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Element-wise interval conformity check

조회 수: 2 (최근 30일)
tompaine
tompaine 2015년 5월 4일
편집: Stephen23 2015년 5월 4일
Hello I have a Matrix
A=[1.0,2.0,0.7;1.2,3.0,1.0;0.8,1.1,0];
and an Interval
lower = 0.75;
upper = 1.25;
Matrix R should give me either 1 if the corresponding element of A is within the interval, or else 0:
R=[1,0,0;1,0,1;1,1,0];
Is there any way to do this without a loop?

채택된 답변

Stephen23
Stephen23 2015년 5월 4일
편집: Stephen23 2015년 5월 4일
Do not use lower and upper as variable names, as these are both names of inbuilt functions.
MATLAB lets you perform basic operations like this on the entire array (which is called code vectorization), so this comparison can be done directly using the relational operators, without any loops:
>> A = [1.0,2.0,0.7;1.2,3.0,1.0;0.8,1.1,0];
>> lwr = 0.75;
>> upr = 1.25;
>> R = lwr<=A & A<=upr
R =
1 0 0
1 0 1
1 1 0
This is covered in the introduction tutorials to MATLAB:

추가 답변 (1개)

Purushottama Rao
Purushottama Rao 2015년 5월 4일
Without looping, this can be done by accessing each element of a matrix and processing with an if condition. That makes lot of hand written code compared to looping.
  댓글 수: 1
Stephen23
Stephen23 2015년 5월 4일
편집: Stephen23 2015년 5월 4일
This is misleading and very poor advice that shows a basic misunderstanding of how to use MATLAB. Simply reading the documentation for the relational operators makes it clear that they operate on whole arrays at once, for example le states " A <= B returns a logical array with elements set to logical 1 (true) where A is less than or equal to B..." note the word array in that sentence. The examples in the documentation also illustrate this:
A = [1 12 18 7 9 11 2 15];
A <= 12
ans =
1 1 0 1 1 1 1 0
This means that no if statements or "lot of hand written code compared to looping" is required. Please learn about code vectorization.

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