How to resolve issue while dealing with larger matrices?

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Ammy
Ammy 2022-3-25
评论： Arif Hoq 2022-3-25
A=randi(16,[4,4]);
B=A(:)';
P=randi(16,[1 4]);
V=randi(16,[1 4]);
B(P)=V; % replace entries at position P with values in V
C=B(1,P)
C=V satisfied in this case
But I'm not getting the right answer in case of larger size matrices.
A=randi(255,[512,512]);
>> B=A(:)';
>> P=randi(255,[1 256]);
>> V=randi(256,[1 256]);
>> B(P)=V; %
>> C=B(1,P);
C should be equal to B , but I'm not getting it right,
Request for help regarding this.
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Ammy
Ammy 2022-3-25
@Arif Hoq Got the point why I'm facing the problem, thank you very much. How can I vote your answers?

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回答（2 个）

Voss
Voss 2022-3-25
Ran in:
As @Arif Hoq points out, the problem is not with the size of the matrices. The problem is that the elements of P are not guaranteed to be unique. Look at what happens when a number occurs more than once in P, in this case 8 appears twice in P:
A=randi(16,[4,4]);
B=A(:)'
B = 1×16
8 9 11 8 9 4 5 11 1 14 13 5 1 15 13 5
P=randi(16,[1 4])
P = 1×4
7 1 8 8
V=randi(16,[1 4])
V = 1×4
2 11 13 4
B(P)=V % replace entries at position P with values in V
B = 1×16
11 9 11 8 9 4 2 4 1 14 13 5 1 15 13 5
C=B(1,P)
C = 1×4
2 11 4 4
C==V
ans = 1×4 logical array
1 1 0 1
In this case, B(P)=V is B([7 1 8 8]) = [2 11 13 4]; which means that B(8) is assigned the value 13 and then immediately assigned the value 4, so that B(8) is 4 after that step.
Then C = B(1,P) is C = B(1,[7 1 8 8]) = [2 11 4 4] not [2 11 13 4] = V
Any time P has repeated values, then V will not be recoverable from B and P.
Arif Hoq
Arif Hoq 2022-3-25
A=randi(16,[4,4]);
B=A(:)';
P=randi(16,[1 4]);
V=randi(16,[1 4]);
[C out]=ismember(V,B);
if numel(out(out~=0)) ==4
disp(out)
else
disp('V is not the member of B')
end

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