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.
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the cyclist
2011년 11월 2일
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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Image Analyst
2011년 11월 2일
This is how I did it:
% Set up parameters.
rows = 10;
columns = 15;
onesPerColumn = 4;
% Initialize matrix.
m = zeros(rows, columns, 'int32')
for col = 1 : columns
% Get random order of rows.
randRows = randperm(rows);
% Pick out "onesPerColumn" rows that will be set to 1.
rowsWithOne = randRows(1:onesPerColumn);
% Set those rows only to 1 for this column.
m(rowsWithOne, col) = 1;
end
% Display m
m
댓글 수: 3
Raghwan Chitranshu
2019년 1월 24일
hello,
How if I also want to keep number of ones in rows to be equal.
what changes shall I perform .
can you help in this
추가 답변 (4개)
Sven
2011년 11월 2일
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
댓글 수: 2
Sven
2011년 11월 2일
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
2011년 11월 2일
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
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Walter Roberson
2011년 11월 2일
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:
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