Question

我一直在学习函数式编程，我开始思考，组装数学运算符。 counting -> addition -> multiplication -> power -> ...自然而然地出现了简单且最天真的代码来表达这一点并且它有效！问题是我真的不知道为什么它能如此好地工作并且输出如此大的输出。

问题是：这个函数的复杂性是什么？

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（立即）：

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

Answer 1

反汇编告诉你这里没有应用任何魔法优化，它实际上只是减少了genexpr。 Python似乎完全可以完成这项任务，即使它让你感到惊讶。

>>> 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f5(2,4)电话，它实际上并没有执行太多操作：

>>> 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

增加65k，更不用说297的取幂，甚至不值得谈谈搞笑优化的现代CPU，所以难怪这在眨眼之间就完成了。尝试增加其中一个参数，看看它如何迅速达到快速评估非常的边界。

顺便说一句，operator是一个内置模块，你不应该像这样命名自己的函数。

递归闭包（函数发生器）

1 个答案: