How to find elements of a vector falling between minimum and maximum of an other vector without loop.

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Levente Gellért
Levente Gellért 2021-7-4
评论： Levente Gellért 2021-7-6
Dear Community,
is there other way, than a loop to find elements in a vector b falling between the minimum and the maximum of vector a?
Let's say:
a=(1:1:10);
b=[5.5 11];
for i=1:length(b)
if b(:,i)>min(a) && b(:,i)<max(a)
c(:,i)=1;
else
c(:,i)=0;
end
end
Thanks for your suggestions! lg

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采纳的回答

Sulaymon Eshkabilov
Logical indexing is the best option, e.g.:
a=(1:1:10);
b=[5.5 11; 13, 3; 10.5 10];
IDX = find(b>min(a) & b<max(a));
C(IDX)=1;
  2 个评论
Stephen23
Stephen23 2021-7-5
编辑：Stephen23 2021-7-5
This answer actually shows linear indexing (the output from the superfluous FIND), not logical indexing.
Remove the superfluous FIND to use simpler and more efficient logical indexing.
Also note that because C is not preallocated, it could have fewer elements than a.

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更多回答（3 个）

Yazan
Yazan 2021-7-4
c = zeros(size(b));
c(b>min(a(:)) & b<max(a(:))) = 1;
dpb
dpb 2021-7-4
>> iswithin(b,min(a),max(a))
ans =
1×2 logical array
1 0
>>
is a common-enough idiom I have a utility function for the purpose--
>> function flg=iswithin(x,lo,hi)
% returns T for values within range of input
% SYNTAX:
% [log] = iswithin(x,lo,hi)
% returns T for x between lo and hi values, inclusive
flg= (x>=lo) & (x<=hi);
end
It isn't any different than writing the logical expression in line except as a function it has the advantage of moving the test to a lower level that is often very helpful in writing concise, legible expressions at the user level.
Matt J
Matt J 2021-7-5
Ran in:
a=(1:1:10);
b=[5.5 11];
[~,~,c]=histcounts([0,5.5,10,11],[min(a),max(a)+eps(max(a))])
c = 1×4
0 1 1 0

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