Suppose a row vector x=1:8 and another vector of same size z=[1 1 0 1 1 0 1 1].I want to swap element in x according to element in z if z==1 then swap the postion of element of x else z==0 element of x should be in same position in vector.

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amit chatterjee
amit chatterjee 2017-5-24
评论： Jan 2017-6-19
The number of swap should be equal to number of one in vector z divide by 2 if even ones else plus one divide by 2.The value of z change after each iteration so logic should be universal.
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amit chatterjee
amit chatterjee 2017-5-24
x=[1 2 3 4 5 6 7 8] and element of x randomly swap for example z=[1 1 0 1 1 1 1 0] then x[1] should swap because z[1] is equal to one and x[3] will have same element because z[3] is 0 and swapping is such that no element is repeated in x.e.g x[1] become x[2] and x[2] become x[4].
nitin jain
nitin jain 2017-6-10
The expected output should be like x=[4 6 3 1 7 2 5 8].whenever z==1 Swap the x element at that postion of x randomly while index of x correspond to x==0 should not be swapped.

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Jan
Jan 2017-5-24
编辑：Jan 2017-5-24
Guessing that you want to swap the elements randomly:
x = rand(1, 8); % Test data
z = [1 1 0 1 1 0 1 1];
xf = x(z == 1);
n = sum(z); % Or: numel(xf)
p = randperm(n, n); % Faster than randperm(n)
x(z) = xf(p);
This does not guarantee, that the elements at z==1 are changed. If you need this, try: FEX: Shuffle:
p = Shuffle(n, 'derange', n)
Or if you have installed Shuffle already:
x(z==1) = Shuffle(x(z==1));
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