CODEWITHSUNDEEP
pythonpython-3.xfindkeymax
n00bprogrammer22
n00bprogrammer22

Reputation: 187

Finding key(s) with most values in dictionary

I'm working on a function where I need to find the keys (of artists names) in a dictionary that have the most values. Sometimes two keys have the same number of values and in that case I need to return a list of the artists names.

Example Dictionary:

{'M':[("One",1400,30.0, 20.5,"oil paint","Austria"),("Three",1430,100.0,102.0,"watercolor","France")],
        'P':[("Eight",1460, 225.0, 200.0, "fresco","Netherlands")],
        'U':[("Nine",1203,182.0, 957.0,"egg tempera","Italy"), ("Twelve",1200,76.2,101.6,"egg tempera","France")]
        }

For this dictionary since M and U have the most values (M has 2 and U has 2 while P only has 1) the function should return

artists_with_most_work(dictionary1())

['M', 'U']

How could I search for the number of values of each keys and return the ones that have the most? I figured using max() would be a good idea but I don't think I'm using it correctly in my current attempt below. Thanks to anyone who can help

Code:

def artist_with_most_work(db):
    matches = []
    for key, record_list in db.items():
        for record in record_list:
            if item in record: 
                max(db) = themax
            matches.append(themax)
        return matches

Upvotes: 2

Views: 6033

Answers (2)

zwol
zwol

Reputation: 140549

Because of the requirement to return all members of a tie, you have to keep track of all the possibilities.

def artist_with_most_work(db):
    keys_by_size = collections.defaultdict(list)
    maxsize = 0
    for key, recordlist in db.items():
        nitems = len(recordlist)
        keys_by_size[nitems].append(key)
        maxsize = max(maxsize, nitems)

    return keys_by_size[maxsize]

Upvotes: 0

Martijn Pieters
Martijn Pieters

Reputation: 1121654

You'll have to first find the maximum length, then return all keys that reference a list with that length:

def artist_with_most_work(db):
    maxcount = max(len(v) for v in db.values())
    return [k for k, v in db.items() if len(v) == maxcount]

Demo:

>>> def artist_with_most_work(db):
...     maxcount = max(len(v) for v in db.values())
...     return [k for k, v in db.items() if len(v) == maxcount]
...
>>> d1 = {'M':[("One",1400,30.0, 20.5,"oil paint","Austria"),("Three",1430,100.0,102.0,"watercolor","France")],
...       'P':[("Eight",1460, 225.0, 200.0, "fresco","Netherlands")],
...       'U':[("Nine",1203,182.0, 957.0,"egg tempera","Italy"), ("Twelve",1200,76.2,101.6,"egg tempera","France")]
...      }
>>> artist_with_most_work(d1)
['M', 'U']

Upvotes: 6

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