python - How to get priorly-unknown array as the output of a function in Fortran

Question

In Python:

def select(x):
    y = []
    for e in x:
        if e!=0:
            y.append(e)
    return y

that works as:

x = [1,0,2,0,0,3]
select(x)
[1,2,3]

to be translated into Fortran:

function select(x,n) result(y)
    implicit none
    integer:: x(n),n,i,j,y(?)
    j = 0
    do i=1,n
        if (x(i)/=0) then
            j = j+1
            y(j) = x(i)
        endif
    enddo
end function

The questions are in Fortran:

1. how to declare y(?)?
2. how to declare predefined values for x
3. how to avoid dimension info n

for 1 if it is defined as y(n) the output will be:

x = (/1,0,2,0,0,3/)
print *,select(x,6)
1,2,3,0,0,0

which is not desired!
!-------------------------------
Comments:
1- All given answers are useful in this post. Specially M.S.B and eryksun's.
2- I tried to adapt the ideas for my problem and compile with F2Py however it was not successful. I had already debugged them using GFortran and all were successful. It might be a bug in F2Py or something that I don't know about using it properly. I will try to cover this issue in another post.

Update: A linked question could be found at here.

Answer 1

I hope a real Fortran programmer comes along, but in the absence of better advice, I would only specify the shape and not the size of x(:), use a temporary array temp(size(x)), and make the output y allocatable. Then after the first pass, allocate(y(j)) and copy the values from the temporary array. But I can't stress enough that I'm not a Fortran programmer, so I can't say if the language has a growable array or if a library exists for the latter.

program test
    implicit none
    integer:: x(10) = (/1,0,2,0,3,0,4,0,5,0/)
    print "(10I2.1)", select(x)

contains

    function select(x) result(y)
        implicit none
        integer, intent(in):: x(:) 
        integer:: i, j, temp(size(x))
        integer, allocatable:: y(:)

        j = 0
        do i = 1, size(x)
            if (x(i) /= 0) then
                j = j + 1
                temp(j) = x(i)
            endif
        enddo

        allocate(y(j))
        y = temp(:j)
    end function select

end program test

---

Edit:

Based on M.S.B.'s answer, here's a revised version of the function that grows temp y with over-allocation. As before it copies the result to y at the end. It turns out i's not necessary to explicitly allocate a new array at the final size. Instead it can be done automatically with assignment.

    function select(x) result(y)
        implicit none
        integer, intent(in):: x(:) 
        integer:: i, j, dsize
        integer, allocatable:: temp(:), y(:)

        dsize = 0; allocate(y(0))

        j = 0
        do i = 1, size(x)
            if (x(i) /= 0) then
                j = j + 1

                if (j >= dsize) then         !grow y using temp
                    dsize = j + j / 8 + 8 
                    allocate(temp(dsize))
                    temp(:size(y)) = y
                    call move_alloc(temp, y) !temp gets deallocated
                endif

                y(j) = x(i)
            endif
        enddo
        y = y(:j)
    end function select