Pythonic way for a conditional return in a function

jorijnsmit :

I have a couple of functions that return either a number or None. I want my wrapper function to return the first result that is not None. Is there any other way to do this than as following?

def func1():
    return None
def func2():
    return 3
def func3():
    return None

def wrapper():
    result = func1()
    if result is not None:
        return result
    result = func2()
    if result is not None:
        return result
    result = func3()
    if result is not None:
        return result

I am aware of return func1() or alternative; which returns the outcome of func1() unless it is None, then alternative is returned. In the most optimal situation I do something like (pseudo code):

return func1() or continue
return func2() or continue
return func3() or continue

chepner :

You need a loop which will return the first non-None value it finds.

def wrapper():
    for f in [func1, func2, func3]:
        result = f()
        if result is not None:
            return result

If every function returns None, wrapper will as well by reaching the end of the body without an explicit return.

Shortened slightly in Python 3.8 or later,

def wrapper():
    for f in [func1, func2, func3]:
        if (result := f()) is not None:
            return result

You can also make use of the any function:

def wrapper():
    if any((result := f()) is not None for f in [func1, func2, func3]):
        return result

(I prefer the explicit loop; YMMV.)