Try to mix as much as possible elements inside an array

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Given a vector
V = [1 1 1 1 1 1 1 4 4 4 4 4 1 1 1 1 1 4 4 4 4 4 4 5 5 5 5 5 5 8 8 8 8 8 8 5 5 5 5 5 5 5 5 6 6 6 6 6 6 8 8 8 8 5 5 5 5 6 6 7 7 7 7 7 9 9 9 9 9 7 7 7]
I would like to destroy the consecutivity of the same number and mix the elements. In particular I want to mix 1 just with 4, 5 just with 6-8 and 7 just with 9.
obtaining for example:
S = [1 1 4 1 4 1 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 4 5 5 8 5 8 5 8 5 8 5 8 6 5 5 6 5 5 5 8 5 6 8 6 5 6 5 8 6 5 8 5 6 5 5 8 6 7 9 7 7 7 9 7 9 7 9 7 9 7]
It's not important the way I mix but it's important the consecutivity: first 1-4 then 5-8-6 and finally 7-9
May someone help me to find a way to obtain S?
  3 个评论
Aquatris
Aquatris 2019-9-17
编辑:Adam Danz 2019-9-23
Check the function randperm(). Then you can create an index vector and shuffle the elements to obtain what you want.
luca
luca 2019-9-18
编辑:luca 2019-9-18
I've changed the structure in order to avoid this problem, but I'm still interested in knowing if there is a solution for that
Thanks

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采纳的回答

Adam Danz
Adam Danz 2019-9-18
The 'exchangeOpts' is a list of exchange options such that any value in V that matches a value in exchangeOpts{n} can be replaced with a random value from exchangeOpts{n}.
V = [1 1 1 1 1 1 1 4 4 4 4 4 1 1 1 1 1 4 4 4 4 4 4 5 5 5 5 5 5 8 8 8 8 8 8 5 5 5 5 5 5 5 5 6 6 6 6 6 6 8 8 8 8 5 5 5 5 6 6 7 7 7 7 7 9 9 9 9 9 7 7 7];
exchangeOpts = {[1,4],[5,6],[7,9]}; %identify exchange groups (in cell array to allow for mixed vector sizes)
% loop through each exchange option
S = V;
for i = 1:numel(exchangeOpts)
idx = ismember(V,exchangeOpts{i}); %identifies which elements will be randomized
S(idx) = exchangeOpts{i}(randi(numel(exchangeOpts{i}),1,sum(idx)));
end

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