How do I remove duplicates from a list, while preserving order? Using a set to remove duplicates destroys the original order. Is there a built-in or a Pythonic idiom?
Related question: In Python, what is the fastest algorithm for removing duplicates from a list so that all elements are unique while 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.)
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 (when it gained a C implementation; prior to 3.5 it's still the clearest solution, though not the fastest).
>>> from collections import OrderedDict
>>> list(OrderedDict.fromkeys('abracadabra'))
['a', 'b', 'r', 'c', 'd']
from itertools import groupby
[ key for key,_ in groupby(sortedList)]
The list doesn't even have to be sorted, the sufficient condition is that equal values are grouped together.
Edit: I assumed that "preserving order" implies that the list is actually ordered. If this is not the case, then the solution from MizardX is the right one.
Community edit: This is however the most elegant way to "compress duplicate consecutive elements into a single element".
here is a simple way to do it:
list1 = ["hello", " ", "w", "o", "r", "l", "d"]
sorted(set(list1 ), key=list1.index)
that gives the output:
["hello", " ", "w", "o", "r", "l", "d"]
pandas users should check out pandas.unique.
>>> import pandas as pd
>>> lst = [1, 2, 1, 3, 3, 2, 4]
>>> pd.unique(lst)
array([1, 2, 3, 4])
The function returns a NumPy array. If needed, you can convert it to a list with the tolist method.
1. These solutions are fine…
For removing duplicates while preserving order, the excellent solution(s) proposed elsewhere on this page:
seen = set()
[x for x in seq if not (x in seen or seen.add(x))]
and variation(s), e.g.:
seen = set()
[x for x in seq if x not in seen and not seen.add(x)]
are indeed popular because they are simple, minimalistic, and deploy the correct hashing for optimal efficency. The main complaint about these seems to be that using the invariant None "returned" by method seen.add(x) as a constant (and therefore excess/unnecessary) value in a logical expression—just for its side-effect—is hacky and/or confusing.
2. …but they waste one hash lookup per iteration.
Surprisingly, given the amount of discussion and debate on this topic, there is actually a significant improvement to the code that seems to have been overlooked. As shown, each "test-and-set" iteration requires two hash lookups: the first to test membership x not in seen and then again to actually add the value seen.add(x). Since the first operation guarantees that the second will always be successful, there is a wasteful duplication of effort here. And because the overall technique here is so efficient, the excess hash lookups will likely end up being the most expensive proportion of what little work remains.
3. Instead, let the set do its job!
Notice that the examples above only call set.add with the foreknowledge that doing so will always result in an increase in set membership. The set itself never gets an chance to reject a duplicate; our code snippet has essentially usurped that role for itself. The use of explicit two-step test-and-set code is robbing set of its core ability to exclude those duplicates itself.
4. The single-hash-lookup code:
The following version cuts the number of hash lookups per iteration in half—from two down to just one.
seen = set()
[x for x in seq if len(seen) < len(seen.add(x) or seen)]
Eliminating the duplicate values in a sequence, but preserve the order of the remaining items. Use of general purpose generator function.
# for hashable sequence
def remove_duplicates(items):
seen = set()
for item in items:
if item not in seen:
yield item
seen.add(item)
a = [1, 5, 2, 1, 9, 1, 5, 10]
list(remove_duplicates(a))
# [1, 5, 2, 9, 10]
# for unhashable sequence
def remove_duplicates(items, key=None):
seen = set()
for item in items:
val = item if key is None else key(item)
if val not in seen:
yield item
seen.add(val)
a = [ {'x': 1, 'y': 2}, {'x': 1, 'y': 3}, {'x': 1, 'y': 2}, {'x': 2, 'y': 4}]
list(remove_duplicates(a, key=lambda d: (d['x'],d['y'])))
# [{'x': 1, 'y': 2}, {'x': 1, 'y': 3}, {'x': 2, 'y': 4}]
One liner list comprehension:
values_non_duplicated = [value for index, value in enumerate(values) if value not in values[ : index]]
x = [1, 2, 1, 3, 1, 4]
# brute force method
arr = []
for i in x:
if not i in arr:
arr.insert(x[i],i)
# recursive method
tmp = []
def remove_duplicates(j=0):
if j < len(x):
if not x[j] in tmp:
tmp.append(x[j])
i = j+1
remove_duplicates(i)
remove_duplicates()
This method is quadratic, because we have a linear lookup into the list for every element of the list (to that we have to add the cost of rearranging the list because of the del s).
