如何减少C API和Python可执行文件之间的执行时间差异?

python3 或者通过使用 libpython3 提供不同的执行时间。

$ time PYTHONPATH=. ./simple
real    0m6,201s
user    1m3,680s
sys     0m0,212s

$ time PYTHONPATH=. python3 -c 'import test; test.run()'
real    0m5,193s
user    0m53,349s
sys     0m0,164s

(删除 __pycache__ 运行间隔似乎没有影响)

蟒蛇3 用脚本比较快;在我的实际用例中,与从嵌入式解释器中运行的相同脚本相比,该因子快1.5。

我想(1)了解差异从何而来,(2)是否可以使用嵌入式解释器获得相同的性能?(例如使用cython目前不是一个选项)。

#include <Python.h>

int main()
{
        Py_Initialize();
        const char* pythonScript = "import test; test.run()";
        int result = PyRun_SimpleString(pythonScript);
        Py_Finalize();
        return result;
}

 g++ -std=c++11 -fPIC $(python3-config --cflags) simple.cpp \
 $(python3-config --ldflags) -o simple

import sys
sys.stdout = open('output.bin', 'bw')
import mandel
def run():
    mandel.mandelbrot(4096)

调整版本从 benchmarks-game's Mandlebrot (见 License )

from contextlib import closing
from itertools import islice
from os import cpu_count
from sys import stdout

def pixels(y, n, abs):
    range7 = bytearray(range(7))
    pixel_bits = bytearray(128 >> pos for pos in range(8))
    c1 = 2. / float(n)
    c0 = -1.5 + 1j * y * c1 - 1j
    x = 0
    while True:
        pixel = 0
        c = x * c1 + c0
        for pixel_bit in pixel_bits:
            z = c
            for _ in range7:
                for _ in range7:
                    z = z * z + c
                if abs(z) >= 2.: break
            else:
                pixel += pixel_bit
            c += c1
        yield pixel
        x += 8

def compute_row(p):
    y, n = p

    result = bytearray(islice(pixels(y, n, abs), (n + 7) // 8))
    result[-1] &= 0xff << (8 - n % 8)
    return y, result

def ordered_rows(rows, n):
    order = [None] * n
    i = 0
    j = n
    while i < len(order):
        if j > 0:
            row = next(rows)
            order[row[0]] = row
            j -= 1

        if order[i]:
            yield order[i]
            order[i] = None
            i += 1

def compute_rows(n, f):
    row_jobs = ((y, n) for y in range(n))

    if cpu_count() < 2:
        yield from map(f, row_jobs)
    else:
        from multiprocessing import Pool
        with Pool() as pool:
            unordered_rows = pool.imap_unordered(f, row_jobs)
            yield from ordered_rows(unordered_rows, n)

def mandelbrot(n):
    write = stdout.write

    with closing(compute_rows(n, compute_row)) as rows:
        write("P4\n{0} {0}\n".format(n).encode())
        for row in rows:
            write(row[1])