Extracting all possible vectors from a big vector

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

[Image Analyst]

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]

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

Answer 1

[DGM]

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.

[Mehmet Fatih]

Thank you for your answer. It really helped me. I will not use it for long vectors.

Answer 2

[SALAH ALRABEEI]

check this

https://www.mathworks.com/matlabcentral/fileexchange/24185-partitions