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:
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.
Most FORTRAN compilers pass arrays by passing the address of the array, in this case all subprograms that use the array will then work on the same (and only) copy. The FORTRAN standard allows array passing by another method called copying in/out or copy/restore.
Unlike C and C++, Fortran can handle returning arrays and composite types with no difficulty.
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
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