Consider the following functions:
def fact1(n):
if n < 2:
return 1
else:
return n * fact1(n-1)
def fact2(n):
if n < 2:
return 1
return n * fact2(n-1)
They should be equivalent. But there's a performance difference:
>>> T(lambda : fact1(1)).repeat(number=10000000)
[2.5754408836364746, 2.5710129737854004, 2.5678811073303223]
>>> T(lambda : fact2(1)).repeat(number=10000000)
[2.8432059288024902, 2.834425926208496, 2.8364310264587402]
The version without the else
is 10% slower. This is pretty significant. Why?
yes, your code will be slower. if A … else if B … else if C … the answer is not dependent on the number of conditions but the order of conditions.
So, the short answer is NO, public does not slow down your code. A simple function call has negligible cost. If you're not programming some number-crunchink code looping over and over million and million of times, the cost of some function call is no concern.
For me, they are virtually the same speed: (Python 2.6.6 on Debian)
In [4]: %timeit fact1(1)
10000000 loops, best of 3: 151 ns per loop
In [5]: %timeit fact2(1)
10000000 loops, best of 3: 154 ns per loop
The byte code is also very similar:
In [6]: dis.dis(fact1)
2 0 LOAD_FAST 0 (n)
3 LOAD_CONST 1 (2)
6 COMPARE_OP 0 (<)
9 JUMP_IF_FALSE 5 (to 17)
12 POP_TOP
3 13 LOAD_CONST 2 (1)
16 RETURN_VALUE
>> 17 POP_TOP
5 18 LOAD_FAST 0 (n)
21 LOAD_GLOBAL 0 (fact)
24 LOAD_FAST 0 (n)
27 LOAD_CONST 2 (1)
30 BINARY_SUBTRACT
31 CALL_FUNCTION 1
34 BINARY_MULTIPLY
35 RETURN_VALUE
36 LOAD_CONST 0 (None)
39 RETURN_VALUE
In [7]: dis.dis(fact2)
2 0 LOAD_FAST 0 (n)
3 LOAD_CONST 1 (2)
6 COMPARE_OP 0 (<)
9 JUMP_IF_FALSE 5 (to 17)
12 POP_TOP
3 13 LOAD_CONST 2 (1)
16 RETURN_VALUE
>> 17 POP_TOP
4 18 LOAD_FAST 0 (n)
21 LOAD_GLOBAL 0 (fact)
24 LOAD_FAST 0 (n)
27 LOAD_CONST 2 (1)
30 BINARY_SUBTRACT
31 CALL_FUNCTION 1
34 BINARY_MULTIPLY
35 RETURN_VALUE
The only difference is that the version with the else
includes code to return None
in case control reaches the end of the function body.
What is happening here is that fact2
has a hash conflict with __name__
in your module globals. That makes the lookup of the global fact2
ever so slightly slower.
>>> [(k, hash(k) % 32) for k in globals().keys() ]
[('__builtins__', 8), ('__package__', 15), ('fact2', 25), ('__name__', 25), ('fact1', 26), ('__doc__', 29)]
i.e. The same answer as for Why is early return slower than else? except that there the hash conflict was with __builtins__
I question the timings. The two functions aren't recursing to themselves. fact1 and fact2 both call fact which isn't shown.
Once that is fixed, the disassembly (in both Py2.6 and Py2.7) shows that both are running the same op codes except for the name of the recursed into function. The choice of name trigger a small difference in timings because fact1 may insert in the module dictionary with no name collisions while *fact2) may have a hash value that collides with another name in the module.
In other words, any differences you see in timings are not due to the choice of whether the else-clause is present :-)
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With