为什么 numba 执行 numpy 计算比执行正常 python 代码花费更长的时间？

Question

parallel.py 是一个 python 文件，它使用 numba 和 numpy 来计算两个矩阵的对角线之和。这里的主要目的是找到使用 numba 的执行速度。 parallel.py 大约需要 0.55 秒才能完成执行，而另一个文件 (sequencial.py) 中的相同代码，用纯 python 编写需要 0.00 秒才能完成解决同样的问题，就是这样讽刺的。我不确定我是否很好地利用了 numba，有人可以建议我需要做什么来实现我的 objective.

parallel.py 从 numba 导入 jit，njit 将 numpy 导入为 np 导入时间

@jit(nopython=True)
def create_matrix(row, col):
    arr = np.zeros((row, col))
    for i in range(row):
        for j in range(1, col + 1):
            arr[i, j - 1] = j + (col * i)
    return arr


print("FIND THE SUM OF PRIMARY DIAGONALS OF ANY TWO MATRICES: ")

start = time.perf_counter()

# calculate the sum of primary diagonals of matrix1
m1 = create_matrix(4, 4)  # you can adjust the size of the matrix by changing the row and column in brackets
print(f"Matrix 1 : {m1}")
print(f"Matrix 1 diagonal: {np.diagonal(m1)}")
print(f"Matrix 1 sum of primary diagonal is : {np.trace(m1)}")
mat1_sum = np.trace(m1)

# calculate the sum of primary diagonals of matrix2
m2 = create_matrix(4, 4)  # you can adjust the size of the matrix by changing the row and column in brackets
print(f"Matrix 2 : {m2}")
print(f"Matrix 2 diagonal : {np.diagonal(m2)}")
print(f"Matrix 2 Sum of diagonal is : {np.trace(m2)}")
mat2_sum = np.trace(m2, dtype='i')

sum_of_two_diagonals = mat1_sum + mat2_sum
print(f"THE SUM IS :  {sum_of_two_diagonals}")

finish = time.perf_counter()
print(f"Finished in {round(finish - start, 2)} seconds(s)")

sequencial.py

import numpy as np
import time

def create_matrix(row, col):
    arr = np.zeros((row, col))
    for i in range(row):
        for j in range(1, col + 1):
            arr[i, j - 1] = j + (col * i)
    return arr

print("FIND THE SUM OF PRIMARY DIAGONALS OF ANY TWO MATRICES: ")

start =  time.perf_counter()

# calculate the sum of primary diagonals of matrix1
mat_1 = create_matrix(4, 4) # you can adjust the size of the matrix by changing the row and column in brackets
print(f"Matrix 1 : {mat_1}")
mat1_sum_of_primary_diagonal = 0
for i in range(len(mat_1)):
    for j in range(len(mat_1[i])):
        if i == j:
             print(mat_1[i][j])
             mat1_sum_of_primary_diagonal = mat1_sum_of_primary_diagonal + mat_1[i][j]

print(f"Matrix 1 sum of diagnals is: {mat1_sum_of_primary_diagonal}")

 # calculate the sum of primary diagonals of matrix2
mat_2 = create_matrix(4, 4) # you can adjust the size of the matrix by changing the row and column in brackets
print(f"Matrix 1 : {mat_2}")
mat2_sum_of_primary_diagonal = 0
for i in range(len(mat_2)):
    for j in range(len(mat_2[i])):
        if i == j:
             print(mat_2[i][j])
             mat2_sum_of_primary_diagonal = mat2_sum_of_primary_diagonal + mat_2[i][j]

print(f"Matrix 1 sum of diagnals is: {mat2_sum_of_primary_diagonal}")

diagonals_total = mat1_sum_of_primary_diagonal + mat2_sum_of_primary_diagonal
print(f"THE SUM IS :  {diagonals_total}")

finish = time.perf_counter()
print(f"Finished in {round(finish - start, 2)} seconds(s)")

Answer 1

Numba 函数的编译时间包含在基准测试中，因为 Numba 使用 lazy compilation。您可以只指定函数参数的类型来急切地编译它。或者，您可以运行两次基准测试，只考虑第二次运行.

这是一个例子：

import numba as nb

@nb.njit('float64[:,::1](int_, int_)')
def create_matrix(row, col):
    arr = np.zeros((row, col))
    for i in range(row):
        for j in range(1, col + 1):
            arr[i, j - 1] = j + (col * i)
    return arr

此外，请注意，最好不要在基准计时中包含 print 调用（因为时间可能不稳定，这可能不是您想要衡量的）。更不用说打印东西通常很慢（与基本计算相比）。

最后，请注意该脚本名为“parallel.py”，但不应并行执行任何操作，因为默认情况下 Numba 不会并行化代码（并且由于开销，它在您的情况下会更慢创建线程）。

为什么 numba 执行 numpy 计算比执行正常 python 代码花费更长的时间？

Why is numba taking longer time to execute numpy calculations than executing normal python code?

python

arrays

numpy

numba