How to create a random binary matrix with equal number of ones in each column?
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Hi All,
I want to create a random binary matrix with equal number of ones in each column.
Appreciate if anyone have an idea to implement this in Matlab.
Thanks.
1 Kommentar
the cyclist
am 2 Nov. 2011
To avoid folks providing answers, and then you saying, "No, that's not what I meant", can you please provide more detail? For example, should each column have equal numbers of zeros and ones? What if there are an odd number of rows? Etc.
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Sven
am 2 Nov. 2011
Try:
% Define your matrix size and randomly pick the number of ones
matSz = [20, 40];
numOnes = randi(matSz(1));
% Make your matrix
myMat = false(matSz);
for i = 1:matSz(2)
myMat(randperm(matSz(1),numOnes), i) = true;
end
% Check that all went as planned
sum(myMat,1)==numOnes
2 Kommentare
Anne
am 2 Nov. 2011
Sven
am 2 Nov. 2011
Really? The variable "myMat" is your answer. The last line that prints out a series of ones was just confirmation that all of your columns had "numOnes" true elements in them.
Setting
matSz = [10, 15];
and
numOnes = 4;
gives the exact same output as what you agreed with below.
Naz
am 2 Nov. 2011
Since it's a RANDOM matrix, you are not guaranteed to have the same amount of one's and zero's (at least it seems logical to me). In order to get about 1/2 probability you need large matrix. You can try this:
a=rand(1000,1000);
a=round(a);
Check out help file for rand vs. randn
Anne
am 2 Nov. 2011
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Walter Roberson
am 2 Nov. 2011
0 Stimmen
Anne, you need to define what it means to take an inverse for you binary matrix. You can treat the binary matrix as being composed of the real numbers 0 and 1 and then do an arithmetic inverse on the array, ending up with a non-binary array. Or you can treat the binary matrix as being composed of boolean values over a field with the '*' being equivalent to 'or' and '+' being equivalent to xor, and the task is then to find a second binary matrix such that matrix multiplication using those operations produces the identity matrix.
If you want the inverse to be a binary matrix instead of a real-valued matrix, please see this earlier Question:
1 Kommentar
Anne
am 3 Nov. 2011
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