I would like to write a bit of code that calls a function specified by a given argument. EG:
def caller(func):
return func()
However what I would also like to do is specify optional arguments to the 'caller' function so that 'caller' calls 'func' with the arguments specified (if any).
def caller(func, args):
# calls func with the arguments specified in args
Is there a simple, pythonic way to do this?
You can do this by using arbitrary argument lists and unpacking argument lists.
>>> def caller(func, *args, **kwargs):
... return func(*args, **kwargs)
...
>>> def hello(a, b, c):
... print a, b, c
...
>>> caller(hello, 1, b=5, c=7)
1 5 7
Not sure why you feel the need to do it, though.
This already exists as the apply function, though it is considered obsolete due to the new *args and **kwargs syntax.
>>> def foo(a,b,c): print a,b,c
>>> apply(foo, (1,2,3))
1 2 3
>>> apply(foo, (1,2), {'c':3}) # also accepts keyword args
However, the * and ** syntax is generally a better solution. The above is equivalent to:
>>> foo(*(1,2), **{'c':3})
