how to compare two consecutive values in a matrix

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deal all,
how to compare two consecutive values in a matrix called 'temp' row-wise, the rule and the expecting result is as the following:
if first value>=second value , then bit=1;
else if first value<second value, then bit=0;
Store the bit generated every time in a matrix ‘M’.
This step will generate a matrix ‘M’ (logical) of size same a temp.
say for instance let's say matrix 'temp' be temp=[1,2,2,1;3,4,4,3;5,6,6,5;7,8,8,7]
can anyone give me some hints, thx a lot !

采纳的回答

Voss
Voss 2022-7-29
temp=[1,2,2,1;3,4,4,3;5,6,6,5;7,8,8,7]
temp = 4×4
1 2 2 1 3 4 4 3 5 6 6 5 7 8 8 7
In each row, you have one fewer comparisons than you have elements, so M is of size one less than temp.
M = logical(temp(:,1:end-1) >= temp(:,2:end))
M = 4×3 logical array
0 1 1 0 1 1 0 1 1 0 1 1
Maybe you want to append a column of false values to M.
M(:,end+1) = false
M = 4×4 logical array
0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0
Another way to do the same:
temp2 = [temp NaN(size(temp,1),1)];
M = logical(temp2(:,1:end-1) >= temp2(:,2:end))
M = 4×4 logical array
0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0
  5 个评论
Voss
Voss 2022-7-30
In that case, you can do it like this:
temp=[1,2,2,1;3,4,4,3;5,6,6,5;7,8,8,7]
temp = 4×4
1 2 2 1 3 4 4 3 5 6 6 5 7 8 8 7
% make a row vector of the elements in temp in the proper order for the comparison
temp_row = reshape(temp.',1,[]);
% add the first element to the end for comparing the last and first elements
temp_row(end+1) = temp(1);
disp(temp_row)
1 2 2 1 3 4 4 3 5 6 6 5 7 8 8 7 1
% perform the comparison
M = temp_row(1:end-1) >= temp_row(2:end)
M = 1×16 logical array
0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1
% reshape the result back to the right size
M = reshape(M,size(temp,2),[]).'
M = 4×4 logical array
0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 1
qing li
qing li 2022-7-31
great that is what i want ! thank you sooo much sir ! that is really helpful !

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更多回答(1 个)

Matt J
Matt J 2022-7-29
M=(diff(temp2,1,2)<=0);

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