Python numba.jit types

Question

I have been trying to deduce how the types are set from the numba documentation all day. I have gotten a bit of the way, but now I want to make a function which returns a one-dimensional array, and a two-dimensjonal array, and take a bunch of args, and I struggle to get any further:

@jit
class name(object)
    @double[:,:], double[:](double[:], double, double, int64)
    def solve(self, u0, a, b, n):
        self.t = linspace(a, b, n+1)
        dt = abs((b-a)/float(n))
        u = zeros(n+1, len([u0]))
        u[0] = u0
        u = advance(u, t, n, dt)
        return u.transpose(), t.transpose()   

The above throws these exceptions:

Traceback (most recent call last):
  File "/home/marius/dev/python/inf1100/test_ODE.py", line 2, in <module>
    from DE import *
  File "/home/marius/dev/python/inf1100/DE.py", line 13
    @double[:,:], double[:](double[:], double, double, int64)
           ^
SyntaxError: invalid syntax

It would be good if you could tell me what is going wrong, however it would be even better if you could recommend a document which rigorously explains these syntaxes once and for all.

Thank you for your time.

Kind regards, Marius

Answer 1

Here is a simpler version of a method that returns a tuple. This works for me using Numba 0.11.1 on OS X:

import numba
import numpy as np

@numba.jit
class name(object):
    @numba.object_(numba.double[:], numba.double)
    def solve(self, x, a):
        y = np.empty(x.shape[0], dtype=np.float64)
        z = np.empty(x.shape[0], dtype=np.float64)
        for k in xrange(x.shape[0]):
            y[k] = x[k] * a
            z[k] = x[k] + a

        return y, z 

Then using it:

C = name()
a, b = C.solve(np.arange(5, dtype=np.float64), 3.0)

Where a and b are:

In [24]: a
Out[24]:
array([  0.,   3.,   6.,   9.,  12.])
In [22]: b
Out[22]:
array([ 3.,  4.,  5.,  6.,  7.])