Suppose you have a list in python that looks like this:

# or like this:

and you want to remove all duplicates so you get this result:

# or

How do you do that? ...the fastest way? I wrote a couple of alternative implementations and did a quick benchmark loop on the various implementations to find out which way was the fastest. (I haven't looked at memory usage). The slowest function was 78 times slower than the fastest function.

However, there's one very important difference between the various functions. Some are order preserving and some are not. For example, in an order preserving function, apart from the duplicates, the order is guaranteed to be the same as it was inputted. Eg, uniqify([1,2,2,3])==[1,2,3]

Here are the functions:

def f1(seq):
   # not order preserving
   set = {}
   map(set.__setitem__, seq, [])
   return set.keys()

def f2(seq): 
   # order preserving
   checked = []
   for e in seq:
       if e not in checked:
   return checked

def f3(seq):
   # Not order preserving
   keys = {}
   for e in seq:
       keys[e] = 1
   return keys.keys()

def f4(seq): 
   # order preserving
   noDupes = []
   [noDupes.append(i) for i in seq if not noDupes.count(i)]
   return noDupes

def f5(seq, idfun=None): 
   # order preserving
   if idfun is None:
       def idfun(x): return x
   seen = {}
   result = []
   for item in seq:
       marker = idfun(item)
       # in old Python versions:
       # if seen.has_key(marker)
       # but in new ones:
       if marker in seen: continue
       seen[marker] = 1
   return result

def f6(seq):
   # Not order preserving    
   set = Set(seq)
   return list(set)

And what you've all been waiting for (if you're still reading). Here are the results:

* f2 13.24
* f4 11.73
* f5 0.37
f1 0.18
f3 0.17
f6 0.19

(* order preserving)

Clearly f5 is the "best" solution. Not only is it really really fast; it's also order preserving and supports an optional transform function which makes it possible to do this:

>>> a=list('ABeeE')
>>> f5(a)
>>> f5(a, lambda x: x.lower())

Download the benchmark script here


From the comments I've now added a couple of more functions to the benchmark. Some which don't support uniqify a list of objects that can't be hashed unless passed with a special hashing method. So see all the functions download the file

Here are the new results:

* f5 10.1
* f5b 9.99
* f8 6.49
* f10 6.57
* f11 6.6
f1 4.28
f3 3.55
f6 4.03
f7 2.59
f9 2.58

(f2 and f4) were too slow for this testdata.

Lawrence Oluyede - 14 August 2006 [«« Reply to this]
Keep in mind you can also use:

>>> lst = [1, 1, 3, 4, 4, 5, 6, 7, 6]
>>> set(lst)
set([1, 3, 4, 5, 6, 7])
Paul Boddie - 14 August 2006 [«« Reply to this]
Isn't that what f6 does, apart from the final conversion to a list again?
Lawrence Oluyede - 14 August 2006 [«« Reply to this]
Right. I totally missed f6()
none - 14 August 2006 [«« Reply to this]
It would have been interesting to tests against more complex objects which redefines __cmp__.

- Sylvain

PS: Your comment system seems to be broken as I currently have the name and email fields filled up with Lawrence details.
Peter Bengtsson - 14 August 2006 [«« Reply to this]
Thanks for the bug report. I'm working on fixing it. It's due to the fact that I've started caching the pages on a proxying server. Thanks.
Peter Bengtsson - 14 August 2006 [«« Reply to this]
Should be fixed now.
M Lauer - 14 August 2006 [«« Reply to this]
problem with the previous comment:
>>> lst = ['a','a','b',1,1,2,3,4,5]
>>> set(lst)
set(['a', 1, 2, 'b', 4, 5, 3])
Justin - 14 August 2006 [«« Reply to this]
looks like you are using Set from python2.3 and not set from python2.4
djo - 14 August 2006 [«« Reply to this]
Peter - worth reading this:

Short version: map() can be trivially parallelized, for() can't. Functional-tasting solutions are going to run faster in multi-processor environments. Worth considering.
Martijn Faassen - 15 August 2006 [«« Reply to this]
So how would you parallelize uniqify, trivially or not?
djo - 15 August 2006 [«« Reply to this]
That's a good point; I just jumped in with something I thought was useful when I saw map(), the comparison implicit in __setitem__ passed me by. Using map() to generate side-effects is pretty alien to me.

