Get unique values from a list in python

Question

I want to get the unique values from the following list:

['nowplaying', 'PBS', 'PBS', 'nowplaying', 'job', 'debate', 'thenandnow']

The output which I require is:

['nowplaying', 'PBS', 'job', 'debate', 'thenandnow']

This code works:

output = []
for x in trends:
    if x not in output:
        output.append(x)
print(output)

is there a better solution I should use?

Answer 1

First declare your list properly, separated by commas. You can get the unique values by converting the list to a set.

mylist = ['nowplaying', 'PBS', 'PBS', 'nowplaying', 'job', 'debate', 'thenandnow']
myset = set(mylist)
print(myset)

If you use it further as a list, you should convert it back to a list by doing:

mynewlist = list(myset)

Another possibility, probably faster would be to use a set from the beginning, instead of a list. Then your code should be:

output = set()
for x in trends:
    output.add(x)
print(output)

As it has been pointed out, sets do not maintain the original order. If you need that, you should look for an ordered set implementation (see this question for more).

Answer 2

To be consistent with the type I would use:

mylist = list(set(mylist))

Answer 3

If we need to keep the elements order, how about this:

used = set()
mylist = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
unique = [x for x in mylist if x not in used and (used.add(x) or True)]

And one more solution using reduce and without the temporary used var.

mylist = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
unique = reduce(lambda l, x: l.append(x) or l if x not in l else l, mylist, [])

UPDATE - Dec, 2020 - Maybe the best approach!

Starting from python 3.7, the standard dict preserves insertion order.

Changed in version 3.7: Dictionary order is guaranteed to be insertion order. This behavior was an implementation detail of CPython from 3.6.

So this gives us the ability to use dict.fromkeys() for de-duplication!

NOTE: Credits goes to @rlat for giving us this approach in the comments!

mylist = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
unique = list(dict.fromkeys(mylist))

In terms of speed - for me its fast enough and readable enough to become my new favorite approach!

UPDATE - March, 2019

And a 3rd solution, which is a neat one, but kind of slow since .index is O(n).

mylist = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
unique = [x for i, x in enumerate(mylist) if i == mylist.index(x)]

UPDATE - Oct, 2016

Another solution with reduce, but this time without .append which makes it more human readable and easier to understand.

mylist = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
unique = reduce(lambda l, x: l+[x] if x not in l else l, mylist, [])
#which can also be writed as:
unique = reduce(lambda l, x: l if x in l else l+[x], mylist, [])

NOTE: Have in mind that more human-readable we get, more unperformant the script is. Except only for the dict.fromkeys() approach which is python 3.7+ specific.

import timeit

setup = "mylist = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']"

#10x to Michael for pointing out that we can get faster with set()
timeit.timeit('[x for x in mylist if x not in used and (used.add(x) or True)]', setup='used = set();'+setup)
0.2029558869980974

timeit.timeit('[x for x in mylist if x not in used and (used.append(x) or True)]', setup='used = [];'+setup)
0.28999493700030143

# 10x to rlat for suggesting this approach!   
timeit.timeit('list(dict.fromkeys(mylist))', setup=setup)
0.31227896199925453

timeit.timeit('reduce(lambda l, x: l.append(x) or l if x not in l else l, mylist, [])', setup='from functools import reduce;'+setup)
0.7149233570016804

timeit.timeit('reduce(lambda l, x: l+[x] if x not in l else l, mylist, [])', setup='from functools import reduce;'+setup)
0.7379565160008497

timeit.timeit('reduce(lambda l, x: l if x in l else l+[x], mylist, [])', setup='from functools import reduce;'+setup)
0.7400134069976048

timeit.timeit('[x for i, x in enumerate(mylist) if i == mylist.index(x)]', setup=setup)
0.9154880290006986

ANSWERING COMMENTS

Because @monica asked a good question about "how is this working?". For everyone having problems figuring it out. I will try to give a more deep explanation about how this works and what sorcery is happening here ;)

So she first asked:

I try to understand why unique = [used.append(x) for x in mylist if x not in used] is not working.

Well it's actually working

>>> used = []
>>> mylist = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
>>> unique = [used.append(x) for x in mylist if x not in used]
>>> print used
[u'nowplaying', u'PBS', u'job', u'debate', u'thenandnow']
>>> print unique
[None, None, None, None, None]

The problem is that we are just not getting the desired results inside the unique variable, but only inside the used variable. This is because during the list comprehension .append modifies the used variable and returns None.

So in order to get the results into the unique variable, and still use the same logic with .append(x) if x not in used, we need to move this .append call on the right side of the list comprehension and just return x on the left side.

But if we are too naive and just go with:

>>> unique = [x for x in mylist if x not in used and used.append(x)]
>>> print unique
[]

We will get nothing in return.

Again, this is because the .append method returns None, and it this gives on our logical expression the following look:

x not in used and None

This will basically always:

1. evaluates to False when x is in used,
2. evaluates to None when x is not in used.

And in both cases (False/None), this will be treated as falsy value and we will get an empty list as a result.

But why this evaluates to None when x is not in used? Someone may ask.

Well it's because this is how Python's short-circuit operators works.

The expression x and y first evaluates x; if x is false, its value is returned; otherwise, y is evaluated and the resulting value is returned.

So when x is not in used (i.e. when its True) the next part or the expression will be evaluated (used.append(x)) and its value (None) will be returned.

But that's what we want in order to get the unique elements from a list with duplicates, we want to .append them into a new list only when we they came across for a fist time.

