Usage of parallel option in numba.jit decoratior m

Question

Given two opposite corners of a rectangle (x1, y1) and (x2, y2) and two radii r1 and r2, find the ratio of points that lie between the circles defined by the radii r1 and r2 to the total number of points in the rectangle.

Simple NumPy approach:

def func_1(x1,y1,x2,y2,r1,r2,n):
     x11,y11 = np.meshgrid(np.linspace(x1,x2,n),np.linspace(y1,y2,n))
     z1 = np.sqrt(x11**2+y11**2)
     a = np.where((z1>(r1)) & (z1<(r2)))
     fill_factor = len(a[0])/(n*n)
     return fill_factor

Next I tried to optimize this function with the jit decorator from numba. When I use:

nopython = True

The function is faster and gives the right output. But when I also add:

parallel = True

The function is faster but gives the wrong result. I know that this has something to do with my z matrix since that is not being updated properly.

@jit(nopython=True,parallel=True)
def func_2(x1,y1,x2,y2,r1,r2,n):
    x_ = np.linspace(x1,x2,n)
    y_ = np.linspace(y1,y2,n)
    z1 = np.zeros((n,n))
    for i in range(n):
        for j in range(n):
            z1[i][j] = np.sqrt((x_[i]*x_[i]+y_[j]*y_[j]))
    a = np.where((z1>(r1)) & (z1<(r2)))
    fill_factor = len(a[0])/(n*n)
    return fill_factor

Test values :

x1 = 1.0
x2 = -1.0
y1 = 1.0
y2 = -1.0
r1 = 0.5
r2 = 0.75
n = 25000

Additional info : Python version : 3.6.1, Numba version : 0.34.0+5.g1762237, NumPy version : 1.13.1

Answer 1

The problem with parallel=True is that it's a black-box. Numba doesn't even guarantee that it will actually parallelize anything. It uses heuristics to find out if it's parallelizable and what could be done in parallel. These can fail and in your example they do fail, just like in my experiments with parallel and numba. That makes parallel untrustworthy and I would advise against using it!

In newer versions (0.34) prange was added an you could have more luck with that. It can't be applied in this case because prange works like range and that's different from np.linspace...

Just a note: You can avoid building z and doing the np.where in your function completely, you could just do the checks explicitly:

import numpy as np
import numba as nb

@nb.njit   # equivalent to "jit(nopython=True)".
def func_2(x1,y1,x2,y2,r1,r2,n):
    x_ = np.linspace(x1,x2,n)
    y_ = np.linspace(y1,y2,n)
    cnts = 0
    for i in range(n):
        for j in range(n):
            z = np.sqrt(x_[i] * x_[i] + y_[j] * y_[j])
            if r1 < z < r2:
                cnts += 1
    fill_factor = cnts/(n*n)
    return fill_factor

That should also provide some speedup compared to your function, maybe even more than using parallel=True (if it would work correctly).