lambda比python中的函数调用慢，为什么 [英] lambda is slower than function call in python, why

问题描述

我认为lambda比函数调用快，但是经过测试，我发现我错了.函数调用绝对比lambda调用快.

I think lambda is faster than function call, but after testing, I find out that I am wrong. Function call is definitely faster than lambda call.

有人可以告诉我为什么吗?

Can anybody tell me why?

以及如何加快Python中的函数调用速度?

And how to speed up function call in Python?

我正在使用Ubuntu 14.04和Python 2.7.6

I'm using Ubuntu 14.04 and Python 2.7.6

>>> timeit('def a():return 222*333 ;a()')
0.08195090293884277
>>> timeit('a=lambda:222*333 ;a()')
0.11071300506591797

>>> timeit('a=lambda: [].extend(range(10)) ;a()')
0.40241098403930664
>>> timeit('a=lambda: [].extend(range(10)) ;a()')
0.4011270999908447
>>> timeit('a=lambda: [].extend(range(10)) ;a()')
0.4064619541168213
>>> timeit('def a(): return [].extend(range(10)) ;a()')
0.07965493202209473
>>> timeit('def a(): return [].extend(range(10)) ;a()')
0.08039593696594238
>>> timeit('def a(): return [].extend(range(10)) ;a()')
0.08103609085083008
>>> timeit('def a(): return [].extend(range(10)) ;a()')
0.08639097213745117

---

对不起，我没有错.正确的测试:

---

Sorry for my mistake, there is no difference. Correct testing:

>>> timeit('a()', setup="def a():return 222*333")
0.07061290740966797
>>> timeit('a()', setup="a=lambda: 222*333")
0.06967616081237793

推荐答案

timeit('def a(): return [].extend(range(10)) ;a()')没有调用a();对a()的调用是a定义的一部分:

timeit('def a(): return [].extend(range(10)) ;a()') is not calling a(); The call to a() is part of the definition of a:

In [34]: def a(): return [].extend(range(10)) ;a()

In [35]: import dis

In [36]: dis.dis(a)
  1           0 BUILD_LIST               0
              3 LOAD_ATTR                0 (extend)
              6 LOAD_GLOBAL              1 (range)
              9 LOAD_CONST               1 (10)
             12 CALL_FUNCTION            1
             15 CALL_FUNCTION            1
             18 RETURN_VALUE        
             19 LOAD_GLOBAL              2 (a)
             22 CALL_FUNCTION            0       #<-- a is called
             25 POP_TOP             

---

如果分别测试各个部分，则差异可以忽略不计:

---

If you test the pieces separately, the difference is negligible:

In [24]: %timeit a=lambda: [].extend(range(10))
10000000 loops, best of 3: 68.6 ns per loop

In [25]: %timeit def a2(): return [].extend(range(10))
10000000 loops, best of 3: 68.8 ns per loop

In [22]: %timeit a()
1000000 loops, best of 3: 445 ns per loop

In [23]: %timeit a2()
1000000 loops, best of 3: 442 ns per loop

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