How to shuffle the order of if statements in a function in Python?

user_15601868 0
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I have a functions in Python which has a series of if statements.

def some_func():
    if condition_1:
        return result_1
    if condition_2:
        return result_2
    ... #other if statements

But I want that the order of these if statements is changed every time I call the function. Like if I call the function there can be a case when condition_1 and condition_2 both are true but since condition_1 is checked first thus I will get result_1 but I want that when I call the function second time some other random condition to be checked. (since I have only five if statements in the function therefore any one of the five conditions).
So is there any way to do this like storing the conditions in a list or something.
Any help will be appreciated.

python-3.x python
3 Answers

You can create list of condition and result then shuffle them like below:

import random
def some_func(x):
    lst_condition_result = [((x>1), True), ((x>2), False)]
    random.shuffle(lst_condition_result)
    
    for condition, result in lst_condition_result:
        if condition:
            return result

Output:

>>> some_func(20)
True

>>> some_func(20)
False

>>> some_func(20)
False
user_17405770

This one only evaluates as much as necessary, i.e., just as much as literally shuffled if-statements would:

def some_func():
    conditions_and_results = [
        (lambda: condition_1, lambda: result_1),
        (lambda: condition_2, lambda: result_2),
        ...
    ]
    random.shuffle(conditions_and_results)
    for condition, result in conditions_and_results:
        if condition():
            return result()

Or more efficiently (at least for large lists), reusing the same list every time and only shuffling as much as needed:

ifs = [
    (lambda: condition_1, lambda: result_1),
    (lambda: condition_2, lambda: result_2),
    ...
]
def some_func():
    for i, if_i in enumerate(ifs):
        j = random.randrange(i, len(ifs))
        condition, result = ifs[i] = ifs[j]
        ifs[j] = if_i
        if condition():
            return result()

Demo (Try it online!):

import random

ifs = [
    (lambda: x > 0, lambda: '> 0'),
    (lambda: x > 1, lambda: '> 1'),
    (lambda: x > 2, lambda: '> 2'),
    (lambda: x > 3, lambda: '> 3'),
]
def some_func():
    for i, if_i in enumerate(ifs):
        j = random.randrange(i, len(ifs))
        condition, result = ifs[i] = ifs[j]
        ifs[j] = if_i
        if condition():
            return result()

x = 2
print([some_func() for _ in range(10)])

Sample output:

['> 1', '> 0', '> 1', '> 0', '> 0', '> 0', '> 0', '> 1', '> 0', '> 0']
user_167591160

If you want to shuffle at run-time, then put your conditions and results as functions in a list of functions (once) and shuffle it (each time). Then evaluate those functions in that shuffled order until you get a non-falsy value (assuming the return statements always return a non-falsy value).

At the time of writing you did not provide a concrete example, so I'll work with this one:

def odd_or_multipleof3(n):
    if n % 2:
        return "odd"
    if n % 3 == 0:
        return "multiple of 3"
    return "neither"

Here odd_or_multipleof3(9) will return "odd", but when the if blocks are swapped, it would return "multiple of 3". To let the evaluation of blocks happen in a random order, so that you could get either of these return values, you could do this:

from random import shuffle 

funcs = [
    lambda n: "odd" if n % 2 else None,
    lambda n: "multiple of 3" if n % 3 == 0 else None
]
def odd_or_multipleof3(n):
    shuffle(funcs) 
    return any((result := func(n)) for func in funcs)  and result

print(odd_or_multipleof3(9))

This way of working will guarantee that conditions are only evaluated for as long as they don't return a truthy value, not more.

user_54598390
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