Python - 为什么这段代码需要这么长时间？

My concern is the part between the two hash-lines. The following code runs too long for me to wait for its output. When I replace the problematic part by another chunk of code, the programme runs in a few seconds (see the end of this post). my aim is to generate 90 data points that are uniformly distributed in a unit square (var1, var2), and then generate 12 points that are randomly placed in a circle of radius 1/8 that lies entirely within the unit square (cir1, cir2), and finally join these two sets (sph1, sph2).

我担心的是两个哈希行之间的部分。以下代码运行时间太长，我无法等待其输出。当我用另一块代码替换有问题的部分时，程序会在几秒钟内运行（参见本文末尾）。我的目标是生成90个均匀分布在单位正方形（var1，var2）中的数据点，然后生成12个点，这些点随机放置在半径为1/8的圆中，该圆完全位于单位正方形内（cir1，cir2） ），最后加入这两组（sph1，sph2）。

I've been digging in this code since yesterday ad I'm really sure it is correct (apparently it is not). I'm probably missing something really obvious...

我从昨天起就开始研究这段代码，我确信它是正确的（显然它不是）。我可能错过了一些非常明显的东西......

#!/usr/bin/env python
# -*- coding: utf-8 -*-

import random
import math
import numpy as np
import sys
from heapq import merge
import multiprocessing as mp
from multiprocessing import Pool

boot = 2
RRpoints = 90
rrpoints = 12
step = np.arange(-8.0 , 0.5 , 0.5)

def distance_between_points(ite, jte):
    xi, yi = ite
    xj, yj = jte
    x2 = (xi - xj) ** 2
    y2 = (yi - yj) ** 2
    d = math.sqrt(x2 + y2)
    return d

def heaviside_step(n):
    return int(n >= 0)

def myscript(iteration_number):
    RRfile_name = "MC_out_cluster_1/output%d.txt" % iteration_number
    with open(RRfile_name, "w") as RRf:
#############################################################################
        np.random.seed()
        var1 = np.random.uniform(0, 1 , RRpoints)
        var2 = np.random.uniform(0, 1 , RRpoints)
        cir1 = []
        cir2 = []
        x0 = np.random.uniform(0.125 , 0.875)
        y0 = np.random.uniform(0.125 , 0.875)
        while ( len(cir1) < rrpoints and len(cir2) < rrpoints ):
            np.random.seed()
            col1 = np.random.uniform(x0 - 0.125 , x0 + 0.125)
            col2 = np.random.uniform(y0 - 0.125 , y0 + 0.125)
            if (x0 - col1) ** 2 + (y0 - col2) ** 2 <= 1/64:
                cir1.append(col1)
                cir2.append(col2)
        sph1 = list(merge(var1 , cir1))
        sph2 = list(merge(var2 , cir2))
############################################################################
        corr = []
        for k in xrange(0, len(step)):
            h = 0
            for i in xrange(0, RRpoints):
                for j in xrange(1 + i, RRpoints):
                    ite = sph1[i] , sph2[i]
                    jte = sph1[j] , sph2[j]
                    dbp = distance_between_points(ite, jte)
                    h += heaviside_step(math.exp( step[k] ) - dbp)
            corr.append([math.exp(step[k]) , h])

        for item in corr:
            RRf.write("{0}\t{1}\n".format(item[0], item[1]))

x = xrange(boot)
p = mp.Pool()

y = p.imap(myscript, x)
list(y)

This is another (working) chunk that I modified to create the above code:

这是我修改为创建上述代码的另一个（工作）块：

 #############################################################################
        var1 = []
        var2 = []
        while (len(var1) < RRpoints):
            np.random.seed()
            col1 = np.random.uniform(0 , 1)
            col2 = np.random.uniform(0 , 1)
            if ( col1 ** 2 + (col2 - 0.5) ** 2 > 1/16 and (col1 - 1) ** 2 + (col2 - 0.5) ** 2 > 1/16 ):
                var1.append(col1)
                var2.append(col2)
        cir1 = []
        cir2 = []
        while (len(cir1) < rrpoints and len(cir2) < rrpoints):
            np.random.seed()
            new1 = np.random.uniform(0.125 , 0.875)
            new2 = np.random.uniform(0.125 , 0.875)
            if ( new1 ** 2 + (new2 - 0.5) ** 2 >= 0.140625 and (new1 - 1) ** 2 + (new2 - 0.5) ** 2 >= 0.140625 ):
                cir1.append(new1)
                cir2.append(new2)
        sph1 = list(merge(var1 , cir1))
        sph2 = list(merge(var2 , cir2))
 #############################################################################

1 个解决方案

#1

You are using Python 2.x so 1/64==0 (in Python 2, division of two integers gives an integer result). If you can, upgrade to Python 3.x, or failing that change the test to use 1./64

您正在使用Python 2.x，因此1/64 == 0（在Python 2中，两个整数的除法给出整数结果）。如果可以，请升级到Python 3.x，否则会将测试更改为使用1./64