python - Get a list/tuple/dict of the arguments passed to a function?

86
votes

Given the following function:

def foo(a, b, c):
    pass

How would one obtain a list/tuple/dict/etc of the arguments passed in, without having to build the structure myself?

Specifically, I'm looking for Python's version of JavaScript's arguments keyword or PHP's func_get_args() method.

What I'm not looking for is a solution using *args or **kwargs; I need to specify the argument names in the function definition (to ensure they're being passed in) but within the function I want to work with them in a list- or dict-style structure.

9
pythonfunctionarguments
Please Do Not Do This. The folks that maintain your code will be absolutely baffled by creating a useless dictionary out of perfectly good keyword arguments. They will be forced to rewrite to remove the dictionary. - S.Lott
Possible duplicate of Getting method parameter names in python - blahdiblah
@blahdiblah This question is asking how to obtain the parameters themselves, as opposed to the names of the parameters. I don't think it's a duplicate. - Anderson Green
@S.Lott: That's a rather narrow-minded view. Sometimes one wants to build a dictionary inside a function, initialised with some compulsory parameters. The function argument list makes it clear what those are, and believe me, they're not necessarily 'useless'. Please refer to Python's DRY principle, and don't be like some junior devs I've worked with, who whinge 'why would anybody want to do THAT?' when I suggest their API could be more versatile. - Michael Scheper

9 Answers

100
votes

You can use locals() to get a dict of the local variables in your function, like this:

def foo(a, b, c):
    print locals()

>>> foo(1, 2, 3)
{'a': 1, 'c': 3, 'b': 2}

This is a bit hackish, however, as locals() returns all variables in the local scope, not only the arguments passed to the function, so if you don't call it at the very top of the function the result might contain more information than you want:

def foo(a, b, c):
    x = 4
    y = 5
    print locals()

>>> foo(1, 2, 3)
{'y': 5, 'x': 4, 'c': 3, 'b': 2, 'a': 1}

I would rather construct a dict or list of the variables you need at the top of your function, as suggested in the other answers. It's more explicit and communicates the intent of your code in a more clear way, IMHO.

8
votes

You can use the inspect module:

def foo(x):
    return x

inspect.getargspec(foo)
Out[23]: ArgSpec(args=['x'], varargs=None, keywords=None, defaults=None)

This is a duplicate of this and this.

5
votes

I would use *args or **kwargs and throw an exception if the arguments are not as expected

If you want to have the same errors than the ones checked by python you can do something like

def check_arguments(function_name,args,arg_names):
    missing_count = len(arg_names) - len(args)
    if missing_count > 0:
        if missing_count == 1:
            raise TypeError(function_name+"() missing 1 required positionnal argument: "+repr(arg_names[-1]))
        else:
            raise TypeError(function_name+"() missing "+str(missing_count)+" required positionnal argument: "+", ".join([repr(name) for name in arg_names][-missing_count:-1])+ " and "+repr(arg_names[-1]))

using with somethin like 

def f(*args):
    check_arguments("f",args,["a","b","c"])
    #whatever you want
    ...
4
votes

Here's a function you can call to get the kwargs of the current function. Or if you want to use those lines directly in your own function instead of calling get_kwargs() just remove the .f_back

import inspect


def get_kwargs():
    frame = inspect.currentframe().f_back
    keys, _, _, values = inspect.getargvalues(frame)
    kwargs = {}
    for key in keys:
        if key != 'self':
            kwargs[key] = values[key]
    return kwargs


def test(x, y=100):
    z = 7
    print(get_kwargs())


test(5, 6)
# Output: {'x': 5, 'y': 6}
test(5)
# Output: {'x': 5, 'y': 100}
3
votes

From the accepted answer from a duplicate (older??) question https://stackoverflow.com/a/582206/1136458 :

    frame = inspect.currentframe()
    args, _, _, values = inspect.getargvalues(frame)
1
votes

One solution, using decorators, is here.

0
votes
class my_class():
        compulsory_default_kwargs = {'p':20,'q':30,'r':40}
        formal_args_names = ['var1','var2']
        def __init__(self,*args,**kwargs):
            for key , value in locals().items():
                if((key == 'args')):
                    if(len(args) == len(my_class.formal_args_names)):
                        for key_name , arg_value in zip(my_class.formal_args_names,args):
                            setattr(self,key_name,arg_value)
                    else:
                        raise Exception ('Error - Number of formal arguments passed mismatched required {} whereas passed {}'.format(len(my_class.formal_args_names),len(args)))

                elif((key == 'kwargs') & (len(kwargs)!=0)):
                    for kw_key , kw_value in kwargs.items():
                        if kw_key in my_class.compulsory_default_kwargs.keys():
                            setattr(self,kw_key,kw_value)
                        else:
                            raise Exception ('Invalid key-word argument passed {}'.format(kw_key))
                #elif((key!='self') & (key!='kwargs') ):
                   # setattr(self,key,value)
            for key , value in my_class.compulsory_default_kwargs.items():
                if key not in kwargs.keys():
                    setattr(self,key,value)


    this_value = 'Foo'    
    my_cl = my_class(3,this_value,p='B',r=10)
    my_cl.__dict__
-1
votes

You've specified the parameters in the header?

Why don't you simply use that same info in the body?

def foo(a, b, c):
   params = [a, b, c]

What have I missed?

-1
votes

You can create a list out of them using:

args = [a, b, c]

You can easily create a tuple out of them using:

args = (a, b, c)