从函数列表返回函数的python函数

Question

I want to make following function:

 1)input is a number. 
 2)functions are indexed, return a function whose index matches given number

here's what I came up with:

def foo_selector(whatfoo):
    def foo1():
        return
    def foo2():
        return
    def foo3():
        return
    ...

    def foo999():
        return

    #something like return foo[whatfoo]

the problem is, how can I index the functions (foo#)? I can see functions foo1 to foo999 by dir(). however, dir() returns name of such functions, not the functions themselves. In the example, those foo-functions aren't doing anything. However in my program they perform different tasks and I can't automatically generate them. I write them myself, and have to return them by their name.

Answer 1

Use a decorator to accumulate a list of functions.

func_list = []

def listed_func(func):
    func_list.append(func)
    return func

@listed_func
def foo1():
   pass

@listed_func
def foo2():
   pass

Now you can easily access the functions by index in a list.

You could also create a dictionary if you want to access the functions by name:

func_dict = {}

def collected_func(func):
    func_dict[func.__name__] = func
    return func

Or extract the index from the name, and use that as the dict key (since dicts are not ordered, you'll want to sort the keys if you want to iterate over them in some order):

func_dict = {}

def collected_func(func):
    key = int("".join(c for c in func.__name__ if c.isdigit()))
    func_dict[key] = func
    return func

Or explicitly pass the index number to the decorator:

func_dict = {}

def collected_func(key):
    def decorator(func):
        func_dict[key] = func
        return func
    return decorator

@collected_func(12)
def foo():
    pass

Answer 2

You could simply place all of your functions into an array, something like:

def foo_selector(whatfoo):
    def foo1():
        return
    def foo2():
        return
    def foo3():
        return
    ...

    def foo999():
        return

    foo_list = [
        foo1,
        foo2,
        ...
        foo999
    ]

    return foo_list[whatfoo]

Answer 3

Here are a couple other ways you could do it:

eval("foo{}()".format(whatfoo))

or

locals()['foo{}'.format(whatfoo)]