How do I make matrix of ones and zeros alternating depending on size and elements of an array?

5 次查看（过去 30 天）
In-chan Kim
In-chan Kim 2019-9-26
回答： Guillaume 2019-9-26
I have an array, A, and I want to make a matrix, B, that has the size dependent on A, and has 1s and 0s alternating, depending on the value of elements in A
E.g. A=[1;2;3;4;5]
B should be of size: ((length(A)+1)/2,sum(A)]
In this case B would be a matrix of 3x15
Then the values for B needs to be be thus:
row 1:[ones(A(1)),zeros(A(2)),zeros(A(3)),zeros(A(4)),zeros(A(5))]
row2: [zeros(A(1)),zeros(A(2)),ones(A(3)),zeros(A(4)),zeros(A(5))]
row3: [zeros(A(1)),zeros(A(2)),zeros(A(3)),zeros(A(4)),ones(A(5))]
B should look like:
[1 0 0 0 0 0 0 0 0 0 0 0 0 0 0;
0 0 0 1 1 1 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0 0 1 1 1 1 1]
I've been trying to formulate something with a nested for look for each dimension of B, but have not been succesful!
Would really appreciate any help thanks!

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采纳的回答

Fabio Freschi
Fabio Freschi 2019-9-26
% initialize
iRow = [];
jCol = [];
% preallocate
B = zeros((length(A)+1)/2,sum(A));
% indices for columns
idx = [0; cumsum(A)]+1;
for i = 1:length(A)
if mod(i,2) ~= 0
iRow = [iRow; repmat((i+1)/2,A(i),1)];
jCol = [jCol; (idx(i):idx(i+1)-1).'];
end
end
B(sub2ind(size(B),iRow,jCol)) = 1;

更多回答（1 个）

Guillaume
Guillaume 2019-9-26
Here's a fairly simple way:
B = zeros((numel(A)+1)/2, sum(A));
for row = 1:2:numel(A)
B(ceil(row/2), :) = repelem([0, 1, 0], [sum(A(1:row-1)), A(row), sum(A(row+1:end))]);
end

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