python中的阶乘递归和迭代

1 python recursion

请帮助我了解我哪里出错了:

问题是:创建一个称为recursive_factorial的递归函数和一个称为iterative_factorial的迭代函数,以执行以下操作

Accepts as parameter an Integer n
Computes the factorial of n
Returns the factorial of n
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这是我用于此问题的测试:

import unittest

class RecursiveTestCase(unittest.TestCase):

  def test_recursive_factorial_one(self):
    result = recursive_factorial(4)
    self.assertEqual(result, 24, msg="Inaccurate value")

  def test_recursive_factorial_two(self):
    result = recursive_factorial(0)
    self.assertEqual(result, 1, msg="Inaccurate value")

  def test_iterative_factorial_one(self):
    result = iterative_factorial(5)
    self.assertEqual(result, 120, msg="Inaccurate value")

  def test_iterative_factorial_two(self):
    result = iterative_factorial(0)
    self.assertEqual(result, 1, msg="Inaccurate value")
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这是我写的代码:

def recursive_factorial(n):
    if n == 0:
        return 1
    else:
        return n * recursive_factorial(n-1)
def iterative_factorial(n):
    x = 1
    li = list(range(1, n + 1))
    for each in li:
        x = x * each
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这是我得到的错误:

1。test_iterative_factorial_one

Failure in line 21, in test_iterative_factorial_one self.assertEqual(result, 120, msg="Inaccurate value") AssertionError: Inaccurate value
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2。test_iterative_factorial_two

Failure in line 25, in test_iterative_factorial_two self.assertEqual(result, 1, msg="Inaccurate value") AssertionError: Inaccurate value 
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请帮助我了解我出了错。

NPE*_*NPE 5

您忘记了return xfrom iterative_factorial(),因此函数隐式返回None

顺便说一句,您可以range()直接迭代以下结果:

for each in range(1, n + 1):
   ...
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最后,这可能是学习Python reduce()函数的好机会。

import operator

def another_iterative_factorial(n):
   return reduce(operator.mul, range(1, n + 1))
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