C++,python,热爱算法和机器学习
全部博文(1214)
分类: Python/Ruby
2020-09-11 17:23:16
python 程序性能优化的套路一般有两种:1)jit, 即just in time compiler, 即时编译器,在运行时将某些函数编译成二进程代码,使用这种方式的有:numba 和pypy;2)将python代码转换成c++/c代码,然后编译执行,这种方式有:cython和nuitka。总而言之,转换成c++/c代码以后编译成二进制文件执行的效率比用numba和pypy即时编译执行的效率要高。
1. 首先看一下python写的求质数的函数 以及 用 numba的jit优化的函数
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# main.py
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# 纯python语言写的求质数的代码
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def primes_python(nb_primes):
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p = []
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n = 2
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while len(p) < nb_primes:
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# Is n prime?
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for i in p:
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if n % i == 0:
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break
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# If no break occurred in the loop
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else:
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p.append(n)
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n += 1
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return p
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# 使用numba的jit优化的代码,只需要在上面的函数加一行代码
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from numba import jit
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@jit
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def primes_jit(nb_primes):
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p = []
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n = 2
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while len(p) < nb_primes:
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# Is n prime?
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for i in p:
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if n % i == 0:
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break
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# If no break occurred in the loop
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else:
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p.append(n)
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n += 1
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return p
2. 新建一个primes.pyx文件,写一个cython函数,其中声明了变量的类型
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# primes.pyx
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def primes(int nb_primes):
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cdef int n, i, len_p
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cdef int p[1000]
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if nb_primes > 1000:
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nb_primes = 1000
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len_p = 0 # The current number of elements in p.
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n = 2
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while len_p < nb_primes:
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# Is n prime?
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for i in p[:len_p]:
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if n % i == 0:
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break
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# If no break occurred in the loop, we have a prime.
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else:
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p[len_p] = n
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len_p += 1
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n += 1
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# Let's return the result in a python list:
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result_as_list = [prime for prime in p[:len_p]]
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return result_as_list
再建立一个primes_python.pyx文件,新建一个和之前python里面写的一样的函数,作为对比。
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# primes_python.pyx
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def primes_python(nb_primes):
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p = []
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n = 2
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while len(p) < nb_primes:
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# Is n prime?
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for i in p:
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if n % i == 0:
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break
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# If no break occurred in the loop
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else:
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p.append(n)
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n += 1
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return p
新建setup.py文件,用来编译.pyx函数
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from distutils.core import setup
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from Cython.Build import cythonize
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setup(
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ext_modules=cythonize(["primes.pyx", "primes_python.pyx"],
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annotate=True)
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)
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# 编译命令用这个
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# python setup.py build_ext --inplace
使用python setup.py build_ext --inplace编译后可以得到.pyd文件,就是可以导入的python库了。
3. 修改一下main.py, 加入函数调用和度量
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# main.py 的完整内容
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import primes
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import primes_python
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import timeit
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from numba import jit
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def primes_python(nb_primes):
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p = []
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n = 2
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while len(p) < nb_primes:
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# Is n prime?
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for i in p:
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if n % i == 0:
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break
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# If no break occurred in the loop
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else:
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p.append(n)
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n += 1
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return p
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@jit
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def primes_jit(nb_primes):
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p = []
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n = 2
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while len(p) < nb_primes:
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# Is n prime?
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for i in p:
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if n % i == 0:
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break
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# If no break occurred in the loop
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else:
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p.append(n)
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n += 1
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return p
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if __name__ == "__main__":
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repeat_times = 1000
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t1 = timeit.timeit(stmt="primes_python(1000)",
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setup="from __main__ import primes_python", number=repeat_times)
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print(f"run in python: {t1}s")
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t2 = timeit.timeit(stmt="primes.primes(1000)",
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setup="import primes", number=repeat_times)
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print(f"run cython with cdef: {t2}s")
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t3 = timeit.timeit(stmt="primes_jit(1000)",
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setup="from __main__ import primes_jit", number=repeat_times)
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print(f"run in python with numba jit: {t3}s")
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t4 = timeit.timeit(stmt="primes_python.primes_python(1000)",
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setup="import primes_python", number=repeat_times)
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print(f"run cython without cdef: {t4}s")
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运行一下,得到的结果如下:
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run in python: 28.519053545829927s
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run cython with cdef: 1.6289360376895452s
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run in python with numba jit: 2.0565857326599577s
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run cython without cdef: 13.221758278866588s
4. 测试一下pypy的结果,建立primes_pypy.py文件:
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# primes_pypy.py
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import timeit
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def primes_python(nb_primes):
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p = []
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n = 2
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while len(p) < nb_primes:
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# Is n prime?
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for i in p:
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if n % i == 0:
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break
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# If no break occurred in the loop
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else:
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p.append(n)
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n += 1
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return p
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if __name__ == "__main__":
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repeat_times = 1000
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t1 = timeit.timeit(stmt="primes_python(1000)",
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setup="from __main__ import primes_python", number=repeat_times)
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print(f"run in pypy: {t1}s")
使用pypy3 primes_pypy.py 运行文件, 得到结果如下:
run in pypy: 3.0445395345987682s
5. nuitka的暂时没弄出来, 总体的运行结果如下:
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run in python: 28.519053545829927s
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run cython with cdef: 1.6289360376895452s
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run in python with numba jit: 2.0565857326599577s
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run cython without cdef: 13.221758278866588s
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run in pypy: 3.0445395345987682s
基本上jit的效果很明显,也不用改动python代码。