Equivalent of ruby obj.send in python

Question

In ruby if I have an object obj, with a method called funcname, I can call the method using the following syntax obj.send(funcname)

Is there something similar in python.

The reason I want to do this, that I have a switch statement where I set the funcname, and want to call it at the end of the switch statement.

Python doesn't have a switch statement. Moreover, generating function names at runtime is almost always a bad idea. — Rafe Kettler, Mar 08 '11 at 21:53
@Rafe: It's not commonly needed, but there's nothing inherently evil about it. — Glenn Maynard, Mar 08 '11 at 22:02
Yes it doesn't but I still want to do it in and if-elif-else loop. I think I got my answer below. — Puneet Paul, Mar 08 '11 at 22:02
You can always just assign the function: "f = obj.func", and call `f(args)` later on. Python's method binding is good for that, when needed. — Glenn Maynard, Mar 08 '11 at 22:12

score 26 · Answer 1 · answered Mar 08 '11 at 21:55

26

getattr(obj, "name")(args)

answered Mar 08 '11 at 21:55

Glenn Maynard

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score 10 · Answer 2 · answered Mar 08 '11 at 21:58

10

hmmm... getattr(obj, funcname)(*args, **kwargs) ?

>>> s = "Abc"

>>> s.upper()
'ABC'

>>> getattr(s, "upper")()
'ABC'

>>> getattr(s, "lower")()
'abc'

answered Mar 08 '11 at 21:58

Georges Martin

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score 1 · Answer 3 · answered Mar 08 '11 at 23:17

Apart from the already mentioned getattr() builtin:

As an alternative to an if-elif-else loop you can use a dictionary to map your "cases" to the desired functions/methods:

# given func_a, func_b and func_default already defined
function_dict = {
    'a': func_a,
    'b': func_b  
}
x = 'a'
function_dict(x)() # -> func_a()
function_dict.get('xyzzy', func_default)() # fallback to -> func_c

Concerning methods instead of plain functions:

you can just turn the above example into method_dict using for example lambda obj: obj.method_a() instead of function_a etc., and then do method_dict[case](obj)
you can use getattr() as other answers have already mentioned, but you only need it if you really need to get the method from its name.
operator.methodcaller() from the stdlib is a nice shortcut in some cases: based on a method name, and optionally some arguments, it creates a function that calls the method with that name on another object (and if you provided any extra arguments when creating the methodcaller, it will call the method with these arguments)

score 1 · Answer 4 · answered Mar 09 '11 at 00:05

Or, if you prefer, operator.methodcaller('foo') is a function which calls foo on whatever you pass it. In other words,

import operator
fooer = operator.methodcaller("foo")
fooer(bar)

is equivalent to

bar.foo()

(I think this is probably the adjoint or dual in a suitable category. If you're mathematically inclined.)

Equivalent of ruby obj.send in python

4 Answers4