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Extracting all possible vectors from a big vector

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Mehmet Fatih
Mehmet Fatih el 12 de Jun. de 2021
Editada: DGM el 12 de Jun. de 2021
Hi guys,
İ want to expalin with an example. Let's say i have a vector [1 2 3 4] i want to extract these [1] [2] [3] [4] [1 2] [1 3] [1 4] [2 3] [2 4] [3 4] [1 2 3] [1 2 4] [1 3 4] [2 3 4] [1 2 3 4] i found a way to do that but the problem is that they are not vectors.
How can i do that?
I know that [2 3] and [3 2] are different vectors but order is not important.
v=[1 2 3 4];
b=length(v);
i=1;
while i<=b
a = uint16(v);
c = nchoosek(a,uint16(i));
disp(c)
i=i+1;
end
  2 comentarios
Image Analyst
Image Analyst el 12 de Jun. de 2021
Why do you want to do this? What is the use case? The number of output vectors would blow up incredibly fast for more than a length of a few. Like if you had a vector that was hundreds long, it would be impractical.
Mehmet Fatih
Mehmet Fatih el 12 de Jun. de 2021
I am trying to make a code for this: Given a vector x, return the indices to elements that will sum to exactly half of the sum of all elements.
To use "sum" ı need them to be vector.
I will use it for vectors that are not long. At most 6 elements

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Respuesta aceptada

DGM
DGM el 12 de Jun. de 2021
Editada: DGM el 12 de Jun. de 2021
Sure they're vectors.
v=[1 2 3 4];
v = uint16(v); % there's no point doing this repeatedly in the loop
C = {};
for k = 1:numel(v)
c = nchoosek(v,k); % class is inherited from v
C = [C; num2cell(c,2)]; % one result per cell
end
format compact
celldisp(C)
C{1} = 1 C{2} = 2 C{3} = 3 C{4} = 4 C{5} = 1 2 C{6} = 1 3 C{7} = 1 4 C{8} = 2 3 C{9} = 2 4 C{10} = 3 4 C{11} = 1 2 3 C{12} = 1 2 4 C{13} = 1 3 4 C{14} = 2 3 4 C{15} = 1 2 3 4
(scroll to see the rest of the output)
Of course, IA is right. If v is very long at all, the problem becomes impractical. If the range of values in v allows, you might be able to save some weight by using uint8(). Still, even when using uint8(v), the result for a vector of length 25 occupies 4.2GB in RAM.
EDIT:
For what it's worth, we can save some time (might be valuable if n gets larger than about 20) by actually calculating how big C needs to be. I just let the array grow last time, but that was just me being lazy.
% calculate how large C needs to be
n = numel(v);
s = factorial(n)./(factorial((1:n)).*factorial(n-(1:n)));
sc = [0; cumsum(s)]; % this makes indexing easier
C = cell(sum(s),1); % allocate output
for k = 1:n
c = nchoosek(v,k);
C(sc(k)+(1:s(k))) = num2cell(c,2);
end
The time savings are maybe 10-20%, though that probably varies with version, hardware, and environment.
  1 comentario
Mehmet Fatih
Mehmet Fatih el 12 de Jun. de 2021
Thank you for your answer. It really helped me. I will not use it for long vectors.

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Más respuestas (1)

SALAH ALRABEEI
SALAH ALRABEEI el 12 de Jun. de 2021
Editada: SALAH ALRABEEI el 12 de Jun. de 2021

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