Element-wise interval conformity check

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tompaine
tompaine 2015-5-4
编辑： Stephen23 2015-5-4
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?

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Stephen23
Stephen23 2015-5-4
编辑：Stephen23 2015-5-4
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:
http://www.mathworks.com/help/matlab/getting-started-with-matlab.html

更多回答（1 个）

Purushottama Rao
Purushottama Rao 2015-5-4
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.
  1 个评论
Stephen23
Stephen23 2015-5-4
编辑：Stephen23 2015-5-4
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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