how to compare two consecutive values in a matrix

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 !

Answer 1

[Voss]

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

[qing li]

thank you so much sir ! That is really helpful !

[Voss]

You're welcome!

[qing li]

Sorry sir, I has another problem regarding the codes above...

In your code, you append a column of false values to M for ensuring matrix M and temp have same size, that is good. But could I ask how to let elements in red circle involve in comparison as the rule (just as the pic shows below), thank you so much for your attention and time for this issue !

and let the expecting result be

M =

0 1 1 0

0 1 1 0

0 1 1 0

0 1 1 1

[Voss]

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]

great that is what i want ！ thank you sooo much sir ！ that is really helpful !

Answer 2

[Matt J]

M=(diff(temp2,1,2)<=0);

[qing li]

thank you so much sir ! That is really helpful !