That said, it is possible to operate in place if we start from the end of the list and proceed toward the origin removing each term that is present in the sub-list at its left
This idea in code is simply
for i in range(len(l)-1,0,-1):
if l[i] in l[:i]: del l[i]
A simple test of the implementation
In [91]: from random import randint, seed
In [92]: seed('20080808') ; l = [randint(1,6) for _ in range(12)] # Beijing Olympics
In [93]: for i in range(len(l)-1,0,-1):
...: print(l)
...: print(i, l[i], l[:i], end='')
...: if l[i] in l[:i]:
...: print( ': remove', l[i])
...: del l[i]
...: else:
...: print()
...: print(l)
[6, 5, 1, 4, 6, 1, 6, 2, 2, 4, 5, 2]
11 2 [6, 5, 1, 4, 6, 1, 6, 2, 2, 4, 5]: remove 2
[6, 5, 1, 4, 6, 1, 6, 2, 2, 4, 5]
10 5 [6, 5, 1, 4, 6, 1, 6, 2, 2, 4]: remove 5
[6, 5, 1, 4, 6, 1, 6, 2, 2, 4]
9 4 [6, 5, 1, 4, 6, 1, 6, 2, 2]: remove 4
[6, 5, 1, 4, 6, 1, 6, 2, 2]
8 2 [6, 5, 1, 4, 6, 1, 6, 2]: remove 2
[6, 5, 1, 4, 6, 1, 6, 2]
7 2 [6, 5, 1, 4, 6, 1, 6]
[6, 5, 1, 4, 6, 1, 6, 2]
6 6 [6, 5, 1, 4, 6, 1]: remove 6
[6, 5, 1, 4, 6, 1, 2]
5 1 [6, 5, 1, 4, 6]: remove 1
[6, 5, 1, 4, 6, 2]
4 6 [6, 5, 1, 4]: remove 6
[6, 5, 1, 4, 2]
3 4 [6, 5, 1]
[6, 5, 1, 4, 2]
2 1 [6, 5]
[6, 5, 1, 4, 2]
1 5 [6]
[6, 5, 1, 4, 2]
In [94]:
zmk's approach uses list comprehension which is very fast, yet keeps the order naturally. For applying to case sensitive strings it can be easily modified. This also preserves the original case.
def DelDupes(aseq) :
seen = set()
return [x for x in aseq if (x.lower() not in seen) and (not seen.add(x.lower()))]
Closely associated functions are:
def HasDupes(aseq) :
s = set()
return any(((x.lower() in s) or s.add(x.lower())) for x in aseq)
def GetDupes(aseq) :
s = set()
return set(x for x in aseq if ((x.lower() in s) or s.add(x.lower())))
Credit to @wjandrea for dict.fromdict method idea:
def solve(arr):
return list(dict.fromkeys(arr[::-1]))[::-1]
This will reverse input and output to iterate properly
In case you use pydantic, you can use unique_list:
Make a list unique while maintaining order. We update the list if another one with the same name is set (e.g. root validator overridden in subclass)
Example:
from pydantic.utils import unique_list
print(unique_list([1, 2, 3, 1, 2, 3, 5, 6, 7, 8]))
# [1, 2, 3, 5, 6, 7, 8]
def remove_duplicates_thenSort():
t = ['b', 'c', 'd','d','a','c','c']
t2 = []
for i,k in enumerate(t):
index = t.index(k)
if i == index:
t2.append(t[i])
return sorted(t2)
print(remove_duplicates_thenSort())