Python - Integer Factorization into Primes

Question

我写了一个整数分解函数,但是在弄乱它之后,我意识到它有几个数字的问题......

>>> pFactors(99) # it does work for numbers with multiple of one prime factor
[3, 3, 11]
>>> pFactors(999) # however, sometimes, it doesn't
[3, 37] # the actual prime factorization of 999 is [3, 3, 3, 37]. 
>>> pFactors(88)
[2, 11]
>>> pFactors(888)
[2, 3, 37]

我的代码有什么问题？

def pFactors(n):
   """Finds the prime factors of 'n'"""
   from primes import isPrime
   from math import sqrt
   pFact, limit, check, num = [], int(round(sqrt(n), 2)) + 1, 2, n
   if isPrime(n):
      return [n]
   for check in range(2, limit):
      if isPrime(check) and num % check == 0:
         pFact.append(check)
         num /= check
         if isPrime(num):
            pFact.append(num)
            break
   pFact = sorted(pFact)
   return pFact

Answer 1

修改如下:

def pFactors(n): 
        """Finds the prime factors of 'n'""" 
        from math import sqrt 
        pFact, limit, check, num = [], int(sqrt(n)) + 1, 2, n 
        if n == 1: return [1] 
        for check in range(2, limit): 
             while num % check == 0: 
                pFact.append(check) 
                num /= check 
        if num > 1: 
          pFact.append(num) 
        return pFact 

for i in range(1,1000):
        print pFactors(i)

虽然我喜欢你最初编写的代码,但有几点:

1. 你不需要isPrime.原因是,除了num之外,范围内的任何素数也将是除以num的任何复合的最小除数,因此当你将这些素数分开时,你将阻止它们组成的复合被发现为后来在范围内的除数,只留下素数因子.

2. 您不需要对数组进行排序,它已经按check顺序升序排序.

3. 添加的while循环确保只要它们继续划分num,就可以正确找到重复因子.

4. 您可以使用同余来过滤掉少于限制的所有数字中的2/3以检查为除数,您能看到如何吗？

上面结果的最后几行是:

[11, 89]
[2, 2, 5, 7, 7]
[3, 3, 109]
[2, 491]
[983]
[2, 2, 2, 3, 41]
[5, 197]
[2, 17, 29]
[3, 7, 47]
[2, 2, 13, 19]
[23, 43]
[2, 3, 3, 5, 11]
[991]
[2, 2, 2, 2, 2, 31]
[3, 331]
[2, 7, 71]
[5, 199]
[2, 2, 3, 83]
[997]
[2, 499]
[3, 3, 3, 37]