In that case, I guess I'd do something based loosely on quicksort. It's nearly there already, it simply needs to throw away duplicate results when concatenating the child arrays (easy to do when the child arrays are guaranteed in-order and unique), and because it's a divide and conquer algorithm it should lend itself to a parallel environment.
ferringb - 15 August 2006 [«« Reply to this]
not much point in parallelizing sorting unless # of elements are high enough that the splitting is actually worth it; plus at least for python, there still is the gil to worry about...
Brantley Harris - 15 August 2006 [«« Reply to this]
This comes up often enough for me that I think there should be a built-in function.
Andrew Dalke - 15 August 2006 [«« Reply to this]
Your benchmark code likely doesn't dowhat you think it does for the current case. The 'X' instances always compare different so everything is unique.

Here's my example, with a bit of a cheat to make the "idfun=None' case faster. "Using "." for leading spaces because I can't figure out how to put Python code in this comment system

def f7(seq, idfun=None):
....return list(_f7(seq, idfun))
def _f7(seq, idfun=None):
....seen = set()
....if idfun is None:
........for x in seq:
............if x in seen:
............yield x
........for x in seq:
............x = idfun(x)
............if x in seen:
............yield x

Since your benchmark didn't test that case I figured I could ignore it. :) The timing numbers I get are

* f2 66.65
* f4 66.13
* f5 2.19
* f7 1.91
f1 1.06
f3 0.97
f6 0.99

and these are in line with my own benchmark. Function call overhead in Python is high. Most of the performance difference comes from calling idfun.

I also don't incur the overhead of doing the list.append lookup for each new element. The list making is all in C code. There is overhead for the generator but that's pretty fast. you may also end up prefering an iterator solution over making full lists.

Regarding the parallelization of 'map'. That assumes a pure functional environment. Consider

>>> class Counter(object):
... def __init__(self): self.count = 0
... def __call__(self, x):
... i = self.count; self.count += 1; return i
>>> map(Counter(), [9, 8, 3, 6, 1])
[0, 1, 2, 3, 4]
Manuzhai - 15 August 2006 [«« Reply to this]
I'd say that list(set([...])) is short enough? Or you could just work from the set directly. And I think the built-in set got a lot faster going from 2.3 to 2.4, due to it being rewritten in C.
Fuzzyman - 16 August 2006 [«« Reply to this]
Even in Python 2.4 the built-in set is based on the Python dictionary. This is why the speed results were so similar.

In Python 2.5 it has been optimised further with a custom internal data-structure, and *should* be faster.

Peter Bengtsson - 16 August 2006 [«« Reply to this]
but not ordered unfortunately.
Dave Kirby - 15 August 2006 [«« Reply to this]
firstly "set" is a built in type in 2.4, and it is bad form to name variables the same as existing types or modules - it can lead to confusing code and subtle bugs.

Secondly, the sets.Set class used in f6 is implemented in Python, both in 2.3 and 2.4, while the set builtin type is implemented in C so should be significantly faster.

Here is alternative order-preserving function. I have not timed it, but it should be pretty fast:

def f7(seq):
seen = set()
return [ x for x in seq if x not in seen and not seen.add(x)]
Michael Urman - 15 August 2006 [«« Reply to this]
What's most interesting is how changing the data seriously changes the actual performance. So if you know what your data looks like, you might be able do better than f5. If you don't know, f5 is probably a safe bet.

