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Question

现在<script src="http://code.angularjs.org/1.2.10/angular.js"></script> <script src="http://bouil.github.io/angular-google-chart/ng-google-chart.js"></script> <div ng-app='myApp'><div ng-controller="MainCtrl"> <div style="float:right;position:relative;z-index:100;background:#f1f1f1;padding-top:40px"> <p> Software: <input type=range min=10000 max=100000 value={{aa}} ng-model="aa" ng-change="chart.data[1][1]=1*aa;chart.data[1][2]=1*aa"> {{aa}} </p> <p> Hardware: <input type=range min=10000 max=100000 value={{bb}} ng-model="bb" ng-change="chart.data[2][1]=1*bb;chart.data[2][2]=1*bb">{{bb}} </p> <p> Services: <input type=range min=10000 max=100000 value={{cc}} ng-model="cc" ng-change="chart.data[3][1]=1*cc;chart.data[3][2]=1*cc">{{cc}} </p> </div> <div google-chart chart="chart"> </div> </div></div>确实比ad-hoc实施更快，这是确凿的吗？

根据旧答案Statistics: combinations in Python，此自制函数在计算组合scipy.misc.comb时比scipy.misc.comb快：

nCr

但是在我自己的机器上运行一些测试之后，使用这个脚本似乎并非如此：

def choose(n, k):
    """
    A fast way to calculate binomial coefficients by Andrew Dalke (contrib).
    """
    if 0 <= k <= n:
        ntok = 1
        ktok = 1
        for t in xrange(1, min(k, n - k) + 1):
            ntok *= n
            ktok *= t
            n -= 1
        return ntok // ktok
    else:
        return 0

我得到以下输出：

from scipy.misc import comb
import random, time

def choose(n, k):
    """
    A fast way to calculate binomial coefficients by Andrew Dalke (contrib).
    """
    if 0 <= k <= n:
        ntok = 1
        ktok = 1
        for t in xrange(1, min(k, n - k) + 1):
            ntok *= n
            ktok *= t
            n -= 1
        return ntok // ktok
    else:
        return 0

def timing(f):
    def wrap(*args):
        time1 = time.time()
        ret = f(*args)
        time2 = time.time()
        print '%s function took %0.3f ms' % (f.__name__, (time2-time1)*1000.0)
        return ret
    return wrap

@timing
def test_func(combination_func, nk):
    for n,k in nk:
        combination_func(n, k)

nk = []
for _ in range(1000):
    n = int(random.random() * 10000)
    k = random.randint(0,n)
    nk.append((n,k))

test_func(comb, nk)
test_func(choose, nk)

时间分析测试是否显示新$ python test.py /usr/lib/python2.7/dist-packages/scipy/misc/common.py:295: RuntimeWarning: overflow encountered in exp vals = exp(lgam(N+1) - lgam(N-k+1) - lgam(k+1)) 999 test_func function took 32.869 ms 999 test_func function took 1859.125 ms $ python test.py /usr/lib/python2.7/dist-packages/scipy/misc/common.py:295: RuntimeWarning: overflow encountered in exp vals = exp(lgam(N+1) - lgam(N-k+1) - lgam(k+1)) 999 test_func function took 32.265 ms 999 test_func function took 1878.550 ms比ad-hoc scipy.misc.comb函数更快？我的测试脚本是否有任何错误导致时间安排不准确的？

为什么choose()现在更快？这是因为一些scipy.misc.comb / cython包裹技巧？

EDITED

在@WarrenWeckesser评论之后：

使用c时使用默认浮点近似值，由于浮点溢出，计算中断。

（有关文档，请参阅http://docs.scipy.org/doc/scipy-0.16.0/reference/generated/scipy.misc.comb.html）

当使用scipy.misc.comb()进行测试时，使用下面的函数计算长整数而不是浮点数，计算1000种组合时速度要慢得多：

exact=True

[OUT]：

@timing
def test_func(combination_func, nk):
    for i, (n,k) in enumerate(nk):
        combination_func(n, k, exact=True)

Answer 1

参考scipy.misc.comb的源代码，结果的更新例程是：

    val = 1
    for j in xrange(min(k, N-k)):
        val = (val*(N-j))//(j+1)
    return val

而您建议的更新例程是：

    ntok = 1
    ktok = 1
    for t in xrange(1, min(k, n - k) + 1):
        ntok *= n
        ktok *= t
        n -= 1
    return ntok // ktok

我猜测SciPy实现速度较慢的原因是由于子程序在每次迭代时都涉及整数除法，而你的子句只在return语句中调用一次除法。

`scipy.misc.comb`比ad-hoc二项式计算更快吗？

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1 个答案: