python - python 3 的 weave 替代品

Question

我有一个来自 here 的函数它使用weave。我是否可以在 Python 3 中运行此代码而无需重写？

代码:

def _thinningIteration(im, iter):
    I, M = im, np.zeros(im.shape, np.uint8)
    expr = """
    for (int i = 1; i < NI[0]-1; i++) {
        for (int j = 1; j < NI[1]-1; j++) {
            int p2 = I2(i-1, j);
            int p3 = I2(i-1, j+1);
            int p4 = I2(i, j+1);
            int p5 = I2(i+1, j+1);
            int p6 = I2(i+1, j);
            int p7 = I2(i+1, j-1);
            int p8 = I2(i, j-1);
            int p9 = I2(i-1, j-1);
            int A  = (p2 == 0 && p3 == 1) + (p3 == 0 && p4 == 1) +
                     (p4 == 0 && p5 == 1) + (p5 == 0 && p6 == 1) +
                     (p6 == 0 && p7 == 1) + (p7 == 0 && p8 == 1) +
                     (p8 == 0 && p9 == 1) + (p9 == 0 && p2 == 1);
            int B  = p2 + p3 + p4 + p5 + p6 + p7 + p8 + p9;
            int m1 = iter == 0 ? (p2 * p4 * p6) : (p2 * p4 * p8);
            int m2 = iter == 0 ? (p4 * p6 * p8) : (p2 * p6 * p8);
            if (A == 1 && B >= 2 && B <= 6 && m1 == 0 && m2 == 0) {
                M2(i,j) = 1;
            }
        }
    } 
    """

    weave.inline(expr, ["I", "iter", "M"])
    return (I & ~M)

Answer 1

Numba可以作为替代解决方案。它即时编译Python代码，并且用法非常简洁。

上面的代码可以用Python用装饰器重写:

from numba import jit

@jit
def _thinningIteration(im, iter_):
    M = np.zeros(im.shape, np.uint8)
    h, w = im.shape
    for i in range(1, h - 1):
        for j in range(1, w - 1):
            p2 = im[i - 1, j]
            p3 = im[i - 1, j + 1]
            p4 = im[i, j + 1]
            p5 = im[i + 1, j + 1]
            p6 = im[i + 1, j]
            p7 = im[i + 1, j - 1]
            p8 = im[i, j - 1]
            p9 = im[i - 1, j - 1]
            A = (p2 == 0 and p3 == 1) + (p3 == 0 and p4 == 1) + \
                (p4 == 0 and p5 == 1) + (p5 == 0 and p6 == 1) + \
                (p6 == 0 and p7 == 1) + (p7 == 0 and p8 == 1) + \
                (p8 == 0 and p9 == 1) + (p9 == 0 and p2 == 1)
            B = p2 + p3 + p4 + p5 + p6 + p7 + p8 + p9
            m1 = (p2 * p4 * p6) if (iter_ == 0) else (p2 * p4 * p8)
            m2 = (p4 * p6 * p8) if (iter_ == 0) else (p2 * p6 * p8)
            if A == 1 and B >= 2 and B <=6 and m1 == 0 and m2 == 0:
                M[i, j] = 1

    return im & ~M

它可以在Python-2.x和Python-3.x中执行，性能接近C。查看this了解更多详情。