For instance, by changing the data to list('abcab'), and running the test 1000 times instead of 10, f2 becomes second fastest, and almost twice as fast as f5:
* f2 0.14
* f4 0.22
* f5 0.28
f1 0.16
f3 0.12
f6 0.45
Jack Diederich - 15 August 2006 [«« Reply to this]
Kirby had the right idea but his implementation is more complicated than it needs to be.

def f7(seq): return list(set(seq))

You can drop the wrapping list() if you just need a sequence and not a real list. In that case the function is just

f7 = set
Charlie Gunyon - 16 August 2006 [«« Reply to this]
Alright, so my list of changes:

Don't use sets.Set, use set built-in for all functions.

Use Andrew Dalke's f5() (though returning a generator is not what was required, reworked to return a list using list.append)

Use my f7(), which is:

def f7(seq):
....return {}.fromkeys(seq).keys()

Use Dave Kirby's f7() as f8().

Modify the test case to be huge:

blahlist = range(100000)
testdata = map(lambda a: a % 3, blahlist) + blahlist

Stop using f2() and f4(), they lose every time.

Results (old test case on the right):

* f5 54.688 --- 0.328
* f8 43.797 --- 0.297
f1 32.047 --- 0.156
f3 24.14 --- 0.157
f6 13.531 --- 0.14
f7 14.251 --- 0.156

So it looks like using the set built-in beats both sets.Set() and dict.

The other thing is there's probably very few cases where you'd want to unique-ify a list AND care about the order of that list at the same time. Removing duplicate transmissions is in that category (probably ought to fix your protocol if that's the case, though), but if there's others I can't think of them.
Anonymous - 16 August 2006 [«« Reply to this]
If you recode the block

if marker in seen: continue
seen[marker] = 1


if marker not in seen:
____seen[marker] = 1

the performance of f5 doubles!
Anonymous - 17 August 2006 [«« Reply to this]
fastest non preserving (?):

def f7(seq):
____S,L = set, list
____return S(L(seq))
Jiri Vit - 18 August 2006 [«« Reply to this]
Speed tip:
make local references in functions for builtins

def foo():
____L = []
____AppendToListL = L.append
thesamet - 20 August 2006 [«« Reply to this]
The results just confirm a known strategy to optimize python code. Replace python logic with C implementation that does the same. If it's built-in, the better. :)
Mike Watkins - 21 August 2006 [«« Reply to this]
if the nicely simple list(set(seq)) appeals, then this variation may as well as its fast and preserves order:

def f13(seq):
# order preserving
....uniq = set(seq)
....return [item for item in seq if item in uniq]
Mike Watkins - 21 August 2006 [«« Reply to this]
oops. never mind the above... should really have a coffee first (and write a test case too ...) Dave Kirby's version gets a +1 from me.
Mike Watkins - 21 August 2006 [«« Reply to this]
The coffee-corrected return for f13 (above) should be:

return [item for item in seq if item in uniq and not uniq.remove(item)]

Essentially the inverse of Dave Kirby's approach but just a hair faster for whatever reason set() only knows.
nn - 22 August 2006 [«« Reply to this]
My bet is on the memory allocation strategy. Adding to a set repeatedly allocates more memory for the set. I guess removing from the set does not deallocate memory until the end of the function.
Sven Berkvens-Matthijsse - 12 October 2011 [«« Reply to this]
The (small) disadvantage over Mike Watkins' version when compared to Dave Kirby's version is that Mike's version walks the given sequence twice. That means that it cannot be a generator, while Dave's version will handle generators just fine.
Maxime Biais - 14 February 2007 [«« Reply to this]
I always use this one (order preserving, generative, concise, but don't work on no hashable).

from itertools import ifilter
def _f14(iterable):
....m = set()
....return ifilter(lambda x: not (m.__contains__(x) or m.add(x)), iterable)

def f14(lst):
....return list(_f14(lst))

My results:
* f5 1.94
* f5b 1.89
* f8 1.18
* f10 1.24
* f11 1.23
* f13 1.31
* f14 1.87
f1 0.67
f3 0.64
f6 0.68
f7 0.38
f9 0.41

Not so bad ;)
MizardX - 22 January 2009 [«« Reply to this]
I tried optimizing the top competitors:

* f8 5.239
* f8b 4.415 <--
* f8c 4.971
* f11 5.692
* f11b 4.882
* f11c 4.713 <--
* f12 5.76
* f12b 4.911 <--

f12 is Mike Watkins' function.

f8b, f11b, f12b is unchanged except for storing a reference to seen.add/uniq.remove

f8c also tried to store a reference to seen.__contains__, which back-fired.

