# GENERATE CONDITIONS BETWEEN TWO ARRAYS AND RESULT IN A VALUE

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I have two matrices named AA and BB that are attached. I want to create conditions that relate the lines of the two and give me the result.
For example:
Condition 1
Values of AA>= 0 to AA<=5 and values of BB>=5 A to BB<=10, for all rows of both columns
Condition 2
AA values>= 5 to AA<=8 and BB values>=3 A to BB<=7 for all rows of both columns
Condition 3
AA values>= 3 to AA<=0 and BB values>=5 A to BB<=10 for all rows of both columnsThe result of these conditions will be in a matrix called AABB.
Being for condition 1 equal to 0.10, for condition 2 equal to 0.20 and for condition 3 equal to 0.30
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Below is explaining better how I want. I have my 3 conditions involving columns A and B, so I want a result in column AABB with the values 0.10, 0.20, 0.30. How do I solve the problem??
for AA >= 0 & AA <= 5 & BB >=5 & BB <= 10 % condition 1
AA >= 5 & AA <= 8 & BB >=3 & BB <= 7 % condition 2
AA >= 3 & AA <= 0 & BB >=5 & BB <= 10 % condition 3
AABB == 0.10 % results for the condition 1
AABB == 0.20 % results for the condition 2
AABB == 0.30 % results for the condition 3
end
Best Regard,
AP
##### 2 CommentsShowHide 1 older comment
No, how do I do it?

Voss on 19 Nov 2022
Something like this?
disp([AA BB]) % check the data
4 -9 8 -7 -1 9 -9 -7 5 9 3 1 2 10 0 13 9 12 -5 13
AABB = zeros(size(AA));
AABB(AA >= 0 & AA <= 5 & BB >=5 & BB <= 10) = 0.1; % condition 1
AABB(AA >= 5 & AA <= 8 & BB >=3 & BB <= 7) = 0.2; % condition 2
AABB(AA >= 3 & AA <= 0 & BB >=5 & BB <= 10) = 0.3; % condition 3
disp(AABB); % check the result
0 0 0 0 0.1000 0 0.1000 0 0 0
Notice that condition 2 and 3 are never true and that condition 3 can never be true because it includes AA >= 3 & AA <= 0 and there are no numbers that would simultaneously satisfy those inequalities.
It works perfectly, it's simpler than I thought. Thank you very much.

Walter Roberson on 19 Nov 2022
condition1 = all(AA>=0 & AA<=5 & BB>=5 & BB<=10, 'all')
The result of these conditions will be in a matrix called AABB
You defined the conditions as only being true if they hold "for all rows of both columns" . Therefore condition1 and condition2 and condition3 are all scalars, so the result in AABB could only be a scalar, not a "matrix".
Thanks

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