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How to select complementary elements from a vector?

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Shadman Shahid
Shadman Shahid on 7 Apr 2021 at 19:01
Answered: Bruno Luong on 20 Apr 2021 at 19:29
I have a vector
d = [33 20 4 5 6 75 8 9 0];
and another vector containing the indices whose complement i have to select from d.
I = [1 3 7];
so, the output I want is a vector which is - d minus the elements contained in d(I)
i.e
ans = [20 5 6 75 9 0];
  2 Comments
Shadman Shahid
Shadman Shahid on 20 Apr 2021 at 18:00
No. Its correct, I have changed the values to make the example clearer.
I is the vector of the index.
So, d(I) = [3 1 8];
What I want is to get rid of these elements from d. Not the values [1,3,7]. I is the index of the vector d.

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Accepted Answer

Bruno Luong
Bruno Luong on 20 Apr 2021 at 18:14
d = [33 20 4 5 6 75 8 9 0];
I = [1 3 7];
d(setdiff(1:end,I))
ans = 1×6
20 5 6 75 9 0

More Answers (4)


Khalid Mahmood
Khalid Mahmood on 7 Apr 2021 at 19:51
Edited: Khalid Mahmood on 19 Apr 2021 at 0:30
%Another but lengthy way is as follows:
%Vector 1
d = [3 2 1 5 6 7 8 9 0];
%Another vector containing the indices, which must be removed from d .
I = [1 3 7];
%so, the output is a vector which removes those values, i.e output= [2 5 6 9 0];
n1=size(d,2);
n2=size(I,2);
%A=zeros(1,n1-n2)
k=1;i=1;
for n=1:n1
if n~=I(k)
A(i)=d(n);
i=i+1;
else
if k<n2
k=k+1;
end
end
end
A
%same as A=d(~ismember(d,I))

John D'Errico
John D'Errico on 7 Apr 2021 at 20:19
Edited: John D'Errico on 7 Apr 2021 at 20:23
d = [3 2 1 5 6 7 8 9 0];
I = [1 3 7];
setdiff(d,I)
ans = 1×6
0 2 5 6 8 9
Note that your example is actually incorrect, in that you claim 7 should be in the final result. But since 7 is a member of I, that is not the case.
Also, it depends on if you want elements that remain in the original order. setdiff will return a sorted set, and if any elements of d were repeated, then only one copy will remain in the result.
So if setdiff does not do as you wish, in that case, you need to use ismember, deleting the elements found. Thus...
d(~ismember(d,I))
ans = 1×6
2 5 6 8 9 0

Bruno Luong
Bruno Luong on 20 Apr 2021 at 19:29
d = [33 20 4 5 6 75 8 9 0];
I = [1 3 7];
d(I) = []
d = 1×6
20 5 6 75 9 0

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