In f11c I tried to change if-continue-statements to if-not-statements, as Anonymous sugested.

Optimizing attribute lookup can give 15% speedup.
Removing continue statements can give another 5%.

Changing the in-operator to a call to __contains__ removes 10%.
zoof - 17 June 2009 [«« Reply to this]
I noticed that these functions don't work with lists of lists (i.e., lists are not hashable) -- is there an easy way to implement these functions for lists of lists or does that quickly become intractable (e.g., lists of lists of lists).
adrian - 08 July 2009 [«« Reply to this]
thanks for this! saved me time testing speed on these various options.
Leigh Gordon - 24 July 2009 [«« Reply to this]
I use this function on a script that needs to work on both version 2.3 and higher versions(and doesn't care about preserving order).

def unique(seq):
return list(set(seq))
except NameError:
return {}.fromkeys(seq).keys()
Anonymous - 27 August 2009 [«« Reply to this]
python newbie here..i need to unify an input, could someone help me with this?
i do not understand how to use any of these functions in a program.
testasdf1 - 17 September 2009 [«« Reply to this]
>>>li=['a', 'mpilgrim', 'foo', 'b', 'c', 'b', 'd', 'd']
>>>[x for x in li if x not in locals()['_[1]']]


['a', 'mpilgrim', 'foo', 'b', 'c', 'd']

While, I can't explain how locals()['_[1]'] works....
qman - 05 November 2009 [«« Reply to this]
I could explain it, but then I'd have to kill you...
Rick - 28 September 2009 [«« Reply to this]
Anonymous - 14 November 2009 [«« Reply to this]
Returns a unique list, sorted. (Speed is medium.)

[a for a,b in itertools.groupby(sorted(seq))]

It's f12 in the results below:
* f5 90.08
* f5b 87.62
* f8 15.004
* f10 56.096
* f11 59.153
f1 5.546
f3 21.098
f6 31.611
f7 3.47
f9 4.584
f12 25.881
Anonymous - 16 November 2009 [«« Reply to this]
what about

David - 11 December 2009 [«« Reply to this]
why on earth would you do that when you could do list(set(seq)) ??
Anonymous - 26 January 2010 [«« Reply to this]
Perhaps you would want the ordering?
Lynwood - 29 January 2010 [«« Reply to this]
This is absolutely ingenious! Thank you for sharing!
Eddie - 14 April 2010 [«« Reply to this]
Warning, this method is not guaranteed to be order-preserving. See -- "CPython implementation detail: Keys and values are listed in an arbitrary order which is non-random, varies across Python implementations, and depends on the dictionary’s history of insertions and deletions." -- Note it says "arbitrary" and it is implementation-dependent.
Mark - 13 March 2010 [«« Reply to this]
Thanks for sharing!
J Wolter - 20 April 2010 [«« Reply to this]
In your updated code (, I notice that f11 calls f10. Is this an error? Did you run it this way?
Peter Bengtsson - 20 April 2010 [«« Reply to this]
Must be a typo. Haven't looked at this for a long time.
juanpi - 06 May 2010 [«« Reply to this]
f1 apparently doesn't work on lists of tuples.
It would be nice to have a generalized version.

[(0, 1), (1, 2), (3, 2), (2, 1), (2, 0), (2, 3), (1, 0), (0, 2)]
lapo3399 - 02 June 2010 [«« Reply to this]
This seems to work on any list, including one of lists or tuples:

[[elem,already.append(elem)][0] for elem in thisList if elem not in already]

Now, I'm not sure if it's bad practise to call a function in there sneakily like that, but it works ;).
Davidog - 22 February 2011 [«« Reply to this]
I'm a newbie to Python, learning a whole new set of skills and programming methods. This page was a complete mini- education in itself.

