How to count alternating ones and zeros in a matrix
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Lets just say that I have a matrix v as following:
v =
0 0 0 0 0 0
1 1 0 1 1 0
1 0 1 0 1 0
0 0 1 0 1 0
0 0 0 1 1 1
I want to get rows where alternating ones and zeros are occuring which are 3rd and 4th rows. By alternating ones and zeros, I mean a sequence of [1 0 1 0] at least.
2 件のコメント
Walter Roberson
2021 年 7 月 14 日
How about if a row ends with 0 1 0 1, is that also alternating 0 and 1, or does the sequence have to start with a 1?
採用された回答
Devanuj Deka
2021 年 7 月 14 日
編集済み: Devanuj Deka
2021 年 7 月 14 日
% Get the dimensions of 'v'
sz = size(v); % Gives two numbers. sz(1) is the number of rows; sz(2) is the number of columns
% Initialize a zero vector, having as many elements as the number of rows
% in v. After the code is run, the i'th element of this vector will be 1 if
% the i'th row of 'v' has an alternating pattern
rows = zeros(1,sz(1));
% Look for the pattern
for i=1:sz(1) % Iterating through rows
count = 0; % Count increments if the next element in the row is different
for j=1:sz(2)-1 % Iterating through the elements of each row
if v(i,j)~=v(i,j+1) % If current element and next element are unequal, increment count
count = count+1;
else % If next element is equal to the current element, count is reset
count = 0;
end
if count==3 % Minimum 3 changes back to back are required for your pattern (0->1->0->1, or 1->0->1->0)
rows(j) = 1; % If pattern detected in i'th row, row(i) is set to 1
break; % Since pattern detected, no need to look for more in the same row.
end
end
end
rows_with_pattern = find(rows); % A vector containing the rows which have the alternating pattern
find(X) gives the indices of non-zero elements in an array X, which is why we initialized rows as a vector of zeroes.
その他の回答 (2 件)
Stephen23
2021 年 7 月 15 日
v =[ 0 0 0 0 0 0; ...
1 1 0 1 1 0; ...
1 0 1 0 1 0; ...
0 0 1 0 1 0; ...
0 0 0 1 0 1; ... Walter Roberson's suggestion,
1 0 1 0 0 0; ... and reversed also.
0 1 0 1 0 1; ... Entire row alternating.
0 0 0 1 1 1];
X = any(diff(v,3,2)>3 | diff(~v,3,2)>3, 2)
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