python - 递归闭包(函数发生器)

Question

我一直在学习函数式编程，并且想到了组装数学运算符的想法。counting -> addition -> multiplication -> power -> ... 自然而然地出现了简单和最天真的代码来表达这个并且它有效!问题是我真的不知道为什么它工作得这么好，输出这么大。

问题是:这个函数的复杂度是多少？

代码在 python 中:

def operator(d):
        if d<=1:
                return lambda x,y:x+y
        else:
                return lambda x,y:reduce(operator(d-1),(x for i in xrange(y)))


#test 
f1 = operator(1)       #f1 is adition
print("f1",f1(50,52))  #50+52

f2 = operator(2)      #f2 is multiplication
print("f2",f2(2,20))  #2*20

f3 = operator(3)      #f3 is power, just look how long output can be
print("f3",f3(4,100)) #4**100 

f4 = operator(4)      #f4 is superpower, this one does not work that well
print("f4",f4(2,6))   #((((2**2)**2)**2)**2)**2

f5 = operator(5)      #f5 do not ask about this one, 
print("f5",f5(2,4))   #

输出(即时):

('f1', 102)
('f2', 40)
('f3', 1606938044258990275541962092341162602522202993782792835301376L)
('f4', 4294967296L)
('f5', 32317006071311007300714876688669951960444102669715484032130345427524655138867890893197201411522913463688717960921898019494119559150490921095088152386448283120630877367300996091750197750389652106796057638384067568276792218642619756161838094338476170470581645852036305042887575891541065808607552399123930385521914333389668342420684974786564569494856176035326322058077805659331026192708460314150258592864177116725943603718461857357598351152301645904403697613233287231227125684710820209725157101726931323469678542580656697935045997268352998638215525166389437335543602135433229604645318478604952148193555853611059596230656L)

Answer 1

反汇编告诉您这里没有应用神奇的优化，它实际上只是对 genexpr 的减少。 Python 似乎可以胜任这项任务，即使它会让您感到惊讶。

>>> import dis
>>> dis.dis(f3)
  5           0 LOAD_GLOBAL              0 (reduce)
              3 LOAD_GLOBAL              1 (operator)
              6 LOAD_DEREF               1 (d)
              9 LOAD_CONST               1 (1)
             12 BINARY_SUBTRACT     
             13 CALL_FUNCTION            1
             16 LOAD_CLOSURE             0 (x)
             19 BUILD_TUPLE              1
             22 LOAD_CONST               2 (<code object <genexpr> at 0x7f32d325f830, file "<stdin>", line 5>)
             25 MAKE_CLOSURE             0
             28 LOAD_GLOBAL              2 (xrange)
             31 LOAD_FAST                1 (y)
             34 CALL_FUNCTION            1
             37 GET_ITER            
             38 CALL_FUNCTION            1
             41 CALL_FUNCTION            2
             44 RETURN_VALUE

如果您专门查看您的 f5(2,4) 调用，它实际上并没有执行那么多操作:

>>> counter = 0
>>> def adder(x, y):
...   global counter
...   counter += 1
...   return x + y
... 
>>> def op(d):
...   if d <= 1: return adder
...   return lambda x,y:reduce(op(d-1),(x for i in xrange(y)))
...
>>> op(5)(2,4)
32317006071311007300714876688669951960444102669715484032130345427524655138867890893197201411522913463688717960921898019494119559150490921095088152386448283120630877367300996091750197750389652106796057638384067568276792218642619756161838094338476170470581645852036305042887575891541065808607552399123930385521914333389668342420684974786564569494856176035326322058077805659331026192708460314150258592864177116725943603718461857357598351152301645904403697613233287231227125684710820209725157101726931323469678542580656697935045997268352998638215525166389437335543602135433229604645318478604952148193555853611059596230656L
>>> counter
65035
>>> counter = 0
>>> op(3)(4,100)
>>> counter
297

65k 的加法，更不用说求幂的 297 次了，对于经过滑稽优化的现代 CPU 来说甚至都不值一提，所以眨眼之间就完成也就不足为奇了。尝试增加其中一个参数，看看它如何非常快速地达到快速评估的边界。

顺便说一下，operator 是一个内置模块，您不应该这样命名您自己的函数。