python - Numba 中的笛卡尔积

Question

我的代码使用如下列表的笛卡尔积:

import itertools

cartesian_product = itertools.product(list('ABCDEF'), repeat=n)

n可以是 0 到 4 之间的任何值。
numba当前不支持itertools.product .到目前为止，我一直无法想出一个可行的替代方案。欢迎任何建议!

Answer 1

这是解决您的问题的一种方法。有两个 numba 函数:一个用于常规笛卡尔积(数组列表)，另一个用于简单数组的重复参数。这是对一些代码的改编 from another one of my SO answers .

@nb.njit(nb.int32[:,:](nb.int32[:]))
def cproduct_idx(sizes: np.ndarray):
    """Generates ids tuples for a cartesian product"""
    assert len(sizes) >= 2
    tuples_count  = np.prod(sizes)
    tuples = np.zeros((tuples_count, len(sizes)), dtype=np.int32)
    tuple_idx = 0
    # stores the current combination
    current_tuple = np.zeros(len(sizes))
    while tuple_idx < tuples_count:
        tuples[tuple_idx] = current_tuple
        current_tuple[0] += 1
        # using a condition here instead of including this in the inner loop
        # to gain a bit of speed: this is going to be tested each iteration,
        # and starting a loop to have it end right away is a bit silly
        if current_tuple[0] == sizes[0]:
            # the reset to 0 and subsequent increment amount to carrying
            # the number to the higher "power"
            current_tuple[0] = 0
            current_tuple[1] += 1
            for i in range(1, len(sizes) - 1):
                if current_tuple[i] == sizes[i]:
                    # same as before, but in a loop, since this is going
                    # to get run less often
                    current_tuple[i + 1] += 1
                    current_tuple[i] = 0
                else:
                    break
        tuple_idx += 1
    return tuples

@nb.njit
def cartesian_product(*arrays):
    sizes = [len(a) for a in arrays]
    sizes = np.asarray(sizes, dtype=np.int32)
    tuples_count  = np.prod(sizes)
    array_ids = cproduct_idx(sizes)
    tuples = np.zeros((tuples_count, len(sizes)))
    for i in range(len(arrays)):
        tuples[:, i] = arrays[i][array_ids[:, i]]
    return tuples

@nb.njit
def cartesian_product_repeat(array, repeat):
    sizes = [len(array) for _ in range(repeat)]
    sizes = np.asarray(sizes, dtype=np.int32)
    tuples_count  = np.prod(sizes)
    array_ids = cproduct_idx(sizes)
    tuples = np.zeros((tuples_count, len(sizes)))
    for i in range(repeat):
        tuples[:, i] = array[array_ids[:, i]]
    return tuples

这是最后一个函数执行的示例:

>>> cartesian_product_repeat(np.arange(2), 3)

array([[0., 0., 0.],
       [1., 0., 0.],
       [0., 1., 0.],
       [1., 1., 0.],
       [0., 0., 1.],
       [1., 0., 1.],
       [0., 1., 1.],
       [1., 1., 1.]])