Most efficient way to find the indexes of unique values in a Python 3 list

Question

What is the most efficient way to find the indexes of strings in a list that occur only once?

foo = ['it', 'does', 'it', 'very', 'very', 'well']
bar = ???  # bar = [1, 5]

I already know about sets, dictionaries and list comprehensions. The problem I'm trying to solve here is in my production code I have parallel data lists where the index of one is the index of many which can't be changed for historical reasons.

Why the close votes as "too broad"? I thought it was well-targeted question that currently has only one good answer. — empty, Sep 29 '17 at 15:54

score 3 · Accepted Answer · answered Sep 27 '17 at 17:57

3

With collections.Counter subclass:

import collections

foo = ['it', 'does', 'it', 'very', 'very', 'well']
counts = collections.Counter(foo)
result = [i for i,v in enumerate(foo) if counts[v] == 1]

print(result)

The output:

[1, 5]

answered Sep 27 '17 at 17:57

RomanPerekhrest

88,541
4
65
105

Just curious : Why do you mention it's a subclass? Isn't (almost) every class in Python a subclass? – Eric Duminil Sep 27 '17 at 18:06
@EricDuminil, sometimes, I cite from documentation: *A Counter is a dict subclass for counting hashable objects.*. (of course, every class is a sublcass, except topmost ABC) – RomanPerekhrest Sep 27 '17 at 18:08
Okay, it sure makes sense if you specify Counter's superclass. Nice answer BTW. – Eric Duminil Sep 27 '17 at 18:10

kaushik santosh · Answer 2 · 2017-09-27T18:01:55.687

0

You will get what you want. Dictionaries are faster in python

from collections import Counter
foo = ['it', 'does', 'it', 'very', 'very', 'well']
d = dict(Counter(foo))
[i for i,v in enumerate(foo) if counts[v]  == 1]

You can also use set(foo)

edited Sep 27 '17 at 18:01

answered Sep 27 '17 at 18:00

kaushik santosh

3
3

There's already an answer with counter. Plus, this code outputs the unique words, not their indices. – Eric Duminil Sep 27 '17 at 18:01

score 0 · Answer 3 · answered Sep 29 '17 at 10:12

0

You can try something like this, especially if the size of your foo list is bigger than in your example above and have lots of duplicates.

seen = set()
[i for i,e in enumerate(foo) if not (e in seen or seen.add(e) or e in foo[i+1:])]

answered Sep 29 '17 at 10:12

Bruno Astrolino

401
5
3

score -2 · Answer 4 · answered Sep 27 '17 at 17:59

It depends on the kind of efficiency you would like to get. You could do this directly in a list comprehension, straightforward and readable:

bar = [index for index,el in enumerate(foo) if foo.count(el)==1]

Please see this for info if you would like to use Counter

Most efficient way to find the indexes of unique values in a Python 3 list

4 Answers4