ismember function too slow
22 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
I have to improve the speed of my simulation. Well my code is quite simple. I have a vector A=[1 2 3 4 5 6 7 8 9 10] and another vector B=[3 4 9]. I use ismember to check if every element is into A matrix.
for n=1:length(A)
if ismember(B(n),A)
do-something
end
end
this part of my code is executed so many time in my simulation and my matrices are really big.some idea to improve my code?
1 Kommentar
Matt J
am 17 Okt. 2012
Since this is speed-critical, it would be wise for you to elaborate on "do-something". That could point the way to eliminating the for-loop altogether, or some other kind of performance optimization.
Antworten (5)
Jonathan Epperl
am 17 Okt. 2012
Don't you want to actually INTERSECT A and B? If so, then here's a very fast intersect function that intersects two sets of positive integers only. If you have negative integers you could obviously shift. If you have reals you're SOL. If your integers are greater than ~1e5 the speed gain is negligible.
C = fastintersect(A,B)
if ~isempty(A)&&~isempty(B)
P = zeros(1, max(max(A),max(B)) ) ;
P(A) = 1;
C = B(logical(P(B)));
else
C = [];
end
Obviously you could leave out the check for emtpy sets if you know you won't have empty sets.
4 Kommentare
Shengtao Wang
am 20 Feb. 2013
Note this may produce the same number multiple times if they are not unique in the arrays. not a problem at all though.
Matt Fig
am 17 Okt. 2012
BinA = B(ismember(B,A));
for n=1:length(BinA)
do-something with BinA(n)
end
2 Kommentare
Matt Fig
am 17 Okt. 2012
Bearbeitet: Matt Fig
am 17 Okt. 2012
If your B vector contains ONLY elements of A, then why were YOU checking if each element was a member in your code?? That is just weird.
If you know that every element of B is a member of A, then what is the point of your question?? Why not just get rid of the IF statement in your FOR loop altogether??
Azzi Abdelmalek
am 17 Okt. 2012
Bearbeitet: Azzi Abdelmalek
am 17 Okt. 2012
use ismember out of the loop
A=[1 2 3 4 5 6 7 8 9 10]
B=[3 4 11 9 22]
idx=find(ismember(B,A)==1)
for k=idx
%do
end
2 Kommentare
Robert Cumming
am 17 Okt. 2012
if you want to keep it in the loop you can do:
for n=1:length(B)
if min (abs( A-B(n)) ) == 0;
% do something
end
end;
Or use logical indexing
tic
A=[1 2 3 4 5 6 7 8 9 10];
B=[3 4 11 9 22];
idx=find(ismember(B,A)==1);
toc
tic
flags = false(1,length(B));
for i=1:length(A)
check = B==A(i);
flags(check)=1;
end
idx2 = find(flags==1);
toc
A for loop can be faster!!
0 Kommentare
Omar Ali Muhammed
am 3 Mär. 2021
A=[1 2 3 4 5 6 7 8 9 10];
B=[3 4 9];
tt=ismember(B,A);
if sum(tt)==3
Do Something;
else
Do Something;
end
0 Kommentare
Siehe auch
Kategorien
Mehr zu Logical finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!