I want to shift vector values one by one to the left

Hello everyone,

I have a binary vector with five 0 and three 1.

num=[1 1 1 0 0 0 0 0]

and I want to shift each 1 left, shift one value as

num=[1 1 0 1 0 0 0 0]

untill I get a complete shift of the vector values and printing of each vector shift

num=[0 0 0 0 0 1 1 1]

any helpfull code of the above program with nested for loop will be highly appreciated

Thanks

[Bruno Luong]

=> direction is on the right to my book.

[John D'Errico]

PLEASE STOP POSTING MULTIPLE TIMES. You have posted the exact same question now three times. One I have now closed.

Answer 1

[Chunru]

num=[1 1 1 0 0 0 0 0]

num = 1×8

1 1 1 0 0 0 0 0

for i=1:3
    num = circshift(num, -1)
end

num = 1×8

1 1 0 0 0 0 0 1

num = 1×8

1 0 0 0 0 0 1 1

num = 1×8

0 0 0 0 0 1 1 1

Answer 2

[Abderrahim. B]

What about this:

num = [1 1 1 1 1 0 0 0 0 0]  ;

for ii = 1: nnz(num)
 num = circshift(num, -1)
end

num = 1×10

1 1 1 1 0 0 0 0 0 1

num = 1×10

1 1 1 0 0 0 0 0 1 1

num = 1×10

1 1 0 0 0 0 0 1 1 1

num = 1×10

1 0 0 0 0 0 1 1 1 1

num = 1×10

0 0 0 0 0 1 1 1 1 1

[Muhammad]

Thank you for your reply, but we need to use for loop as circular shift doesnot meet our random data vectors, eg if our vector is

num = [1 1 0 1 0 1 0 1 0 0]

then using circular shift will not move 1s to left and zeros to right, we need to shift every 1 on the left most side

num = [1 1 0 1 0 1 0 0 0 1]

num = [1 1 0 1 0 0 0 0 1 1]

num = [1 1 0 0 0 0 0 1 1 1]

num = [1 0 0 0 0 0 1 1 1 1]

num = [0 0 0 0 0 1 1 1 1 1]

like this. thanks in advance

[Chunru]

Do you get a rule of shifting data?

[Muhammad]

I already had an idea of shifting but it does not fullfill the requirement that i want.

[Abderrahim. B]

if you only need the last vector,then use sort function.

sort(yourVectorHere)

Answer 3

[Bruno Luong]

Is it what you want?

num = [1 1 0 1 0 1 0 1 0 0];
j = find(num);
m = length(j);
l = (-m+1:0)+(length(num))-j;
q = sum(l)+1;
J = zeros(q,m);
i = 1;
J(i,:) = j;
for k=m:-1:1
    for n=1:l(k)
        i = i+1;
        j(k) = j(k)+1;
        J(i,:) = j;
    end
end
I = repmat((1:q)',1,m);
B = accumarray([I(:) J(:)],1)

B = 20×10

1 1 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 1 0 1 1 0 1 0 1 0 0 0 1 1 1 0 1 0 0 1 0 0 1 1 1 0 1 0 0 0 1 0 1 1 1 0 1 0 0 0 0 1 1 1 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 0 1 1 1 1 0 0 0 0 1 0 1 1 1 1 0 0 0 0 0 1 1 1

Answer 4

[Bruno Luong]

May be this?

num = [1 1 0 1 0 1 0 1 0 0];
j = find(num);
m = length(j);
J=repmat(j,m,1);
for i=m:-1:1
    J(m-i+1:end,i) = length(num)+i-m;
end
J = [j; J];
I = repmat((1:m+1)',1,m);
B = accumarray([I(:) J(:)],1)

B = 6×10

1 1 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 0 1 1 1 0 1 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 0 0 0 0 0 1 1 1 1 1