13 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
13 hours ago
$begingroup$
@tiffany you setmandnin the first two lines...
$endgroup$
– Vsevolod A.
12 hours ago
2
$begingroup$
Justrnd := If[RandomReal[{0, 1}] < 1/m, 1, 0];will do, since the function never uses its argument.
$endgroup$
– J. M. is computer-less♦
9 hours ago
add a comment |
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}}
answered 3 hours ago
J. M. is computer-less♦J. M. is computer-less
97k10303463
$endgroup$
add a comment |
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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]
edited 3 hours ago
J. M. is computer-less♦
97k10303463
answered 13 hours ago
bill sbill s
53.7k376153
$endgroup$
$begingroup$
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
12 hours ago
2
$begingroup$
@tiffany, just doWith[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
$endgroup$
– J. M. is computer-less♦
9 hours ago
add a comment |
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]
edited 3 hours ago
J. M. is computer-less♦
97k10303463
answered 13 hours ago
bill sbill s
53.7k376153
$endgroup$
$begingroup$
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
12 hours ago
2
$begingroup$
@tiffany, just doWith[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
$endgroup$
– J. M. is computer-less♦
9 hours ago
add a comment |
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]
edited 3 hours ago
J. M. is computer-less♦
97k10303463
answered 13 hours ago
bill sbill s
53.7k376153
$endgroup$
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]
edited 3 hours ago
J. M. is computer-less♦
97k10303463
answered 13 hours ago
bill sbill s
53.7k376153
edited 3 hours ago
J. M. is computer-less♦
97k10303463
edited 3 hours ago
J. M. is computer-less♦
97k10303463
edited 3 hours ago
J. M. is computer-less♦
97k10303463
97k10303463
answered 13 hours ago
bill sbill s
53.7k376153
answered 13 hours ago
bill sbill s
53.7k376153
answered 13 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
12 hours ago
2
$begingroup$
@tiffany, just doWith[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
$endgroup$
– J. M. is computer-less♦
9 hours ago
add a comment |
$begingroup$
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
12 hours ago
2
$begingroup$
@tiffany, just doWith[{n = 8, m = 3}, (# - DiagonalMatrix[Diagonal[#]]) &[RandomVariate[BernoulliDistribution[1/m], {n, n}]]].
$endgroup$
– J. M. is computer-less♦
9 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
12 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
12 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♦
9 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♦
9 hours ago
add a comment |
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
$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$
answered 12 hours ago
Okkes DulgerciOkkes Dulgerci
5,0991917
$endgroup$
$begingroup$
Altho using theBernoulliDistributionis 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♦
9 hours ago
add a comment |
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
$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$
answered 12 hours ago
Okkes DulgerciOkkes Dulgerci
5,0991917
$endgroup$
$begingroup$
Altho using theBernoulliDistributionis 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♦
9 hours ago
add a comment |
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
$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$
answered 12 hours ago
Okkes DulgerciOkkes Dulgerci
5,0991917
$endgroup$
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
$left(
begin{array}{ccccc}
0 & 1 & 1 & 0 & 1 \
1 & 0 & 1 & 0 & 1 \
1 & 0 & 0 & 0 & 1 \
0 & 0 & 1 & 0 & 0 \
0 & 0 & 0 & 1 & 0 \
end{array}
right)$
answered 12 hours ago
Okkes DulgerciOkkes Dulgerci
5,0991917
answered 12 hours ago
Okkes DulgerciOkkes Dulgerci
5,0991917
answered 12 hours ago
Okkes DulgerciOkkes Dulgerci
5,0991917
answered 12 hours ago
Okkes DulgerciOkkes Dulgerci
5,0991917
5,0991917
$begingroup$
Altho using theBernoulliDistributionis 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♦
9 hours ago
add a comment |
$begingroup$
Altho using theBernoulliDistributionis 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♦
9 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♦
9 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♦
9 hours 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
answered 13 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
13 hours ago
$begingroup$
@tiffany you setmandnin the first two lines...
$endgroup$
– Vsevolod A.
12 hours ago
2
$begingroup$
Justrnd := If[RandomReal[{0, 1}] < 1/m, 1, 0];will do, since the function never uses its argument.
$endgroup$
– J. M. is computer-less♦
9 hours 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
answered 13 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
13 hours ago
$begingroup$
@tiffany you setmandnin the first two lines...
$endgroup$
– Vsevolod A.
12 hours ago
2
$begingroup$
Justrnd := If[RandomReal[{0, 1}] < 1/m, 1, 0];will do, since the function never uses its argument.
$endgroup$
– J. M. is computer-less♦
9 hours ago
add a comment |
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
answered 13 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
answered 13 hours ago
Vsevolod A.Vsevolod A.
478211
answered 13 hours ago
Vsevolod A.Vsevolod A.
478211
answered 13 hours ago
Vsevolod A.Vsevolod A.
478211
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
13 hours ago
$begingroup$
@tiffany you setmandnin the first two lines...
$endgroup$
– Vsevolod A.
12 hours ago
2
$begingroup$
Justrnd := If[RandomReal[{0, 1}] < 1/m, 1, 0];will do, since the function never uses its argument.
$endgroup$
– J. M. is computer-less♦
9 hours ago
add a comment |
$begingroup$
Thank you Vsevolod A. How will the command be if I do not restrict what n and m be?
$endgroup$
– tiffany
13 hours ago
$begingroup$
@tiffany you setmandnin the first two lines...
$endgroup$
– Vsevolod A.
12 hours ago
2
$begingroup$
Justrnd := If[RandomReal[{0, 1}] < 1/m, 1, 0];will do, since the function never uses its argument.
$endgroup$
– J. M. is computer-less♦
9 hours ago
$begingroup$
Thank you Vsevolod A. How will the command be if I do not restrict what n and m be?
$endgroup$
– tiffany
13 hours ago
$begingroup$
Thank you Vsevolod A. How will the command be if I do not restrict what n and m be?
$endgroup$
– tiffany
13 hours ago
$begingroup$
@tiffany you set
m and n in the first two lines...$endgroup$
– Vsevolod A.
12 hours ago
$begingroup$
@tiffany you set
m and n in the first two lines...$endgroup$
– Vsevolod A.
12 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♦
9 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♦
9 hours ago
add a comment |
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}}
answered 3 hours ago
J. M. is computer-less♦J. M. is computer-less
97k10303463
$endgroup$
add a comment |
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}}
answered 3 hours ago
J. M. is computer-less♦J. M. is computer-less
97k10303463
$endgroup$
add a comment |
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}}
answered 3 hours ago
J. M. is computer-less♦J. M. is computer-less
97k10303463
$endgroup$
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}}
answered 3 hours ago
J. M. is computer-less♦J. M. is computer-less
97k10303463
answered 3 hours ago
J. M. is computer-less♦J. M. is computer-less
97k10303463
answered 3 hours ago
J. M. is computer-less♦J. M. is computer-less
97k10303463
answered 3 hours ago
J. M. is computer-less♦J. M. is computer-less
97k10303463
97k10303463
add a comment |
add a comment |
tiffany is a new contributor. Be nice, and check out our Code of Conduct.
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tiffany is a new contributor. Be nice, and check out our Code of Conduct.
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tiffany is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
For a start, your code seems to set the diagonal elements to 1 rather than zero.
$endgroup$
– MarcoB
14 hours ago
$begingroup$
Same question posted here.
$endgroup$
– Rohit Namjoshi
7 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