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.

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.

[Walter Roberson]

What are the elements swapped with

Please give an example showing how the swaps proceed.

[KSSV]

Can you give the expected output for the present example given?

[amit chatterjee]

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]

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.

Answer 1

[Jan]

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));

[amit chatterjee]

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].

[Jan]

@amit: Did you read my suggestion?

[nitin jain]

there is no shuffle function in matlab so how to do swapping in all z==1 condition?

[Stephen23]

@nitin jain: read Jan Simon's answer. The solution is there.

[nitin jain]

@stephen the answer provided by jan simons does not guarantee swaping in x[z==1] every time.As in matlab when i use shuffle it show error as undefined function or variable.

[Stephen23]

@nitin jain: the solution to your question is already in Jan Simon's answer: Shuffle is not an inbuilt function, but you can download it from MATLAB File Exchange. To get the file follow the link that Jan Simon gave FEX: shuffle. Click the big blue Download button. Unzip the zip file into your current directory, and then compile the mex file:

https://www.mathworks.com/help/matlab/matlab_external/what-you-need-to-build-mex-files.html

You will also find plenty of other shuffle type functions on FEX:

https://www.mathworks.com/matlabcentral/fileexchange/?term=shuffle

[nitin jain]

@stephen i don't have c/c++ compiler and unable to build mex file,is there is any other alternative to Shuffle to desired random swapping of element of x?

[Stephen23]

@nitin jain: what you are looking for is called a derangement, where the index values obey idx(j)~=j. It is not trivial to generate a random derangement efficiently and without bias, but here is one brute-force and totally inefficient way to generate a random derangement:

function idx = derange(n)
assert(n>1,'Input n must be greater than one.')
idx = randperm(n);
while any(idx==1:n)
    idx = randperm(n);
end
end

And simply use it with Jan Simon's answer:

p = derange(n)

[nitin jain]

@stephan thank you for your answer, it is working for me.

[Jan]

@Stephen: Just a note: idx = randperm(n,n) is more efficient with the 2nd input.