How to select complementary elements from a vector?
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Shadman Shahid
el 7 de Abr. de 2021
Respondida: Bruno Luong
el 20 de Abr. de 2021
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];
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Respuesta aceptada
Bruno Luong
el 20 de Abr. de 2021
d = [33 20 4 5 6 75 8 9 0];
I = [1 3 7];
d(setdiff(1:end,I))
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Más respuestas (4)
Khalid Mahmood
el 7 de Abr. de 2021
Editada: Khalid Mahmood
el 19 de Abr. de 2021
%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))
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John D'Errico
el 7 de Abr. de 2021
Editada: John D'Errico
el 7 de Abr. de 2021
d = [3 2 1 5 6 7 8 9 0];
I = [1 3 7];
setdiff(d,I)
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))
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