How to delete elements in array efficiently

Hi all,
I'm writing a simple script in Matlab where I compare adjacent element and delete one of them if there difference between them is one.
for i=1:length(Vector) - 1
if Vector(i+1) - Vector(i) == 1
Vector(i) = [];
end
if i == length(Vector)
break
end
However, I'm getting an error that my indices are out of bound. Is there a simpler way of doing this by utilizing internal functions. I think my problem is that my array is constantly decreasing and the Vector(i+1) - Vector(i) are out of bounds.

 Akzeptierte Antwort

Azzi Abdelmalek
Azzi Abdelmalek am 25 Aug. 2013

2 Stimmen

Vector=[1 2 4 5 66 88 100 101]
id=find([0 diff(Vector)]==1)-1
Vector(id)=[]

Weitere Antworten (2)

dpb
dpb am 25 Aug. 2013
Bearbeitet: dpb am 25 Aug. 2013

2 Stimmen

When doing such in loops, start at end and work to beginning -- then the indices of those removed are above where you're headed next.
But, in Matlab use vector operations and supplied functions to do such things...
v([false diff(v)==1])=[];
Azzi Abdelmalek
Azzi Abdelmalek am 25 Aug. 2013

0 Stimmen

You can also do it with while loop
Vector=[1 2 3 4 5 66 88 100 101 14]
i=1;
while i<numel(Vector)
if Vector(i+1) - Vector(i) == 1
Vector(i) = [];
i=i-1;
end
i=i+1
end
Vector

4 Kommentare

Or, variations on a theme, the looping solution...
for i=length(v):-1:2
if v(i)-v(i-1)==1;
v(i)=[];
end
end
Azzi Abdelmalek
Azzi Abdelmalek am 25 Aug. 2013
The result is not the same, because this will keep the first number of the consecutive numbers
Jan
Jan am 25 Aug. 2013
Repeated shrinking of an array suffers from the same inefficiency as iterative growing.
dpb
dpb am 26 Aug. 2013
Posted simply as pedagogical example for OP, not as recommended solution--see earlier posting.

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