# How to count alternating ones and zeros in a matrix

6 ビュー (過去 30 日間)
Awais Saeed 2021 年 7 月 14 日

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 件のコメント表示非表示 1 件の古いコメント
Awais Saeed 2021 年 7 月 14 日
Yes 0 1 0 1 would also be an alternating sequence.

サインインしてコメントする。

### 採用された回答

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.
##### 2 件のコメント表示非表示 1 件の古いコメント
Devanuj Deka 2021 年 7 月 15 日
You're welcome!

サインインしてコメントする。

### その他の回答 (2 件)

Matt J 2021 年 7 月 14 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 1 1];
A=v(:,1:end-3); B=v(:,2:end-2); C=v(:,3:end-1); D=v(:,4:end);
pattern1=A==0 & B==1 & C==0 & D==1;
pattern2=A==1 & B==0 & C==1 & D==0;
rows=find( any(pattern1|pattern2,2) )
rows = 2×1
3 4
##### 1 件のコメント表示非表示 なし
Stephen 2021 年 7 月 15 日
+1 nice use of MATLAB.

サインインしてコメントする。

Stephen 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)
X = 8×1 logical array
0 0 1 1 1 1 1 0

サインインしてコメントする。

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!