Python提升循环速度的高效方法小姐

# 普通循环
squares = []
for i in range(10):
    squares.append(i ** 2)


# 使用列表推导式
squares = [i ** 2 for i in range(10)]

# 列表推导式
squares = [i ** 2 for i in range(10)]

# 生成器表达式
squares_gen = (i ** 2 for i in range(10))

import numpy as np

# 使用NumPy数组进行循环
array = np.arange(10)
squares = array ** 2

import numpy as np

# 使用向量化操作计算平方和
array = np.arange(10)
squares_sum = np.sum(array ** 2)

from joblib import Parallel, delayed

# 使用joblib库进行并行计算
def square(x):
    return x ** 2

result = Parallel(n_jobs=-1)(delayed(square)(i) for i in range(10))

# 使用Cython优化循环代码
# 编写并编译Cython代码，然后在Python中调用

# 使用Numba优化循环代码
from numba import njit@njit
def square(x):
 

  return x ** 2

result = [square(i) for i in range(10)]

import dask.array as da

# 使用Dask库进行并行计算
array = da.arange(10)
squares = array ** 2

import asyncio

# 使用asyncio库进行异步IO操作
async def square(x):
    return x ** 2

async def main():
    tasks = [square(i) for i in range(10)]
    results = await asyncio.gather(*tasks)
    print(results)

asyncio.run(main())

# 使用Cython优化循环代码
# 1. 编写并编译Cython代码
# 在文件 square.pyx 中编写以下代码
# cython: language_level=3
def square_sum(n):
    cdef int i
    cdef int result = 0
    for i in range(n):
        result += i ** 2
    return result

cythonize -i square.pyx

# 2. 在Python中调用编译好的Cython函数
import square

result = square.square_sum(10)
print(result)

# 使用Numba优化循环代码
from numba import jit

@jit
def square_sum(n):
    result = 0
    for i in range(n):
        result += i ** 2
    return result

result = square_sum(10)
print(result)