How to find elements of a vector falling between minimum and maximum of an other vector without loop.
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Levente Gellért
el 4 de Jul. de 2021
Comentada: Levente Gellért
el 6 de Jul. de 2021
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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Respuesta aceptada
Sulaymon Eshkabilov
el 4 de Jul. de 2021
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 comentarios
Stephen23
el 5 de Jul. de 2021
Editada: Stephen23
el 5 de Jul. de 2021
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.
Más respuestas (3)
dpb
el 4 de Jul. de 2021
>> 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
el 5 de Jul. de 2021
a=(1:1:10);
b=[5.5 11];
[~,~,c]=histcounts([0,5.5,10,11],[min(a),max(a)+eps(max(a))])
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