lambda is slower than function call in python, why

Question

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.

Can anybody tell me why

Answer 1

There's no difference in calling a lambda versus a function. A lambda is just a function created with a single expression and no name.

Say we have two identical functions, one created with a function definition, the other with a lambda expression:

def a():
    return 222*333

b = lambda: 222*333

We see that both are the same type of function object and they both share equivalent byte-code:

>>> type(a)
<class 'function'>
>>> type(b)
<class 'function'>

>>> import dis
>>> dis.dis(a)
  2           0 LOAD_CONST               3 (73926)
              2 RETURN_VALUE
>>> dis.dis(b)
  1           0 LOAD_CONST               3 (73926)
              2 RETURN_VALUE

How can you speed that up? You don't. It's Python. It's pre-optimized for you. There's nothing more for you to do with this code.

Perhaps you could give it to another interpreter, or rewrite it in another language, but if you're sticking to Python, there's nothing more to do now.

Timing it

Here's how I would examine the timings.

Timeit's timeit and repeat both take a callable:

import timeit

Note that timeit.repeat takes a repeat argument as well:

>>> min(timeit.repeat(a, repeat=100))
0.06456905393861234
>>> min(timeit.repeat(b, repeat=100))
0.06374448095448315

These differences are too small to be significant.

Answer 2

As pointed out above, your first test only profiles the time it takes to define a. It's actually never called.

Lambda expressions and "normal" functions generate the exact same bytecode, as you can see if you use the dis module:

def a(): return 10
b = lambda: 10

import dis

>>> dis.dis(a)
1           0 LOAD_CONST               1 (10)
            3 RETURN_VALUE
>>> dis.dis(b)
1           0 LOAD_CONST               1 (10)
            3 RETURN_VALUE

Answer 3

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             

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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