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Python Python Collections (2016, retired 2019) Dictionaries Packing and Unpacking Dictionaries

Noah Fields
Noah Fields
13,985 Points
Posted by Noah Fields
Noah Fields
13,985 Points

Program not Working, Troubleshooting Help?

Currently, my program is as follows:

template = "Hi, I'm {name} and I love to eat {food}!"

def string_factory(dict_list):
    new_list = []
    for diction in dict_list:
        unpacked = (**diction)
        if name and food in unpacked:
            string = ("Hi, I'm {} and I love to eat {}!".format(name, food))
            new_list.append(string)

Unpacked is based on code from this page: http://python-reference.readthedocs.io/en/latest/docs/operators/dict_unpack.html

I found this while looking for ways to unpack a dictionary based on a variable name (diction in this case), because I cannot reference it directly without knowing what arguments the user will pass to it.

When running this code, I am told there is a syntax error on line 11. I copied this code into IDLE, which highlights syntax errors, and the highlighted portion was the second asterisk on the line:

unpacked = (**diction)

Most critically, I need to know the proper way to unpack each dict in the list, without exactly specifying each dict. The video used one created on the spot, rather than one that was then referenced by a variable. As such, I'm not sure how to solve this problem.

Lionel Victor
Lionel Victor
2,490 Points

The reason I believe you have the syntax error is because unpacking is only done inside a function call

I was able to get the output by changing a few things on your code

  1. name and food in the IF statement should be in quotes to avoid NameError
  2. The dictionaries in the dict_list can directly be used in the IF statement
  3. Unpacking of 'diction' happens inside the .format() call

def string_factory(dict_list): new_list = [] #print new_list for diction in dict_list: #unpacked=(diction) if "name" and "food" in diction: string = ("Hi, I'm {name} and I love to eat {food}!".format(diction)) new_list.append(string) #print new_list return new_list

Regards, -Lionel