Most efficient way of making an if-elif-elif-else statement when the else is done the most?

问题:

I've got a in if-elif-elif-else statement in which 99% of the time, the else statement is executed:

if something == 'this':     doThis() elif something == 'that':     doThat() elif something == 'there':     doThere() else:     doThisMostOfTheTime() 

This construct is done a lot, but since it goes over every condition before it hits the else I have the feeling this is not very efficient, let alone Pythonic. On the other hand, it does need to know if any of those conditions are met, so it should test it anyway.

Does anybody know if and how this could be done more efficiently or is this simply the best possible way to do it?

回答1:

The code...

options.get(something, doThisMostOfTheTime)() 

...looks like it ought to be faster, but it's actually slower than the if ... elif ... else construct, because it has to call a function, which can be a significant performance overhead in a tight loop.

Consider these examples...

1.py

something = 'something'  for i in xrange(1000000):     if something == 'this':         the_thing = 1     elif something == 'that':         the_thing = 2     elif something == 'there':         the_thing = 3     else:         the_thing = 4 

2.py

something = 'something' options = {'this': 1, 'that': 2, 'there': 3}  for i in xrange(1000000):     the_thing = options.get(something, 4) 

3.py

something = 'something' options = {'this': 1, 'that': 2, 'there': 3}  for i in xrange(1000000):     if something in options:         the_thing = options[something]     else:         the_thing = 4 

4.py

from collections import defaultdict  something = 'something' options = defaultdict(lambda: 4, {'this': 1, 'that': 2, 'there': 3})  for i in xrange(1000000):     the_thing = options[something] 

...and note the amount of CPU time they use...

1.py: 160ms 2.py: 170ms 3.py: 110ms 4.py: 100ms 

...using the user time from time(1).

Option #4 does have the additional memory overhead of adding a new item for every distinct key miss, so if you're expecting an unbounded number of distinct key misses, I'd go with option #3, which is still a significant improvement on the original construct.

回答2:

I'd create a dictionary :

options = {'this': doThis,'that' :doThat, 'there':doThere} 

Now use just:

options.get(something, doThisMostOfTheTime)() 

If something is not found in the options dict then dict.get will return the default value doThisMostOfTheTime

Some timing comparisons:

Script:

from random import shuffle def doThis():pass def doThat():pass def doThere():pass def doSomethingElse():pass options = {'this':doThis, 'that':doThat, 'there':doThere} lis = range(10**4) + options.keys()*100 shuffle(lis)  def get():     for x in lis:         options.get(x, doSomethingElse)()  def key_in_dic():     for x in lis:         if x in options:             options[x]()         else:             doSomethingElse()  def if_else():     for x in lis:         if x == 'this':             doThis()         elif x == 'that':             doThat()         elif x == 'there':             doThere()         else:             doSomethingElse() 

Results:

>>> from so import * >>> %timeit get() 100 loops, best of 3: 5.06 ms per loop >>> %timeit key_in_dic() 100 loops, best of 3: 3.55 ms per loop >>> %timeit if_else() 100 loops, best of 3: 6.42 ms per loop 

For 10**5 non-existent keys and 100 valid keys::

>>> %timeit get() 10 loops, best of 3: 84.4 ms per loop >>> %timeit key_in_dic() 10 loops, best of 3: 50.4 ms per loop >>> %timeit if_else() 10 loops, best of 3: 104 ms per loop 

So, for a normal dictionary checking for the key using key in options is the most efficient way here:

if key in options:    options[key]() else:    doSomethingElse() 

回答3:

Are you able to use pypy?

Keeping your original code but running it on pypy gives a 50x speed-up for me.

CPython:

matt$ python Python 2.6.8 (unknown, Nov 26 2012, 10:25:03) [GCC 4.2.1 Compatible Apple Clang 3.0 (tags/Apple/clang-211.12)] on darwin Type "help", "copyright", "credits" or "license" for more information. >>> >>> from timeit import timeit >>> timeit(""" ... if something == 'this': pass ... elif something == 'that': pass ... elif something == 'there': pass ... else: pass ... """, "something='foo'", number=10000000) 1.728302001953125 

Pypy:

matt$ pypy Python 2.7.3 (daf4a1b651e0, Dec 07 2012, 23:00:16) [PyPy 2.0.0-beta1 with GCC 4.2.1] on darwin Type "help", "copyright", "credits" or "license" for more information. And now for something completely different: ``a 10th of forever is 1h45'' >>>> >>>> from timeit import timeit >>>> timeit(""" .... if something == 'this': pass .... elif something == 'that': pass .... elif something == 'there': pass .... else: pass .... """, "something='foo'", number=10000000) 0.03306388854980469 

回答4:

That's an example of a if with dynamic conditions translated to a dictionary.

selector = {lambda d: datetime(2014, 12, 31) >= d : 'before2015',             lambda d: datetime(2015, 1, 1) <= d < datetime(2016, 1, 1): 'year2015',             lambda d: datetime(2016, 1, 1) <= d < datetime(2016, 12, 31): 'year2016'}  def select_by_date(date, selector=selector):     selected = [selector[x] for x in selector if x(date)] or ['after2016']     return selected[0] 

It is a way, but may not be the most pythonic way to do it because is less readable for whom is not fluent in Python.

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