Is there a built-in that removes duplicates from list in Python, whilst preserving order? I know that I can use a set to remove duplicates, but that destroys the original order. I also know that I can roll my own like this:
def uniq(input): output = [] for x in input: if x not in output: output.append(x) return output
(Thanks to unwind for that code sample.)
But I'd like to avail myself of a built-in or a more Pythonic idiom if possible.
Related question: In Python, what is the fastest algorithm for removing duplicates from a list so that all elements are unique while preserving order?
➤ Go to Data > Remove Duplicates tool in Excel Toolbar under the section Data Tools. Step 2: ➤ Click on Remove Duplicates. ➤ Put a check on all names of the columns you want to remove duplicates from.
sort() does not remove duplicates.
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.)
The best solution varies by Python version and environment constraints:
First introduced in PyPy 2.5.0, and adopted in CPython 3.6 as an implementation detail, before being made a language guarantee in Python 3.7, plain dict
is insertion-ordered, and even more efficient than the (also C implemented as of CPython 3.5) collections.OrderedDict
. So the fastest solution, by far, is also the simplest:
>>> items = [1, 2, 0, 1, 3, 2] >>> list(dict.fromkeys(items)) # Or [*dict.fromkeys(items)] if you prefer [1, 2, 0, 3]
Like list(set(items))
this pushes all the work to the C layer (on CPython), but since dict
s 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 set
s in a listcomp).
Important note: The unique_everseen
solution from more_itertools
(see below) has some unique advantages in terms of laziness and support for non-hashable input items; if you need these features, it's the only solution that will work.
As Raymond pointed out, in CPython 3.5 where OrderedDict
is implemented in C, ugly list comprehension hacks are slower than OrderedDict.fromkeys
(unless you actually need the list at the end - and even then, only if the input is very short). So on both performance and readability the best solution for CPython 3.5 is the OrderedDict
equivalent of the 3.6+ use of plain dict
:
>>> from collections import OrderedDict >>> items = [1, 2, 0, 1, 3, 2] >>> list(OrderedDict.fromkeys(items)) [1, 2, 0, 3]
On CPython 3.4 and earlier, this will be slower than some other solutions, so if profiling shows you need a better solution, keep reading.
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 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.
The module is adapting 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
but unlike the itertools
recipe, it supports non-hashable items (at a performance cost; if all elements in iterable
are non-hashable, the algorithm becomes O(n²)
, vs. O(n)
if they're all hashable).
Important note: Unlike all the other solutions here, unique_everseen
can be used lazily; the peak memory usage will be the same (eventually, the underlying set
grows to the same size), but if you don't list
ify the result, you just iterate it, you'll be able to process unique items as they're found, rather than waiting until the entire input has been deduplicated before processing the first unique item.
You have two options:
Copy and paste in the unique_everseen
recipe to your code and use it per the more_itertools
example above
Use ugly hacks to allow a single listcomp to both check and update a set
to track what's been seen:
seen = set() [x for x in seq if x not in seen and not seen.add(x)]
at the expense of relying on 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 that all of the solutions above are O(n)
(save calling unique_everseen
on an iterable of non-hashable items, which is O(n²)
, while the others would fail immediately with a TypeError
), so all solutions are performant enough when they're not the hottest code path. Which one to use depends on which versions of the language spec/interpreter/third-party modules you can rely on, whether or not performance is critical (don't assume it is; it usually isn't), and most importantly, readability (because if the person who maintains this code later ends up in a murderous mood, your clever micro-optimization probably wasn't worth it).
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