I have a function func :
def func(a):
print a
func.__name_ = "My_Function"
To get the name of the function I will just do func.__name__.
But is there a way to call func() by using its name i.e. "My_Function"?
Edit: There are going to be large number of functions for which this has to be done. Is there any better solution apart from keeping a mapping of function.name to the function. ?
It will be much easier to use a dictionary:
def func():
print('func')
func_dict = {"My_Function": func}
func_dict["My_Function"]()
# 'func'
func_dict[func.__name__]()?
func.__name__.
func is already there why bother accessing it via func_dict?func.__name__. But I think I didn't understand the question correctly, sorry :)func_dict = {func.__name__: func} ; func_dict[func.__name__]() but I don't see the point.Assuming that you want to access functions that are defined in global scope (not part of any class/object):
def foo():
print "foo"
def bar():
print "bar"
foo.__name__ = "1"
bar.__name__ = "2"
# locals().items() will return the current local symbol table, which is an iterable and
# it contains the globally defined functions.
symbol_table = locals().items()
# we will generate a list of functions that their __name__ == "1"
one_functions = [value for key, value in symbol_table if callable(value) and value.__name__ == "1"]
# now we can call the first item on the list - foo()
one_functions[0]()
__name__is just a label, it's not the same as doingMy_Function = func.getattrto get function and invoke it. Have a look at this Question[f for _,f in locals().items() if callable(f) and f.__name__=='My_Function' ][0]('Hello'). Do try this at home, not in production. Use other option as Deepspace/johrsharpe suggested.