Python分解

Question

我想知道列出一个数字的所有整数因子的最佳方法,给出它的素因子及其指数的字典.
例如,如果我们有{2:3,3:2,5:1}(2 ^ 3*3 ^ 2*5 = 360)
那么我可以写:

for i in range(4):
  for j in range(3):
    for k in range(1):
      print 2**i * 3**j * 5**k

但在这里,我有3个可怕的循环.是否可以将此抽象为一个函数,给定任何因子分解作为字典对象参数？

Answer 1

我在博客上写过这篇文章,最快的纯python(没有itertools)来自Tim Peters发布到python列表的帖子,并使用嵌套的递归生成器:

def divisors(factors) :
    """
    Generates all divisors, unordered, from the prime factorization.
    """
    ps = sorted(set(factors))
    omega = len(ps)

    def rec_gen(n = 0) :
        if n == omega :
            yield 1
        else :
            pows = [1]
            for j in xrange(factors.count(ps[n])) :
                pows += [pows[-1] * ps[n]]
            for q in rec_gen(n + 1) :
                for p in pows :
                    yield p * q

    for p in rec_gen() :
        yield p

注意它的编写方式,它需要一个素数因子列表,而不是字典,即[2, 2, 2, 3, 3, 5]代替{2 : 3, 3 : 2, 5 : 1}.

Answer 2

itertools.product从Python 2.6 使用:

#!/usr/bin/env python
import itertools, operator

def all_factors(prime_dict):
    series = [[p**e for e in range(maxe+1)] for p, maxe in prime_dict.items()]
    for multipliers in itertools.product(*series):
        yield reduce(operator.mul, multipliers)

例:

print sorted(all_factors({2:3, 3:2, 5:1}))

输出:

[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 30, 36, 40, 45, 60,
 72, 90, 120, 180, 360]

Answer 3

好吧,不仅你有3个循环,但如果你有超过3个因素,这种方法将无法运作:)

一种可能的方式:

def genfactors(fdict):    
    factors = set([1])

    for factor, count in fdict.iteritems():
        for ignore in range(count):
            factors.update([n*factor for n in factors])
            # that line could also be:
            # factors.update(map(lambda e: e*factor, factors))

    return factors

factors = {2:3, 3:2, 5:1}

for factor in genfactors(factors):
    print factor

此外,您可以避免在内循环中复制一些工作:如果您的工作集是(1,3),并且想要应用于2 ^ 3个因素,我们正在做:

• (1,3) U (1,3)*2 = (1,2,3,6)
• (1,2,3,6) U (1,2,3,6)*2 = (1,2,3,4,6,12)
• (1,2,3,4,6,12) U (1,2,3,4,6,12)*2 = (1,2,3,4,6,8,12,24)

看看我们在第二套中有多少重复？

但我们可以这样做:

• (1,3) + (1,3)*2 = (1,2,3,6)
• (1,2,3,6) + ((1,3)*2)*2 = (1,2,3,4,6,12)
• (1,2,3,4,6,12) + (((1,3)*2)*2)*2 = (1,2,3,4,6,8,12,24)

没有套装,解决方案看起来更好:

def genfactors(fdict):
    factors = [1]

    for factor, count in fdict.iteritems():
        newfactors = factors
        for ignore in range(count):
            newfactors = map(lambda e: e*factor, newfactors)
            factors += newfactors

    return factors