我想使用cython的功能使用高效索引来使一些代码更快：http://docs.cython.org/src/tutorial/numpy.html

基本上,代码表示游戏http://www.hacker.org/cross/index.php游戏板上按钮的依赖性

# file test_so_cy.pyx
import time
import numpy as np
cimport numpy as np

DTYPE = np.uint8
ctypedef np.uint8_t DTYPE_t

def time_fmt(td):
    return "{:.2f} s".format(td)

def derive_equations(np.ndarray[DTYPE_t, ndim=2] field not None):
    cdef unsigned int n, m, i, j, x, y
    t1 = time.time()
    n, m = len(field), len(field[0])
    # generate equations for dimensions n and m
    eqs = []
    block = 2  # as soon as a 2 is hit there isnt any influence
    for i in xrange(n):
        for j in xrange(m):
            eq = 0L
            if field[i][j] == block:
                eqs.append([i*m+j ,field[i][j], eq])
                continue

            # rows upwards
            for x in xrange(i-1, -1, -1):
                if field[x][j] == block: break
                eq ^= 1L << (x*m+j)

            # rows downwards
            for x in xrange(i, n):
                if field[x][j] == block: break
                eq ^= 1L << (x*m+j)

            # cols left
            for y in xrange(j-1, -1, -1):
                if field[i][y] == block: break
                eq ^= 1L << (i*m+y)

            # cols right
            # j+1 to avoid resetting the influence of itself
            for y in xrange(j+1, m):
                if field[i][y] == block: break
                eq ^= 1L << (i*m+y)

            eqs.append([i*m+j, field[i][j], eq])

    t2 = time.time()
    print 'preprocess time:', time_fmt(t2 - t1)
    return n, m, eqs


def main():
    field = np.array(
[[0,1,0,0,0,0,0,0,0,0,0,1,1,1,0,2,1,0,0,2,1,0,1,1,0,0,0,0,0],
 [0,1,0,0,1,1,0,1,0,0,0,1,1,0,0,1,0,1,0,0,1,0,1,1,1,0,1,1,1],
 [1,1,0,1,0,0,0,0,0,0,0,1,1,0,1,0,1,0,0,1,0,1,1,0,0,1,0,0,2],
 [0,0,0,0,1,0,1,1,0,1,1,1,0,1,0,1,1,0,0,0,1,1,0,0,2,1,1,0,1],
 [0,1,0,1,1,1,1,1,2,1,1,0,1,0,0,0,0,0,0,1,0,0,1,0,2,0,1,0,1],
 [0,1,1,0,0,1,1,0,1,0,0,1,1,1,0,1,1,1,0,0,1,1,1,0,1,0,1,1,1],
 [0,0,0,1,0,1,1,0,1,0,0,1,1,1,0,1,0,0,0,0,0,0,0,1,0,1,0,1,1],
 [1,0,1,0,1,1,0,0,0,0,0,1,0,0,2,0,1,1,0,0,0,0,1,0,0,2,1,0,0],
 [1,0,1,0,1,0,1,0,1,1,1,0,1,0,1,1,0,1,1,0,1,0,1,0,1,0,1,1,1],
 [0,0,1,0,0,1,1,0,1,0,0,1,0,0,1,0,0,0,1,0,0,0,0,1,0,1,1,1,2],
 [1,0,1,1,0,0,1,0,1,1,1,0,1,2,1,1,1,2,1,0,1,1,1,0,0,0,0,0,0],
 [0,0,1,0,1,0,0,1,0,1,1,1,1,1,1,0,0,1,1,0,0,1,0,0,0,1,0,0,1],
 [1,1,0,0,0,1,0,0,1,0,0,1,0,1,1,0,0,0,0,1,1,0,0,1,0,0,0,0,0],
 [1,1,1,0,1,1,1,1,0,0,1,0,1,1,0,0,0,0,1,1,1,1,1,0,1,0,1,0,1],
 [1,0,0,0,1,1,0,0,2,0,1,1,2,0,0,1,0,1,0,1,0,2,1,1,1,1,0,0,2],
 [1,0,1,1,1,1,1,0,0,1,1,0,1,1,0,0,1,0,0,0,2,1,0,1,0,1,0,1,1],
 [0,0,1,1,1,0,0,0,0,0,2,1,0,1,0,1,0,1,1,1,1,0,0,1,1,1,1,0,1],
 [0,1,0,1,2,0,0,0,0,0,1,1,0,1,0,1,0,1,0,1,0,0,1,0,1,0,1,1,0],
 [0,1,0,0,2,0,0,0,1,0,1,0,0,1,0,1,1,0,0,1,0,1,1,1,0,1,1,1,1],
 [1,0,0,1,0,0,1,0,1,0,0,2,0,1,1,1,1,1,0,0,1,0,1,0,1,1,0,1,1],
 [0,0,1,0,1,1,0,0,1,0,0,0,1,1,1,0,0,1,0,0,1,0,1,2,0,1,1,0,2],
 [0,1,1,0,1,0,1,1,0,0,1,0,0,0,1,1,0,1,0,1,1,1,1,1,2,0,1,2,0],
 [0,0,0,0,1,0,1,0,0,0,0,1,0,0,1,0,0,1,1,1,2,0,0,1,0,0,1,1,0],
 [0,0,1,1,0,1,1,0,0,1,1,1,1,0,0,1,0,0,1,1,1,0,0,0,1,1,1,0,1],
 [0,2,0,1,1,1,1,0,1,0,0,0,0,0,1,1,1,0,1,1,1,1,0,1,0,0,0,1,1],
 [0,2,1,1,1,1,1,1,1,1,0,1,1,0,0,0,0,0,1,0,1,1,1,1,1,1,0,1,1],
 [0,1,1,1,0,1,0,0,0,1,0,2,0,1,1,1,1,1,0,1,0,1,0,0,1,1,0,1,0],
 [0,1,1,1,1,1,0,1,0,1,0,1,0,0,0,0,0,1,0,0,0,0,1,1,0,0,0,0,0],
 [1,0,0,0,0,1,0,1,0,1,0,1,0,0,1,0,0,1,0,0,1,0,1,0,0,1,2,1,1]], dtype=DTYPE)
    derive_equations(field)

if __name__ == '__main__':
    main()

# file setup_so.py
from distutils.core import setup
from Cython.Build import cythonize
import numpy

setup(
    name = "test_so",
    ext_modules = cythonize('test_so_cy.pyx'),
    include_dirs=[numpy.get_include()]
)

# usage: python setup_so.py build_ext --inplace
# import test_so_cy
# test_so_cy.main()

问题在于cython代码的运行速度比纯python版本慢约3倍. (我使用时间模块来衡量执行时间,因为对于更大的矩阵来说还可以).

cython -a告诉我

if field[x][j] == block: break

线路仍在使用大量python.因此,似乎仍然无法使用快速索引.
任何想法我在做什么错？

解决方法:

原始速度：0.14s

14倍加速(0.01s)：field [i] [j]将首先评估field [i],然后尝试评估所得的python对象.使用field [i,j]表示法可大幅提高速度

5倍加速(0.0018s)：输入eq变量cdef long eq

12X s5eedup(0.00012s)：将列表替换为由np数组组成的堆栈：

cdef np.ndarray[long, ndim=2] eqs=np.zeros((n*m,3),np.long)
cdef int curr_eqn=0

#append to list code
    if field[i,j] == block:
        eqs[curr_eqn,0]=i*m+j
        eqs[curr_eqn,1]=field[i,j]
        eqs[curr_eqn,2]=eq
        curr_eqn+=1
        continue

总加速：1100倍