How do you remove duplicates from a list whilst preserving order? How do you remove duplicates from a list whilst preserving order? python python

How do you remove duplicates from a list whilst preserving order?


Here you have some alternatives: http://www.peterbe.com/plog/uniqifiers-benchmark

Fastest one:

def f7(seq):    seen = set()    seen_add = seen.add    return [x for x in seq if not (x in seen or seen_add(x))]

Why assign seen.add to seen_add instead of just calling seen.add? Python is a dynamic language, and resolving seen.add each iteration is more costly than resolving a local variable. seen.add could have changed between iterations, and the runtime isn't smart enough to rule that out. To play it safe, it has to check the object each time.

If you plan on using this function a lot on the same dataset, perhaps you would be better off with an ordered set: http://code.activestate.com/recipes/528878/

O(1) insertion, deletion and member-check per operation.

(Small additional note: seen.add() always returns None, so the or above is there only as a way to attempt a set update, and not as an integral part of the logical test.)


Edit 2020

As of CPython/PyPy 3.6 (and as a language guarantee in 3.7), plain dict is insertion ordered, and even more efficient than the (also C implemented) collections.OrderedDict. So the fastest solution, by far, is also the simplest:

>>> items = [1, 2, 0, 1, 3, 2]>>> list(dict.fromkeys(items))[1, 2, 0, 3]

Like list(set(items)) this pushes all the work to the C layer (on CPython), but since dicts are insertion ordered, dict.fromkeys doesn't lose ordering. It's slower than list(set(items)) (takes 50-100% longer typically), but much faster than any other order-preserving solution (takes about half the time of hacks involving use of sets in a listcomp).

Edit 2016

As Raymond pointed out, in python 3.5+ where OrderedDict is implemented in C, the list comprehension approach will be slower than OrderedDict (unless you actually need the list at the end - and even then, only if the input is very short). So the best solution for 3.5+ is OrderedDict.

Important Edit 2015

As @abarnert notes, the more_itertools library (pip install more_itertools) contains a unique_everseen function that is built to solve this problem without any unreadable (not seen.add) mutations in list comprehensions. This is also the fastest solution too:

>>> from  more_itertools import unique_everseen>>> items = [1, 2, 0, 1, 3, 2]>>> list(unique_everseen(items))[1, 2, 0, 3]

Just one simple library import and no hacks.This comes from an implementation of the itertools recipe unique_everseen which looks like:

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 filterfalse(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

In Python 2.7+ the accepted common idiom (which works but isn't optimized for speed, I would now use unique_everseen) for this uses collections.OrderedDict:

Runtime: O(N)

>>> from collections import OrderedDict>>> items = [1, 2, 0, 1, 3, 2]>>> list(OrderedDict.fromkeys(items))[1, 2, 0, 3]

This looks much nicer than:

seen = set()[x for x in seq if x not in seen and not seen.add(x)]

and doesn't utilize the ugly hack:

not seen.add(x)

which relies on the fact that set.add is an in-place method that always returns None so not None evaluates to True.

Note however that the hack solution is faster in raw speed though it has the same runtime complexity O(N).


In CPython 3.6+ (and all other Python implementations starting with Python 3.7+), dictionaries are ordered, so the way to remove duplicates from an iterable while keeping it in the original order is:

>>> list(dict.fromkeys('abracadabra'))['a', 'b', 'r', 'c', 'd']

In Python 3.5 and below (including Python 2.7), use the OrderedDict. My timings show that this is now both the fastest and shortest of the various approaches for Python 3.5.

>>> from collections import OrderedDict>>> list(OrderedDict.fromkeys('abracadabra'))['a', 'b', 'r', 'c', 'd']


matomo