CODEWITHSUNDEEP
pythonlistdictionary
Jack Fleeting
Jack Fleeting

Reputation: 24930

Check dictionary keys for presence of list elements, create list of values if present, 'None' otherwise

Forgive the convoluted title, but I couldn't find a more elegant way to express the problem. The closet question I could locate can be found here, but it doesn't quite get me there.

Assume a number of dictionaries of varying lengths:

dict_1 = {'A':4, 'C':5}
dict_2 = {'A':1, 'B':2, 'C':3}
dict_3 = {'B':6}

The common denominator for these dictionaries is that they all share the keys on this list:

my_keys= ['A','B','C']

but some are missing one or more of the keys.

The intent is to create a list of lists, where each list element is a list of all the existing values in each dictionary, or 'None' where a specific key isn't present. They keys themselves, being identical across all dictionaries, can be disregarded.

So in this case, the expected output is:

    final_list = 
    [[4,"None",5],
    [1,2,3],
    ["None",6,"None"]]

I'm not sure exactly how to approach it. You could start with each element in my_keys, and check its presence against each key in each dictionary; if the relevant dict has the key, the value of that key is appended to a temporary list; otherwise, 'None' is appended. Once the my_keys are all iterated over, the temp list is appended to a final list and the cycle start again.

To me at least, it's easier said than done. I tried quite a few things (which I won't bother to post because they didn't get even close). So I was wondering if there is an elegant approach to the problem.

Upvotes: 1

Views: 61

Answers (1)

Ofer Sadan
Ofer Sadan

Reputation: 11932

dict.get can return a default value (for example, None). If we take your examples:

dict_1 = {'A':4, 'C':5}
dict_2 = {'A':1, 'B':2, 'C':3}
dict_3 = {'B':6}
my_keys= ['A','B','C']

Then dict_1.get('B', None) is the way to make sure we get a default None value. We can loop across all keys the same way:

def dict_to_list(d, keys):
    return [d.get(key, None) for key in keys]

Example:

>>> dict_to_list(dict_1, my_keys)
[4, None, 5]
>>> dict_to_list(dict_2, my_keys)
[1, 2, 3]
>>> dict_to_list(dict_3, my_keys)
[None, 6, None]

EDIT: None is the default argument even if it's not explicitly specified, so dict_1.get('B') would work just as well as dict_1.get('B', None)

Upvotes: 2

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