Numba @jit(nopython=True) 函数对重型 Numpy 函数没有提供速度改进

Question

我目前运行test_matrix_speed()想看看我的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。使用 Numba @jit(nopython=True) 装饰器对该函数的性能没有影响。这是因为没有改进，Numpy 在这里运行得尽可能快吗？（我不关心 generate_searches 函数的速度）

这是我的代码运行

import random

import numpy as np
from numba import jit


def 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 False


def 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，没有 jit 版本大约为 7.5s。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，特别是显式 for 循环。在你的代码中有 none，你只使用矢量化操作。因此，我预计 jit 不会优于基线解决方案，但我必须承认，我很惊讶地看到它更糟。如果您希望优化您的解决方案，您可以使用以下代码缩短执行时间（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

并通过将 matrix 初始化为 np.zeros(shape, dtype=np.bool)，而不是默认的 float64。我能够获得大约 3.8 秒的执行时间，比你的 unjited 解决方案提高了 ~50%，比 jited 版本提高了~70%。希望对您有所帮助。

Numba @jit(nopython=True) 函数对重型 Numpy 函数没有提供速度改进

Numba @jit(nopython=True) function offers no speed improvement on heavy Numpy function

python

performance

numpy

numba