Hi, I would like to vectorize the following loop, have tried generating indices and but so far been successful.
a = any 1D vector
N = some value
for i=1:length(a)
for j=(i+1):length(a)
if(a(i)-a(j) > N)
disp('Far');
end
end
end
Does anyone have any ideas on this ?

2 commentaires

Guillaume
Guillaume le 31 Jan 2018
The ability of vectorising loops depends solely on the do something. So if you don't tell us what it is, we can't answer your question.
Balkrishna Patankar
Balkrishna Patankar le 31 Jan 2018
Updated

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Guillaume
Guillaume le 31 Jan 2018

0 votes

Your example is trivially vectorised:
%R2016b or later:
isgreater = (a - a.') > N;
%any version:
isgreater = bsxfun(@minus, a, a.') > N;
isgreater(r, c) is true (1) when a(r)-a(c) > N

3 commentaires

This is not giving the correct answer. For example, if I run the following program :
N=-2;
b=[1 1 1;2 2 2;3 3 3;4 4 4;5 5 5];
a=[1;2;3;4;5];
for i = 1:length(a)
for j = (i+1):length(a)
if(a(i)-a(j) > N)
disp('Far');
end
end
end
This displays Far 4 times which means there ought to be 4 1's in your matrix.
isgreater = (a - a.') > N;
Gives 19 ones.
Guillaume
Guillaume le 1 Fév 2018
Modifié(e) : Guillaume le 1 Fév 2018
Indeed, my answer was equivalent to having the j loop starting at 1 instead of i+1. The upper triangle of that isgreater matrix is the exact equivalent of your loops, so:
isgreater = triu((a - a.') > N, 1)
Balkrishna Patankar
Balkrishna Patankar le 2 Fév 2018
Thanks this worked !

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