My thanks to everyone who participated in the dialog, and especially our host, Peter Bengtsson.

"I am not worthy!" (Mike Myers and Dana Carvey)
Ming Huang - 05 April 2011 [«« Reply to this]
I find a way to get the unique list

A = [1,1,2,2,3,3]
A = [i for i in set(A)]
Joe Jiang - 28 April 2011 [«« Reply to this]
>>> aux = {}; [aux.setdefault(p, p) for p in list('ABeeE') if p not in aux];
['A', 'B', 'e', 'E']
Sergey Shepelev - 25 June 2011 [«« Reply to this]
Thank you very much for this collection of functions. However, i
needed fast version with key (id) function that doesn't preserve
order, here's what i came up with:

def f12(seq, idfun):
return dict((idfun(x), x) for x in seq).values()

timing(f12, 100, testdata, len)

Results on my box:
* f8 2.43
* f10 2.5
* f11 2.6
f3 1.56
f7 1.1
f9 1.21
f12 1.8
Mike Hoy - 17 August 2011 [«« Reply to this]
Thanks for this write up. It was very useful this morning.
Anonymous - 28 September 2011 [«« Reply to this]
f1 0.0
o.O ? It finishes in 0.0 seconds? I think f1 is the fastest :)
Lars Hupfeldt - 03 November 2011 [«« Reply to this]
Hi, I found a bug in the f10 version. When calling with the idfun argument it returns a list of the id objects instead of the original objects.
: else:
:: for x in seq:
::: x = idfun(x)
::: if x in seen:
:::: continue
::: seen.add(x)
::: yield x
should be:
: else:
:: for x in seq:
::: xi = idfun(x)
::: if xi in seen:
:::: continue
::: seen.add(xi)
::: yield x
Dana Woodman - 09 March 2012 [«« Reply to this]
FYI: You seem to have broken the download links for the script when migrating to your new blog.
hasan bardak - 19 April 2012 [«« Reply to this]
Very nice blog and post :D, the download link seems to be broken. Could you check that?
Peter Bengtsson - 19 April 2012 [«« Reply to this]
Fixed now. Thanks for noticing.
James Robert - 26 April 2012 [«« Reply to this]
My personal favorite:
(preserves order)

def uniq(iterable, idfunc=lambda x:x):
. . . . seen = set()
. . . . return [seen.add(idfunc(x)) or x for x in iterable if idfunc(x) not in seen]
Anonymous - 08 May 2012 [«« Reply to this]
* f5 4.846
* f5b 4.93
* f8 3.316
* f10 3.643
* f11 3.638
f1 2.064
f3 1.63
f6 1.935
f7 1.072
f9 1.199
Anonymous - 08 May 2012 [«« Reply to this]
FYI: small spelling error ==> uniqifiers should be uniquifiers
eseprimo - 26 May 2012 [«« Reply to this]
Did anyone address the issues with non hashable types? Note that sets and non hashable types don't go well together---which renders useless most of the functions above, except when dealing with trivial data. For example,

>>> alist=['a',[1,3,4],[3,3,3],[2,'b'],'a']
>>> set(alist)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unhashable type: 'list'
Anonymous - 14 June 2012 [«« Reply to this]
Not that hard. Some of the ordering is changed but that could be understandable since we are removing dups.

tp = lambda x: str(type(x)).split("'")[1].split("'")[0].replace("'",'')
blist = [ i if tp(i) != 'list' else [ j for j in set(i) ] for i in alist ]
print("Alist: {0}\nBList: {1}".format(alist,blist))

Alist: ['a', [1, 3, 4], [3, 3, 3], [2, 'b'], 'a']
BList: ['a', [3, 1, 4], [3], ['b', 2], 'a']
Anonymous - 14 June 2012 [«« Reply to this]
Not too hard. Bit ugly but working.

