0 个投票

Hi, i have an array
n= [1,19,15,18,21,10,16,2,7,3,17,6,1;
1,14,20,8,4,13,9,5,1,0,0,0,0;
1,11,12,1,0,0,0,0,0,0,0,0,0];
What i need to do is to randomly choose 2 elements of each row (but not the ones and the zeros) and rotate the vector in between them .
For example new_n= [ 1,19,15,17,3,7,2,16,10,21,18,6,1;
1,14,13,4,8,20,9,5,1,0,0,0,0;
1,12,11,1,0,0,0,0,0,0,0,0,0]
If it is possible i would like to be able to do it for any array i might have. (the thing is that 1-etc-1 in each row are non equal paths that start at depot 1 and end at it after they go through the other nodes, so i want the rearrangement to happen between the depots)

4 个评论
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Adam Danz
Adam Danz 2020-1-12
You lost me at "rotate the vector in between them".
Nikoleta
Nikoleta 2020-1-12
编辑：Nikoleta 2020-1-12
Like at the example i gave. It is 1,19,15,18,21,10,16,2,7,3,17,6,1 but the two nodes that are chosen (18 and 17) are the edges of a vector that gets rotaded so it becomes 1,19,15,17,3,7,2,16,10,21,18,6,1.
Adam Danz
Adam Danz 2020-1-12
I see. The next point that needs clarifying is ,"'randomly choose 2 elements of each row (but not the ones and the zeros) ",
Does that mean that any value can be chosen except 1s and 0s or does it mean that the vector between the two end points must not include any 1s and 0s?
For example, is it OK if these two points are chosen?
[0 0 0 0 9 8 7 6 5 4 3 2 1 1 1 1 2 3 4]
% ^...........^
Nikoleta
Nikoleta 2020-1-12
No this is not ok, it means that the two end points should be limited by the first and the last 1. So no change will happen on these elements [1 _____1 0 0 0 0].

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0 个投票

This may solve your problem:
rng(42)
n= [1,19,15,18,21,10,16,2,7,3,17,6,1;
1,14,20,8,4,13,9,5,1,0,0,0,0;
1,11,12,1,0,0,0,0,0,0,0,0,0];
nrev = n;
for idx=1:size(n,1)
% Get the position of the initial 0
Ncolums = find(n(idx,:)==0);
if isempty(Ncolums)
Ncolums = size(n,2);
else
Ncolums = Ncolums(1)-1;
end
Ncolums = Ncolums-2; % Remove first and last 1
% Get a random index permutation between valid index
Index = randperm(Ncolums)+1; % Add offset so it starts by two
Index = sort(Index(1:2));
% Invert row
nrev(idx,Index(1):Index(2)) = n(idx,Index(2):-1:Index(1));
end
nrev
nrev =
1 19 15 18 21 10 16 6 17 3 7 2 1
1 14 20 8 13 4 9 5 1 0 0 0 0
1 12 11 1 0 0 0 0 0 0 0 0 0

3 个评论
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Nikoleta
Nikoleta 2020-1-12
This is really helpful,yes, thank you very much!
Nikoleta
Nikoleta 2020-1-15
Could you help me with something on your solution?
After i end up with the nrev, i have to estimate the cost according to another array ( For example from node 1 to 3 the cost is 11, from 2 to 6 it is 32 etc) and compare it to n's cost. The thing is that i want to run this random change and then the comparison,20 times . (like for random 1:20).In which part of the code should i add it?
Adam Danz
Adam Danz 2020-1-15
The description is unclear. Perhaps a full example would be helpful.

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更多回答（1 个）

Adam Danz
Adam Danz 2020-1-12
编辑：Adam Danz 2020-1-12

0 个投票

Simpler
n= [1,19,15,18,21,10,16,2,7,3,17,6,1;
1,14,20,8,4,13,9,5,1,0,0,0,0;
1,11,12,1,0,0,0,0,0,0,0,0,0];
% randomly choose 2 elements of each row
% (but not the ones and the zeros)
randIdx = arrayfun(@(i)sort(randsample(find(~ismember(n(i,:),[1,0])),2)),1:size(n,1),'UniformOutput',false);
% rotate the vector in between each index, per row
for i = 1:size(n,1)
n(i, randIdx{i}) = fliplr(n(i, randIdx{i}));
end

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