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Python装饰器结合递归原理解析

时间:2020-10-04 14:42:43 | 栏目:Python代码 | 点击:

代码如下:

import functools

def memoize(fn):
  print('start memoize')
  known = dict()
  
  @functools.wraps(fn)
  def memoizer(*args):
    if args not in known:
      print('memorize %s'%args)
      # known[args] = fn(*args)
    for k in known.keys():
        print('%s : %s'%(k, known[k]), end = ' ')
    print()
    # return known[args]
  return memoizer


@memoize
def nsum(n):
  print('now is %s'%n)
  assert (n >= 0), 'n must be >= 0'
  return 0 if n == 0 else n + nsum(n - 1)


@memoize
def fibonacci(n):
  assert (n >= 0), 'n must be >= 0'
  return n if n in (0, 1) else fibonacci(n - 1) + fibonacci(n - 2)

if __name__ == '__main__':
  print(nsum(10))
  print(fibonacci(10))

输出如下:

start memoize
start memoize
memorize 10

None
memorize 10

None

对比代码(把注释的地方去掉后)的输出:

start memoize
start memoize
memorize 10
now is 10
memorize 9
now is 9
memorize 8
now is 8
memorize 7
now is 7
memorize 6
now is 6
memorize 5
now is 5
memorize 4
now is 4
memorize 3
now is 3
memorize 2
now is 2
memorize 1
now is 1
memorize 0
now is 0
(0,) : 0
(0,) : 0 (1,) : 1
(0,) : 0 (1,) : 1 (2,) : 3
(0,) : 0 (1,) : 1 (2,) : 3 (3,) : 6
(0,) : 0 (1,) : 1 (2,) : 3 (3,) : 6 (4,) : 10 
(0,) : 0 (1,) : 1 (2,) : 3 (3,) : 6 (4,) : 10 (5,) : 15
(0,) : 0 (1,) : 1 (2,) : 3 (3,) : 6 (4,) : 10 (5,) : 15 (6,) : 21 
(0,) : 0 (1,) : 1 (2,) : 3 (3,) : 6 (4,) : 10 (5,) : 15 (6,) : 21 (7,) : 28
(0,) : 0 (1,) : 1 (2,) : 3 (3,) : 6 (4,) : 10 (5,) : 15 (6,) : 21 (7,) : 28 (8,) : 36
(0,) : 0 (1,) : 1 (2,) : 3 (3,) : 6 (4,) : 10 (5,) : 15 (6,) : 21 (7,) : 28 (8,) : 36 (9,) : 45 
(0,) : 0 (1,) : 1 (2,) : 3 (3,) : 6 (4,) : 10 (5,) : 15 (6,) : 21 (7,) : 28 (8,) : 36 (9,) : 45 (10,) : 55 

通过取消注释的对比,可以得到如下结论:

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