Pythonic way to call a function

Asked Aug 25 '21 at 16:06

Active Aug 17 '22 at 07:56

Viewed 68 times

I was wondering if I could achieve something along these lines using python:

I have these 5 functions:

def compile_a(self):
    pass


def compile_b(self):
    pass


def compile_c(self):
    pass


def compile_d(self):
    pass


def compile_e(self):
    pass

and I have some function which is supposed to wrap these functions, in a sense like this:

def compile_by_command(self, command_type):
    if command_type == 'a':
        self.compile_a()
    elif command_type == 'b':
        self.compile_b()
    elif command_type == 'c':
        self.compile_c()
    elif command_type == 'd':
        self.compile_d()
    elif command_type == 'e':
        self.compile_e()

I was wondering if there is some pythonic magic which looks something like this:

command_pool = ['a', 'b', 'c', 'd', 'e']

def compile_by_command(self, command_type):
    if command_type in command_pool:
        self.compile_{command_type}()

I mean a way to call the function using string formatting or something of that sort.

Thanks.

asked Aug 25 '21 at 16:06

eaRobust

3

A map of commands to functions would be the better way of doing it. – Ryan Schaefer Aug 25 '21 at 16:08
You can do `exec(f'self.compile_{command_type}()')` – Aug 25 '21 at 16:08
2

Why not make `command_pool` a dict of command_type to function? `command_pool = {'a': compile_a, ...}` – Iain Shelvington Aug 25 '21 at 16:08
7

@Sujay `exec` should be a last resort, however. – Ryan Schaefer Aug 25 '21 at 16:08
Also, https://stackoverflow.com/questions/3061/calling-a-function-of-a-module-by-using-its-name-a-string – mkrieger1 Aug 25 '21 at 16:09
Reflection magic: `getattr(self, f"compile_{command_type}")` – Mateen Ulhaq Aug 25 '21 at 16:10
The non-reflection ways of doing this is by keeping a dictionary, and then `.register("a", Class.a)`. – Mateen Ulhaq Aug 25 '21 at 16:11
Thanks guys, tired me did not think about it, lol. – eaRobust Aug 25 '21 at 16:13

Pythonic way to call a function

0 Answers0