tp = lambda x: str(type(x)).split("'")[1].split("'")[0].replace("'",'')
blist = [ i if tp(i) != 'list' else [ j for j in set(i) ] for i in alist ]
clist = list([ i for i in set( [ i if tp(i) != 'list' else '!None' for i in blist ] ) ])[:-1] +[ i for i in blist if tp(i)=='list' ]
print("Alist: {0}\nBList: {1}\nClist: {2}".format(alist,blist, clist))
grd - 03 August 2012 [«« Reply to this]
order set:

def UniqueList(List):
    unique_set = sets.Set()
    unique_list = []
    for n in List:
        if n not in unique_set:

    return unique_list
nasgar - 14 November 2012 [«« Reply to this]
why not?
def f5(self, seq, idfun=lambda x : x):

and remove:

  if idfun is None:
       def idfun(x): return x
Peter Bengtsson - 15 November 2012 [«« Reply to this]
Because that makes it possible to optionally override what the key should be. Some items aren't hashable. E.g. dicts.
Harsh - 04 December 2012 [«« Reply to this]
Peter - nasgar's merely replacing your identity function plug with an inline lambda in the function def. I think that is valid to do (achieves the same thing), but does look harder to read.
kmbt - 05 March 2013 [«« Reply to this]
Hi. You have an error in your benchmark in the function
def f11(seq): # f10 but simpler
    # Order preserving
    return list(_f10(seq))

It should say _f11 instead of _f10 in the last line.
Ely Spears - 11 September 2013 [«« Reply to this]
I'm curious if the recursive solution here: ( is comparable for average-case kinds of lists (where number of unique elements is relatively small compared to total number of elements)?

This was inspired by Haskell's `nub` function. It would be cool to add it to the collection of uniquifiers in the main post and explore some timing about it. Below is the code for good measure:

def unique(lst):
    return [] if lst==[] else [lst[0]] + unique(filter(lambda x: x != lst[0], lst[1:]))

I also think it's interesting that this could be readily generalized to uniqueness by other operations. Like this:

import operator
def unique(lst,
    return [] if lst==[] else [lst[0]] + unique(filter(lambda x: cmp_op(x, lst[0]), lst[1:]), cmp_op)

For example, you could pass in a function that uses the notion of rounding to the same integer as if it was "equality" for uniqueness purposes, like this:

def test_round(x,y):
    return round(x) != round(y)

then unique(some_list, test_round) would provide the unique elements of the list where uniqueness no longer meant traditional equality (which is implied by using any sort of set-based or dict-key-based approach to this problem) but instead meant to take only the first element that rounds to K for each possible integer K that the elements might round to, e.g.:

In [6]: unique([1.2, 5, 1.9, 1.1, 4.2, 3, 4.8], test_round)
Out[6]: [1.2, 5, 1.9, 4.2, 3]
Aaron Swan - 28 January 2014 [«« Reply to this]
I needed the index of the unique elements in an order preserved manner, so I made a small modification to f8:

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

Hope this is helpful to someone.

Great post. Thank you.
toddrjen - 23 March 2014 [«« Reply to this]
Would it be possible to create a github or other online repo for this?

Also, what is the license of the code?
Murali - 20 April 2014 [«« Reply to this]

Thanks a ton for the snippet. f5 works like a charm...and it's fast too!

M & M
Jane - 21 October 2014 [«« Reply to this]
What about lists of dictionaries like this:
all = [ {'id': '001', 'age': '12'}, {'id': '533', 'info': 'student'}, {'id': '001', 'age': '12'} ]
qemi - 23 October 2014 [«« Reply to this]
Asking toddrjens question again. What license is the code listed on this page?
Peter Bengtsson - 21 August 2015 [«« Reply to this]
Free for all.
Satish - 25 November 2014 [«« Reply to this]

Thank you. f5 helped a lot!
Cornelis - 21 August 2015 [«« Reply to this]
Sadly f5 does not work for a list of lists

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