How to identify an odd item in a list of items using python

Question

My goal is to identify the odd element in the list below.

list_1=['taska1', 'taska2', 'taska3', 'taskb2', 'taska7']

The odd item is tasksb2 as the other four items are under taska .

They all have equal length, hence discriminating using the len function will not work. Any ideas? thanks.

Answer 1

If you simply want to find the item that does not start with 'taska', then you could use the following list comprehension :

>>> list_1=['taska1', 'taska2', 'taska3', 'taskb2', 'taska7']
>>> print [l for l in list_1 if not l.startswith('taska')]
['taskb2']

Another option is to use filter + lambda :

>>> filter(lambda l: not l.startswith('taska'), list_1)
['taskb2']

Answer 2

Seems to be an easy problem solved by alphabetical sort.

print sorted(list_1)[-1]

Don't wanna sort? Try an O(n) time-complexity solution with O(1) space complexity:

print max(list_1)

Answer 3

If you know what the basic structure of the items will be, then it's easy.

If you don't know the structure of your items a priori, one approach is to score the items according to their similarity against each other. Using info from this question for the standard library module difflib ,

import difflib
import itertools

list_1=['taska1', 'taska2', 'taska3', 'taskb2', 'taska7']

# Initialize a dict, keyed on the items, with 0.0 score to start
score = dict.fromkeys(list_1, 0.0)

# Arrange the items in pairs with each other
for w1, w2 in itertools.combinations(list_1, 2):
    # Performs the matching function - see difflib docs
    seq=difflib.SequenceMatcher(a=w1, b=w2)
    # increment the "match" score for each
    score[w1]+=seq.ratio()
    score[w2]+=seq.ratio()

# Print the results

>>> score
{'taska1': 3.166666666666667,
 'taska2': 3.3333333333333335,
 'taska3': 3.166666666666667,
 'taska7': 3.1666666666666665,
 'taskb2': 2.833333333333333}

It turns out that taskb2 has the lowest score!