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

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.

---

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()

---

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

---

Here 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.

---

I tried with the match statement, introduced in python 3.10:

5.py

something = 'something'
for i in range(10000000):
    match something:
        case "this":
            the_thing = 1
        case "that":
            the_thing = 2
        case "there":
            the_thing = 3
        case _:
            the_thing = 4

Here are the results I get with 3.10.0: 1.py: 1.4s 2.py: 0.9s 3.py: 0.7s 4.py: 0.7s 5.py: 1.0s I thought I would get something similar to 1.py but it is quite faster.

---

People warn about exec for security reasons, but this is an ideal case for it.
It's an easy state machine.

Codes = {}
Codes [0] = compile('blah blah 0; nextcode = 1')
Codes [1] = compile('blah blah 1; nextcode = 2')
Codes [2] = compile('blah blah 2; nextcode = 0')

nextcode = 0
While True:
    exec(Codes[nextcode])

---

Recently I came across an approach alternative to "nested if else" which reduce running time of my function from 2.5hrs to ~2min..Baam! Let's begin:

Earlier Code bin = lambda x:"Unknown" if x==0 else("High" if x>75 else("Medium" if x>50 and x<=75 else("Medium_Low" if x>25 and x<=50 else "Low")))

col.apply(bin) Time ~2.5hrs

Optimize Code

 def dict_function(*args):
'Pass in a list of tuples, which will be key/value pairs'
ret = {}
for k,v in args:
    for i in k:
        ret[i] = v
return ret
Dict = dict_function(([0],"Unknown"),(range(1,25),"Low"),(range(25,50),"Medium_Low"),(range(50,75),"Medium"),(range(75,100),"High"))

col.apply(lambda x:Dict[x])

dict_function make multiple key_value pairs for given range. Time~2mins

---

I had the same problem recently, though not regarding performance, but I dislike the "API" of creating functions and manually adding them to a dict. I wanted an API similar to functools.singledispatch, but to dispatch based on values not on types. So ...

def value_dispatch(func):
    """value-dispatch function decorator.
    Transforms a function into a function, that dispatches its calls based on the
    value of the first argument.
    """
    funcname = getattr(func, '__name__')
    registry = {}

    def dispatch(arg):
        """return the function that matches the argument"""
        return registry.get(arg, func)

    def register(arg):
        def wrapper(func):
            """register a function"""
            registry[arg] = func
            return func
        return wrapper

    def wrapper(*args, **kwargs):
        if not args:
            raise ValueError(f'{funcname} requires at least 1 positional argument')
        return dispatch(args[0])(*args, **kwargs)

    wrapper.register = register
    wrapper.dispatch = dispatch
    wrapper.registry = registry
    return wrapper

Use like this:

@value_dispatch
def handle_something():
    print("default")

@handle_something.register(1)
def handle_one():
    print("one")

handle_something(1)
handle_something(2)

PS: I created a snippet on Gitlab for reference

---

you could immitate if-elif-else with switch-case type like by using dictionary and lambda function

For example:

x = 5
y = 5
operator = 'add'

def operation(operator, x, y): 
 return {
   'add': lambda: x+y,
   'sub': lambda: x-y,
   'mul': lambda: x*y,
   'div': lambda: x/y
 }.get(operator, lambda: None)()

result = operation(operator, x, y)
print(result)

---