how to avoid repeat numbers in a row
3 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
imola
am 19 Mär. 2015
Kommentiert: imola
am 19 Mär. 2015
Dear All,
I have this row
A=[1 2 3 4 5 6];
I tried to find random permutation on the vector but in many cases I got a row with repetition numbers for example
z=[1 5 2 4 1 5];or z=[1 5 2 4 1 1];
I want to avoid the repetition in this code but it is totally wrong, at first I should assign the missing numbers in such way, here is [3 6].
zc=[3 6 ];
e=size(zc,2);
z=[1 5 2 4 1 5];
for i=1:6
z(i)==z
s = find((z(i)==z) == 1)
ss=size(s,2);
for j=2:ss
for k=1:e
z(j)=zc(k)
end
end
end
I don't want to use a function from matlab to get the permutation, I need to fix my row or any permutation exchange file like (perm file ). If anyone help me for it and save me I will be grateful.
regards,
Imola
2 Kommentare
Guillaume
am 19 Mär. 2015
I don't want to use a function from matlab to get the permutation:
Does that mean that you've written your own uniform number generator? Otherwise, at some point you'll have to use rand, so you just might as well use randperm.
Akzeptierte Antwort
Weitere Antworten (1)
Guillaume
am 19 Mär. 2015
It seems to me that your effort would be better spent on fixing the code that generate these invalid permutation rather than fixing the permutation afterward.
Anyway, one way to do what you want:
z = [1 5 2 4 1 1];
zc = [3 6];
%build a cell array containing the indices of repeated values:
repindices = accumarray(z', 1:numel(z), [], @(v) {v(2:end)'})';
%convert cell array into vector:
repindices = [repindices{:}];
%replace values at indices by replacement:
z(repindices) = zc
Siehe auch
Kategorien
Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!