Numba @jit(nopython=True) 函数对繁重的 Numpy 函数没有提供速度提升

Question

我目前正在运行test_matrix_speed()以查看我的search_and_book_availability功能有多快。使用 PyCharm 分析器，我可以看到每个search_and_book_availability函数调用的平均速度为 0.001 毫秒。拥有 Numba@jit(nopython=True)装饰器对该函数的性能没有影响。这是因为没有任何改进并且 Numpy 在这里运行得尽可能快吗？（我不在乎generate_searches函数的速度）这是我正在运行的代码import randomimport numpy as npfrom numba import jitdef generate_searches(number, sim_start, sim_end):    searches = []    for i in range(number):        start_slot = random.randint(sim_start, sim_end - 1)        end_slot = random.randint(start_slot + 1, sim_end)        searches.append((start_slot, end_slot))    return searches@jit(nopython=True)def search_and_book_availability(matrix, search_start, search_end):    search_slice = matrix[:, search_start:search_end]    output = np.where(np.sum(search_slice, axis=1) == 0)[0]    number_of_bookable_vecs = output.size    if number_of_bookable_vecs > 0:        if number_of_bookable_vecs == 1:            id_to_book = output[0]        else:            id_to_book = np.random.choice(output)        matrix[id_to_book, search_start:search_end] = 1        return True    else:        return Falsedef test_matrix_speed():    shape = (10, 1440)    matrix = np.zeros(shape)    sim_start = 0    sim_end = 1440    searches = generate_searches(1000000, sim_start, sim_end)    for i in searches:        search_start = i[0]        search_end = i[1]        availability = search_and_book_availability(matrix, search_start, search_end)

Answer 1

使用您的函数和以下代码来分析速度

import time

shape = (10, 1440)

matrix = np.zeros(shape)

sim_start = 0

sim_end = 1440

searches = generate_searches(1000000, sim_start, sim_end)

def reset():

    matrix[:] = 0

def test_matrix_speed():

    for i in searches:

        search_start = i[0]

        search_end = i[1]

        availability = search_and_book_availability(matrix, search_start, search_end)

def timeit(func):

    # warmup

    reset()

    func()

    reset()

    start = time.time()

    func()

    end = time.time()

    return end - start

print(timeit(test_matrix_speed))

我发现jited 版本大约为 11.5s，而没有jit. 我不是 numba 方面的专家，但它的目的是优化以非矢量化方式编写的数字代码，尤其是显式for循环。在您的代码中没有，您只使用矢量化操作。因此，我预计jit不会超过基线解决方案，但我必须承认，我很惊讶地看到它更糟。如果您想优化您的解决方案，您可以使用以下代码减少执行时间（至少在我的 PC 上）：

def search_and_book_availability_opt(matrix, search_start, search_end):

    search_slice = matrix[:, search_start:search_end]

    # we don't need to sum in order to check if all elements are 0.

    # ndarray.any() can use short-circuiting and is therefore faster.

    # Also, we don't need the selected values from np.where, only the

    # indexes, so np.nonzero is faster

    bookable, = np.nonzero(~search_slice.any(axis=1))

    # short circuit

    if bookable.size == 0:

        return False

    # we can perform random choice even if size is 1

    id_to_book = np.random.choice(bookable)

    matrix[id_to_book, search_start:search_end] = 1

    return True

并通过初始化matrix为np.zeros(shape, dtype=np.bool)，而不是默认值float64。我能够获得大约 3.8 秒的执行时间，比您的 unjited 解决方案提高了约 50%，比 jted 版本提高了约 70%。希望有帮助。