Trying to modify a vector by removing alternate elements

2 views (last 30 days)
Hi ,
I am learning matlab and I came across a problem on cody:
I need to be able to remove even elements from a 1D vector:
if x= [ 1 8 5 9]
then y = [1 5]
I tried with somethin like this:
n=1;
y=x;
for i = 1:length(x)
if mod(n,2) == 0
y(n)=[];
end
n= n+1;
end
and I got this error:
Matrix index is out of range for deletion.
Error in for2 (line 5)
y(n)=[];

Accepted Answer

Stephen23
Stephen23 on 19 Jan 2020
Edited: Stephen23 on 19 Jan 2020
There is no point in defining n when it always has exactly the same value as the loop iteration variable i. Get rid of one of those variables.
Your code is buggy because you did not consider that when you remove elements from a vector it gets smaller, which is clear if you print the vector on each loop iteration. This is what your code does:
  1. iteration -> remove nothing [1 8 5 9]
  2. iteration -> remove 2nd value [1 5 9]
  3. iteration -> remove nothing [1 5 9]
  4. iteration -> remove 4th value [1 5 9] !!! EEROR: the vector does not have a 4th element !!!
What do you expect to happen when you try to remove the 4th element of a 3-element vector?
You can resolve this by looping over the vector in reverse:
for n = numel(y):-1:1 % reverse!
...
end
Or by getting rid of the loop and writing simpler, more efficient MATLAB code using basic indexing:
y(2:2:end) = []
  4 Comments
Subham Kumar
Subham Kumar on 16 Mar 2021
How do we use the second solution for a 2D array? I want to keep it a 2D array after removing the alternative elements.
Stephen23
Stephen23 on 17 Mar 2021
@Subham Kumar: do you mean like this?:
M = rand(4,3)
M = 4×3
0.2389 0.9783 0.6445 0.2981 0.6712 0.3533 0.8601 0.4258 0.1559 0.3875 0.8520 0.8540
M(2:2:end,:) = []
M = 2×3
0.2389 0.9783 0.6445 0.8601 0.4258 0.1559

Sign in to comment.

More Answers (0)

Categories

Find more on Loops and Conditional Statements in Help Center and File Exchange

Products


Release

R2019b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!