Python的递归程序向总理比化了一些 [英] Python recursive program to prime factorize a number

问题描述

我写了下面的程序，以质比化一些：

I wrote the following program to prime factorize a number:

import math
def prime_factorize(x,li=[]):
    until = int(math.sqrt(x))+1
    for i in xrange(2,until):
    	if not x%i:
    	    li.append(i)
    	    break
    else:                      #This else belongs to for
    	li.append(x)
    	print li               #First print statement; This is what is returned
    	return li
    prime_factorize(x/i,li)

if __name__=='__main__':
    print prime_factorize(300)   #Second print statement, WTF. why is this None

以下是我得到的输出：

Following is the output I get:

 [2, 2, 3, 5, 5]
 None

Altho，返回的值将被打印正常，后返回值似乎没有印刷，所有的时间。我在想什么？

Altho', the returned value is printed properly, the after returned value seems to be printing none, all the time. What am I missing?

另外，我怎么能改善方案（继续使用递归）

Also, how can I improve the program (continuing to use the recursion)

推荐答案

您prime_factorize功能没有在递归情况下return语句 - 你想要调用返回prime_factorize（X / I，LI）在其最后一行。与一个素数（所以递归调用不需要）尝试它，看，它的作品在这种情况下。

Your prime_factorize function doesn't have a return statement in the recursive case -- you want to invoke "return prime_factorize(x/i,li)" on its last line. Try it with a prime number (so the recursive call isn't needed) to see that it works in that case.

另外，你可能想使签名是这样的：

Also you probably want to make the signature something like:

def prime_factorize(x,li=None):
    if li is None: li = []

否则你打电话时，两次或更多次得到错误的结果：

otherwise you get wrong results when calling it two or more times:

>>> prime_factorize(10)
[2, 5]
>>> prime_factorize(4)
[2, 5, 2, 2]
>>> prime_factorize(19)
[2, 5, 2, 2, 19]

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