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

1 回表示 (過去 30 日間)
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

採用された回答

Sulaymon Eshkabilov
Sulaymon Eshkabilov 2021 年 7 月 4 日
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 件のコメント
Stephen23
Stephen23 2021 年 7 月 5 日
編集済み: Stephen23 2021 年 7 月 5 日
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.
Levente Gellért
Levente Gellért 2021 年 7 月 6 日
Dear All, many thanks for the nice comments! lg

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その他の回答 (3 件)

Yazan
Yazan 2021 年 7 月 4 日
c = zeros(size(b));
c(b>min(a(:)) & b<max(a(:))) = 1;

dpb
dpb 2021 年 7 月 4 日
>> 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 年 7 月 5 日
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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