Loop through a vector with changing legth
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Hello everyone. I want to loop through a vector, say z, and at each iteration remove some elements from it, using something like
z = z(abs(z-z(i)>=1))
My problem is, I don't know how to form the loop. I tried using
for member = z
but then again, I need to do calculations with each element so that doesn't work. Can anyone help me?
采纳的回答
KL
2018-3-22
why not use a while loop?
idx = 1;
while idx<=numel(z)
%do something
z = z(abs(z-z(idx)>=1));
idx = idx+1;
end
2 个评论
Alex De C
2018-3-22
"I need to do calculations with each element so that doesn't work"
Neither does this answer. This answer does not "do calculations with each element" as the question requests, it actually misses calculating with elements that follow any removed element that happens to be one position more than the current index (because the element removal and the index increment both conspire against its author).
Compare with this simple example, following the same concept, to remove all even values from z:
z = [1,1,2,2,3,3,4,4,4,5];
idx = 1;
while idx<=numel(z)
if mod(z(idx),2)==0
z(idx)=[];
end
idx = idx+1;
end
but at the end there are still even values in the vector!:
>> z
z =
1 1 2 3 3 4 5
Actually it did not test all elements to check if they are even! In general not all elements get tested by this algorithm although in some special cases they might be.
See Jan Simon's answer for a correct analysis of this problem.
更多回答（1 个）
Jan
2018-3-22
A loop over the elements of a vector cannot work, if you remove elements of the vector, except if you process the elements from the end to the start and remove only elements, which are after the current element.
z = rand(1, 10)
for k = 9:-1:1
if z(k) < 0.5
z(k+1) = [];
end
end
This cannot work, if you run it in the opposite direction from 1 to 9.
So what does "loop through a vector" exactly mean, if you remove elements? It cannot be an "element by element". What should happen exactly with the vector? You have to define this clearly, before it can be implemented.
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