I have a little code that takes a list of objects, and only outputs the items in the list that are unique.
This is my code
def only_once(a):
retu
To clarify, what you want is a set of items that appear once, and only once.
The best option here is to use collections.Counter(), as it means you only count the items once, rather than once per item, greatly increasing performance:
>>> import collections
>>> {key for key, count in collections.Counter(a).items() if count == 1}
{1, 2, 3}
We simply replace the square brackets with curly braces to signify a set comprehension over a list comprehension, to get a set of results.
[I'm assuming that you're also user1744238 and user1744316 -- please pick a username and stick to it, that way it's easier to check to see what variants of a question you've asked and what you've already tried.]
One set-based approach is to use two sets as a counter. You only care about whether you've seen something once or more than once. For example, here's an easy-to-explain approach:
once
and more
. once
. once
and add it to more
. once
. once
set so you don't output the same element twice.This gives me:
In [49]: f([1,4,6,7,3,2,4,5,7,5,6])
Out[49]: [1, 3, 2]
If you need to remove any item that is in the list more than once, not just occurences after the first, you can use:
# without using generators / comprehensions
def only_once(iterable):
seen = set()
duplicates = set()
for item in iterable:
if item in seen:
duplicates.add(item)
seen.add(item)
result = []
for item in iterable:
if item not in duplicates:
result.append(item)
return result
For general order-preserving duplicate elimination, see unique_everseen
in the itertools recipes:
def unique_everseen(iterable, key=None):
"List unique elements, preserving order. Remember all elements ever seen."
# unique_everseen('AAAABBBCCDAABBB') --> A B C D
# unique_everseen('ABBCcAD', str.lower) --> A B C D
seen = set()
seen_add = seen.add
if key is None:
for element in ifilterfalse(seen.__contains__, iterable):
seen_add(element)
yield element
else:
for element in iterable:
k = key(element)
if k not in seen:
seen_add(k)
yield element