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# More Interesting Examples with f2py

Let us now look at some more interesting examples using f2py. We will implement a simple iterative method for solving laplace’s equation in a square. Actually, this implementation is taken from http://www.scipy.org/PerformancePython?highlight=%28performance%29 page on the scipy website. It has lots of information on implementing numerical algorithms in python.

The following fortran code implements a single iteration of a relaxation method for solving Laplace’s equation in a square.

    %fortran
          subroutine timestep(u,n,m,dx,dy,error)
          double precision u(n,m)
          double precision dx,dy,dx2,dy2,dnr_inv,tmp,diff
          integer n,m,i,j
    cf2py intent(in) :: dx,dy
    cf2py intent(in,out) :: u
    cf2py intent(out) :: error
    cf2py intent(hide) :: n,m
          dx2 = dx*dx
          dy2 = dy*dy
          dnr_inv = 0.5d0 / (dx2+dy2)
          error = 0d0
          do 200,j=2,m-1
             do 100,i=2,n-1
                tmp = u(i,j)
                u(i,j) = ((u(i-1,j) + u(i+1,j))*dy2+
         &           (u(i,j-1) + u(i,j+1))*dx2)*dnr_inv
                diff = u(i,j) - tmp
                error = error + diff*diff
     100     continue
     200  continue
          error = sqrt(error)
          end

If you do

    timestep?

You find that you need pass timestep a numpy array u, and the grid spacing dx,dy and it will return the updated u, together with an error estimate. To apply this to actually solve a problem use this driver code

    import numpy
    j=numpy.complex(0,1)
    num_points=50
    u=numpy.zeros((num_points,num_points),dtype=float)
    pi_c=float(pi)
    x=numpy.r_[0.0:pi_c:num_points*j]
    u[0,:]=numpy.sin(x)
    u[num_points-1,:]=numpy.sin(x)
    def solve_laplace(u,dx,dy):
       iter =0
       err = 2
       while(iter <10000 and err>1e-6):
          (u,err)=timestep(u,dx,dy)
          iter+=1
       return (u,err,iter)

Now call the routine using

    (sol,err,iter)=solve_laplace(u,pi_c/(num_points-1),pi_c/(num_points-1))

This solves the equation with boundary conditions that the right and left edge of the square are half an oscillation of the sine function. With a 51x51 grid that we are using I find that it takes around .2 s to solve this requiring 2750 iterations. If you have the gnuplot package installed (use optional packages() to find its name and install package to install it), then you can plot this using

    import Gnuplot
    g=Gnuplot.Gnuplot(persist=1)
    g('set parametric')
    g('set data style lines')
    g('set hidden')
    g('set contour base')
    g('set zrange [-.2:1.2]')
    data=Gnuplot.GridData(sol,x,x,binary=0)
    g.splot(data)

To see what we have gained by using f2py let us compare the same algorithm in pure python and a vectorized version using numpy arrays.

    def slowTimeStep(u,dx,dy):
        """Takes a time step using straight forward Python loops."""
        nx, ny = u.shape
        dx2, dy2 = dx**2, dy**2
        dnr_inv = 0.5/(dx2 + dy2)

        err = 0.0
        for i in range(1, nx-1):
            for j in range(1, ny-1):
                tmp = u[i,j]
                u[i,j] = ((u[i-1, j] + u[i+1, j])*dy2 +
                          (u[i, j-1] + u[i, j+1])*dx2)*dnr_inv
                diff = u[i,j] - tmp
                err += diff*diff

        return u,numpy.sqrt(err)

    def numpyTimeStep(u,dx,dy):
        dx2, dy2 = dx**2, dy**2
        dnr_inv = 0.5/(dx2 + dy2)
        u_old=u.copy()
        # The actual iteration
        u[1:-1, 1:-1] = ((u[0:-2, 1:-1] + u[2:, 1:-1])*dy2 +
                         (u[1:-1,0:-2] + u[1:-1, 2:])*dx2)*dnr_inv
        v = (u - u_old).flat
        return u,numpy.sqrt(numpy.dot(v,v))

You can try these out by changing the timestep function used in our driver routine. The python version is slow even on a 50x50 grid. It takes 70 seconds to solve the system in 3000 iterations. It takes the numpy routine 2 seconds to reach the error tolerance in around 5000 iterations. In contrast it takes the f2py routine around .2 seconds to reach the error tolerance using 3000 iterations. I should point out that the numpy routine is not quite the same algorithm since it is a jacobi iteration while the f2py one is gauss-seidel. This is why the numpy version requires more iterations. Even accounting for this you can see the f2py version appears to be around 5 times faster than the numpy version. Actually if you try this on a 500x500 grid I find that it takes the numpy routine 30 seconds to do 500 iterations while it only takes about 2 seconds for the f2py to do this. So the f2py version is really about 15 times faster. On smaller grids each actual iteration is relatively cheap and so the overhead of calling f2py is more evident, on larger examples where the iteration is expensive, the advantage of f2py is clear. Even on the small example it is still very fast. Note that a 500x500 grid in python would take around half an hour to do 500 iterations.

To my knowledge the fastest that you could implement this algorithm in matlab would be to vectorize it exactly like the numpy routine we have. Vector addition in matlab and numpy are comparable. So unless there is some trick I don’t know about, using f2py you can interactively write code 15 times faster than anything you could write in matlab (Please correct me if I’m wrong). You can actually make the f2py version a little bit faster by using intent(in,out,overwrite) and creating the initial numpy array using order=’FORTRAN’. This eliminates the unnecessary copying that is occurring in memory.

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