Question

这里的第一个问题。我试图通过逐步执行项目euler来学习python，我遇到了障碍。以下方法（返回素数因子列表）适用于单个调用：

def findPrimeFactors(num, primeFactors = []):
    '''Find the prime factors of an arbitrary positive integer

        input: num to factorize
        returns: a list containing the prime factors of the number
    '''
    pIndex = 2

    while (num >= pIndex):
        if num % pIndex == 0:
            num /= pIndex
            primeFactors.append(pIndex)
            return FindPrimes.findPrimeFactors(num, primeFactors)

        else:
            pIndex += 1

    return primeFactors

然而，当我在循环中使用它时，这样的方法（此方法可能尚未完成，目前导致无限循环，因为无法找到更多的素数）：

def countPrimes(n = 1001):
    '''find n amount of unique primes ascending

        input: number of primes to find
        returns: list of n primes starting from 2   '''

    primes = []
    i = 2

    while len(primes) < n:
        primeFactors = FindPrimes.findPrimeFactors(i)
        print(primeFactors) #verify method behavior

        if len(primeFactors) is 1:
            primes.append(primeFactors[0])   
        i += 1

    return primes

结果是第一个循环返回[2]，下一个循环返回[2,3]，依此类推，将新结果附加到我希望在第一个递归调用中为空的列表中。似乎我的名单仍然存在，但我不确定为什么？我也阅读了Python Class scope & lists，它给了我一些线索，但递归使它复杂化了。

递归也意味着我不能简单地为它分配一个空集。来自C ++背景，我的期望是每次从我的程序调用函数时都应该重新初始化primeFactors变量。还是小蛇在这里。

编辑：这是我写的findPrimeFactors的迭代版本。我知道这不是最优的 - 但我希望至少能够有效地满足Project Euler的1分钟规则。任何改进或明确的建议都值得赞赏。

PRIMES = [2,3,5,7,11,13,17,19]
import math

class FindPrimes():

    '''V2 iterative'''
    def findPrimeFactors(n, primeFactors = None):
        '''Find the prime factors of an arbitrary positive integer

            input: num to factorize
            returns: a list containing the prime factors of the number
        '''

        if primeFactors is None:
            primeFactors = []

        num = n
        ceil = math.sqrt(n) #currently unused

        global PRIMES
        knownPrimes = PRIMES

        #check known primes for divisors first, then continue searching for primes by brute force
        while True:

            factorFound = False
            for prime in knownPrimes:   

                if num % prime == 0:
                    primeFactors.append(prime)
                    num /= prime
                    factorFound = True
                    break       #ensure that the list returned has ascending primes

            if not factorFound:
                break

        #once attempts have been made to reduce using known primes
        #search for new primes if the number is not fully reduced

        i = knownPrimes[-1] + 2

        while num != 1:

            if num % i == 0:
                knownPrimes.append(i)
                primeFactors.append(i)
                num /= i

            i += 2          

        return primeFactors


    def countPrimes(n = 10001):
        '''find n amount of unique primes ascending

            input: number of primes to find
            returns: list of n primes starting from 2   '''

        primes = []
        i = 2

        while len(primes) < n:

            primeFactors = FindPrimes.findPrimeFactors(i)

            if len(primeFactors) == 1:
                primes.append(primeFactors[0])
                #print(primeFactors[-1])

            i += 1

        print(len(primes))
        return primes

nth = 10001
print(FindPrimes.countPrimes(nth)[nth-1])   #print the largest prime found

Answer 1

请参阅"Least Astonishment" and the Mutable Default Argument

Answer 2

默认值primeFactors在呼叫之间共享，因此当您更改呼叫时，它会保持更改以供将来呼叫使用。

示例：

def foo(bar = []):
    bar.append(1)
    return bar

print foo()
print foo()

输出：

[1]
[1, 1]

您应该返回一个新列表，而不是更改默认值：

def foo(bar = []):
    return bar + [1]

print foo()
print foo()

输出：

[1]
[1]

Answer 3

如hammar所述，默认值仅在定义函数时创建一次，并在调用之间共享。

通常的方法是使用标记值作为默认值：

def findPrimeFactors(num, primeFactors=None):
    if primeFactors is None:
        primeFactors = []
    ...

偏离主题，但是对于找到的每个素数因子，您的函数findPrimeFactor()将递归一次。 Python不执行尾调用删除，因此您应该使用迭代而不是递归来重写它。

Python中的列表范围 - Project Euler 007

3 个答案: