create subarray from an given array

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Industrial
Industrial el 21 de Mzo. de 2014
Comentada: Azzi Abdelmalek el 22 de Mzo. de 2014
To create few no. of arrays from a given long array, eg. given = [1 2 3 4 5 6 7 8 9 ....]
to create randomly [3 4 9] [1 5 8] [2 6 7] ... if size of group is given between 3 and 3 (between two nos.) by user
or [2 6 9 5] [1 3 4 7 8] etc.. if the size of group is between 4 and 5 given by user (may vary and no of group can also vary) also no number should repeat again, and should be present in array given earlier

Respuestas (2)

Azzi Abdelmalek
Azzi Abdelmalek el 21 de Mzo. de 2014
Editada: Azzi Abdelmalek el 21 de Mzo. de 2014
EDIT
a=[1 2 3 4 5 6 7 8 9]
n=numel(a)
% find divisor of a
c=1:n
d = c(rem(n,c)==0)
% number of groupes
ng=d(randi(numel(d),1))
%------Length of group------------------
lg=n/ng
%-------------Result---------------------
s=randperm(n);
out=num2cell(reshape(a(s),ng,lg),2)
  6 comentarios
Industrial
Industrial el 22 de Mzo. de 2014
Editada: Industrial el 22 de Mzo. de 2014
sir i am usin R2009a version of matlab
Azzi Abdelmalek
Azzi Abdelmalek el 22 de Mzo. de 2014
Ok, try the edited answer

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Jos (10584)
Jos (10584) el 21 de Mzo. de 2014
This problem is somewhat ill-defined. If there are multiple group sizes allowed, should all group sizes be present an equal amount of times? And things might not fit, for instance, when the number of elements in A is odd, and the AllowGroupSize is even …
Here is a suggestion:
A = 1:21
AllowedGroupSize = [3 4]
GroupSizes = repmat(AllowedGroupSize,1, numel(A) ./ sum(AllowedGroupSize))
C = mat2cell(A(randperm(numel(A))),1,GroupSizes)
  1 comentario
Industrial
Industrial el 22 de Mzo. de 2014
Thank you for your kind consideration
But in this problem all sizes of groups might not be present in the answer. Even groups may have same sizes, but may differ when executed number of times in a loop. Actually it is to be used in a Manufacturing cell family formation problem in which a part family may have some parts and another part family may have from rest of the parts.
thanks.

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