How to find most common element in list, and if there's a tie, the one who's last occurance is first?

python

Solution

Approach

Find the most frequently occurring items (Counter.most_common). Then find the item among those candidates that has the minimum index (enumerate into a dictionary of indexes, min of {index: key}.iteritems()).

Code

Stealing liberally from @gnibbler and @Jeff:

from collections import Counter

def most_frequent_first(events):
    frequencies = Counter(events)
    indexes = {event: i for i, event in enumerate(events)}
    most_frequent_with_indexes = {indexes[key]: key for key, _ in frequencies.most_common()}
    return min(most_frequent_with_indexes.iteritems())[1]

events = [123,123,456,456,456,123, 1, 2, 3, 2, 3]
print(most_frequent_first(events))

Result

>>> print(most_frequent_first(events))
456

Code

A better piece of code would provide you with the frequency and the index, showing you that the code is working correctly. Here is an implementation that uses a named_tuple:

from collections import Counter, namedtuple

frequent_first = namedtuple("frequent_first", ["frequent", "first"])

def most_frequent_first(events):
    frequencies = Counter(events)
    indexes = {event: i for i, event in enumerate(events)}
    combined = {key: frequent_first(value, indexes[key]) for key, value in frequencies.iteritems()}
    return min(combined.iteritems(), key=lambda t: (-t[1].frequent, t[1].first))

events = [123,123,456,456,456,123, 1, 2, 3, 2, 3]
print(most_frequent_first(events))

Result

>>> print(most_frequent_first(events))
(456, frequent_first(frequent=3, first=4))

Problem

Basically if given a list ``` events = [123,123,456,456,456,123] ``` I expect it returns 456 because 456 was last seen earlier than 123 was last seen. I made lists comprised of the counts and indices of the initial list of numbers. I also made a dictionary in which the key is the element from events (original part) and hte value is the `.count()` of the key. I don't really know where to go from here and could use some help.

Original source