有没有办法通过浮点精度或限制浮点范围来加速Python？

发布于 2025-01-10 21:20:10 字数 69 浏览 0 评论 0原文

在 glsl 中，可以使用低精度浮点数来加快计算速度。有没有办法限制 python 中浮点数的精度或范围以加快数学计算速度？

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ゝ杯具 2025-01-17 21:20:10

您可以使用 numba.njit(fastmath=True) 并下载 Intel SVML 来加速计算，当然，它使用不安全的浮点。您可以在此处查看纪录片 Numba Just -及时编译

from numba import njit

# Without njit(fastmath=True)
def math_multiplication_power_loop(x,y):
    number = []
    
    for i in range(40):
        number.append((x*y)**i)
    
    return sum(number)


# With njit(fastmath=True)
@njit(fastmath=True)
def math_multiplication_power_loop_njit(x,y):
    number = []
    
    for i in range(40):
        number.append((x*y)**i)
    
    return sum(number)

x = 12344
y = 5567.8


%timeit math_multiplication_power_loop(x,y)
13.1 µs ± 189 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit math_multiplication_power_loop_njit(x,y)
1.29 µs ± 21.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

You can use numba.njit(fastmath=True) and download Intel SVML to speed up the calculation, which of course in return, it uses unsafe floating points. You can check the documentaries here Numba Just-In-Time Compilation

from numba import njit

# Without njit(fastmath=True)
def math_multiplication_power_loop(x,y):
    number = []
    
    for i in range(40):
        number.append((x*y)**i)
    
    return sum(number)


# With njit(fastmath=True)
@njit(fastmath=True)
def math_multiplication_power_loop_njit(x,y):
    number = []
    
    for i in range(40):
        number.append((x*y)**i)
    
    return sum(number)

x = 12344
y = 5567.8


%timeit math_multiplication_power_loop(x,y)
13.1 µs ± 189 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

%timeit math_multiplication_power_loop_njit(x,y)
1.29 µs ± 21.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

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