Fastest way to uniqify a list in Python

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])

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.

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])

looks like you are using Set from python2.3 and not set from python2.4

Peter - worth reading this:

http://www.joelonsoftware.com/items/2006/08/01.html

Short version: map() can be trivially parallelized, for() can't. Functional-tasting solutions are going to run faster in multi-processor environments. Worth considering.

This comes up often enough for me that I think there should be a built-in function.

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:
................continue
............seen.add(x)
............yield x
....else:
........for x in seq:
............x = idfun(x)
............if x in seen:
................continue
............seen.add(x)
............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]
>>>

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.

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)]

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

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

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.

If you recode the block

if marker in seen: continue
seen[marker] = 1
result.append(item)

to

if marker not in seen:
____seen[marker] = 1
____result.append(item)

the performance of f5 doubles!

fastest non preserving (?):

def f7(seq):
____S,L = set, list
____return S(L(seq))

Speed tip:
make local references in functions for builtins

def foo():
____L = []
____AppendToListL = L.append
...
...
____AppendToListL(item)

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. :)

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]

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.

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.

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 ;)

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%.

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).

thanks for this! saved me time testing speed on these various options.

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):
try:
return list(set(seq))
except NameError:
return {}.fromkeys(seq).keys()

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.

>>>li=['a', 'mpilgrim', 'foo', 'b', 'c', 'b', 'd', 'd']
>>>[x for x in li if x not in locals()['_[1]']]

output:

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

While, I can't explain how locals()['_[1]'] works....

cool

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

what about

dict(zip(seq,seq)).keys()

Thanks for sharing!

In your updated code (http://www.peterbe.com/plog/uniqifiers-benchmark/uniqifiers_benchmark.py), I notice that f11 calls f10. Is this an error? Did you run it this way?

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)]

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)

I find a way to get the unique list

A = [1,1,2,2,3,3]
A = [i for i in set(A)]

>>> aux = {}; [aux.setdefault(p, p) for p in list('ABeeE') if p not in aux];
['A', 'B', 'e', 'E']

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

Thanks for this write up. It was very useful this morning.

f1 0.0
o.O ? It finishes in 0.0 seconds? I think f1 is the fastest :)

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

FYI: You seem to have broken the download links for the script when migrating to your new blog.

Hello,
Very nice blog and post :D, the download link seems to be broken. Could you check that?
Cheers

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]

* 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

FYI: small spelling error ==> uniqifiers should be uniquifiers

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'

order set:

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

    return unique_list

why not?
def f5(self, seq, idfun=lambda x : x):

and remove:

  if idfun is None:
       def idfun(x): return x

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.
Cheers!

I'm curious if the recursive solution here: (http://stackoverflow.com/a/18729193/567620) 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, cmp_op=operator.ne):
    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]

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.

Would it be possible to create a github or other online repo for this?

Also, what is the license of the code?

hi,

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

M & M

What about lists of dictionaries like this:
all = [ {'id': '001', 'age': '12'}, {'id': '533', 'info': 'student'}, {'id': '001', 'age': '12'} ]

Asking toddrjens question again. What license is the code listed on this page?

Hi,

Thank you. f5 helped a lot!

Sadly f5 does not work for a list of lists

I am new to python and have run into a serious problem that is causing me to grey prematurely. The problem that I am currently running into is that I am trying to pull unique IP addresses from a list of IPs that often have duplicates. What I have read above seems to be a list populated by the user. How would you do this if the list is prepopulated text file? Right now this is what my script looks like for pulling unique IPs out:

def uniq_count(item):
      seen = set()
      for item in item:
           seen.add(ips)
           yield len(seen)

in small batches it works but when the list is larger it seems to spit out two of each. Any help would be appreciated.

Hi All,

I added a new variant of f5 / f5b (see below) that simply swaps the seen items being a dict to use a set. The thinking amongst myself, and colleagues, is a set() would be faster ... running the tests (simply adding below into the downloadable benchmark py) showed the dict (both f5 and f5b) is faster by 10-20% across python versions 2.6 - 3.1 ... does anyone else find this counter intuitive? and have a reason why this is so? (oh, for py3 testing I did remove f7 and used print function too)

# BEGIN f5c
def f5c(seq, idfun=None): # Alex Martelli ******* order preserving
    if idfun is None:
        def idfun(x): return x
    seen = set()
    result = []
    for item in seq:
        marker = idfun(item)
        # in old Python versions:
        # if seen.has_key(marker)
        # but in new ones:
        if marker not in seen:
            seen.add(marker)
            result.append(item)

    return result
# END f5c

TIA,

Hello guys, the ordered variant can be even faster, if you change the condition:

```
    seen = set()
    return [x for x in seq if not (x in seen or seen.add(x))]
```

and another quite important optimization is to avoid __getattr__ call by storing the actual used function:

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

in the example of the list above it mentioned that function f5 is the best solution. I thought that the fastest is f3 i do not understand why it shows f5. is it a typo? please explain

Wow, just wow. Using a dict{} to check duplicates is brilliant. So fast, so obviously faster than a repeat check in list. Well done!

www.JamesKoss.com

added https://gist.github.com/katakumpo/6931e18dcd0cc4c20f8f8f324cea6056 for python 3 compatibility and also with orderedset package if installed

Hi all, I discovered this page while trying to write a simple one-liner to de-dupe lists with order preservation. I came up with the following using a list comprehension and slicing. The idea is to slice the list one index at a time and only add an element to the de-duped list if it does not yet exist in the slice. I'm not sure about performance but as far as readability goes I like the simplicity of this one.

series = [v for i, v in enumerate(series) if series[0:i].count(v) == 0]

Joking m8

I'd just like to point out that the premise of the test is somewhat flawed, because the various methods have different asymptotic complexity. f5 is fast for short inputs with many duplicates, but it scales poorly, so cannot be said to be universally the best approach. I came across this page while thinking about how to identify (rather than eliminate) duplicates in numeric arrays using NumPy, but after I'd arrived at something several times faster for large inputs. Still, the presentation and comparison between the algorithms is interesting.