列出dict python的值

Question

I am trying to write a function that contains a dict in a way that the value of every key looks in the following order: for exmple if the function got the following list:

list = [(a,b),(a,c)(f,q)] 

it would create this:

dict = {'a':('a',['b','c'],'f':('f',['q'])}


def create_dict(self, link_list):
    """
    creating a dict 
    :param link_list:
    :return:
    """
    collector = {}
    for tup in link_list:
        if tup[0] not in collector:
            collector[str(tup[0])] = (tup[0], [tup[1]])
        else:
            collector[tup[0]][1].append(tup[1])

I'm having some problems with the syntax and how to approach the list in the value of every key.

Answer 1

You've got a typo on the first line.

list = [(a,b),(a,c)(f,q)]  # missing comma
# should be:
list = [(a,b),(a,c),(f,q)]

In your version, (a, c) is a tuple and then without a comma, the (f, q) after it represents calling that as a function with f and q as parameters.

Also, don't use built-in names for variables. Name it list_ instead, at least.

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Edit: wrt "represents calling that as a function " == or rather, as a callable .