So we really want to evaluate used.append(x) only when x is not in used, maybe if there is a way to turn this None value into a truthy one we will be fine, right?

Well, yes and here is where the 2nd type of short-circuit operators come to play.

The expression x or y first evaluates x; if x is true, its value is returned; otherwise, y is evaluated and the resulting value is returned.

We know that .append(x) will always be falsy, so if we just add one or next to him, we will always get the next part. That's why we write:

x not in used and (used.append(x) or True)

so we can evaluate used.append(x) and get True as a result, only when the first part of the expression (x not in used) is True.

Similar fashion can be seen in the 2nd approach with the reduce method.

(l.append(x) or l) if x not in l else l
#similar as the above, but maybe more readable
#we return l unchanged when x is in l
#we append x to l and return l when x is not in l
l if x in l else (l.append(x) or l)

where we:

1. Append x to l and return that l when x is not in l. Thanks to the or statement .append is evaluated and l is returned after that.
2. Return l untouched when x is in l

Answer 4

A Python list:

>>> a = ['a', 'b', 'c', 'd', 'b']

To get unique items, just transform it into a set (which you can transform back again into a list if required):

>>> b = set(a)
>>> print(b)
{'b', 'c', 'd', 'a'}

Answer 5

What type is your output variable?

Python sets are what you need. Declare output like this:

output = set()  # initialize an empty set

and you're ready to go adding elements with output.add(elem) and be sure they're unique.

Warning: sets DO NOT preserve the original order of the list.

Answer 6

Options to remove duplicates may include the following generic data structures:

• set: unordered, unique elements
• ordered set: ordered, unique elements

Here is a summary on quickly getting either one in Python.

Given

from collections import OrderedDict


seq = [u"nowplaying", u"PBS", u"PBS", u"nowplaying", u"job", u"debate", u"thenandnow"]

Code

Option 1 - A set (unordered):

list(set(seq))
# ['thenandnow', 'PBS', 'debate', 'job', 'nowplaying']
    

Python doesn't have ordered sets, but here are some ways to mimic one.

Option 2 - an OrderedDict (insertion ordered):

list(OrderedDict.fromkeys(seq))
# ['nowplaying', 'PBS', 'job', 'debate', 'thenandnow']

Option 3 - a dict (insertion ordered), default in Python 3.6+. See more details in this post:

list(dict.fromkeys(seq))
# ['nowplaying', 'PBS', 'job', 'debate', 'thenandnow']

Note: listed elements must be hashable. See details on the latter example in this blog post. Furthermore, see R. Hettinger's post on the same technique; the order preserving dict is extended from one of his early implementations. See also more on total ordering.

Answer 7

Maintaining order:

# oneliners
# slow -> . --- 14.417 seconds ---
[x for i, x in enumerate(array) if x not in array[0:i]]

# fast -> . --- 0.0378 seconds ---
[x for i, x in enumerate(array) if array.index(x) == i]

# multiple lines
# fastest -> --- 0.012 seconds ---
uniq = []
[uniq.append(x) for x in array if x not in uniq]
uniq

Order doesn't matter:

# fastest-est -> --- 0.0035 seconds ---
list(set(array))

Answer 8

Getting unique elements from List

mylist = [1,2,3,4,5,6,6,7,7,8,8,9,9,10]

Using Simple Logic from Sets - Sets are unique list of items

mylist=list(set(mylist))

In [0]: mylist
Out[0]: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Using Simple Logic

newList=[]
for i in mylist:
    if i not in newList:
        newList.append(i)

In [0]: mylist
Out[0]: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Using pop method ->pop removes the last or indexed item and displays that to user. video

k=0
while k < len(mylist):
    if mylist[k] in mylist[k+1:]:
        mylist.pop(mylist[k])
    else:
        k=k+1

In [0]: mylist
Out[0]: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Using Numpy

import numpy as np
np.unique(mylist)

In [0]: mylist
Out[0]: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

Reference

Answer 9

set - unordered collection of unique elements. List of elements can be passed to set's constructor. So, pass list with duplicate elements, we get set with unique elements and transform it back to list then get list with unique elements. I can say nothing about performance and memory overhead, but I hope, it's not so important with small lists.

list(set(my_not_unique_list))

Simply and short.

Answer 10

If you are using numpy in your code (which might be a good choice for larger amounts of data), check out numpy.unique:

>>> import numpy as np
>>> wordsList = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
>>> np.unique(wordsList)
array([u'PBS', u'debate', u'job', u'nowplaying', u'thenandnow'], 
      dtype='<U10')

(http://docs.scipy.org/doc/numpy/reference/generated/numpy.unique.html)

As you can see, numpy supports not only numeric data, string arrays are also possible. Of course, the result is a numpy array, but it doesn't matter a lot, because it still behaves like a sequence:

>>> for word in np.unique(wordsList):
...     print word
... 
PBS
debate
job
nowplaying
thenandnow

If you really want to have a vanilla python list back, you can always call list().

However, the result is automatically sorted, as you can see from the above code fragments. Check out numpy unique without sort if retaining list order is required.

Answer 11

Same order unique list using only a list compression.

> my_list = [1, 2, 1, 3, 2, 4, 3, 5, 4, 3, 2, 3, 1]
> unique_list = [
>    e
>    for i, e in enumerate(my_list)
>    if my_list.index(e) == i
> ]
> unique_list
[1, 2, 3, 4, 5]

enumerates gives the index i and element e as a tuple.

my_list.index returns the first index of e. If the first index isn't i then the current iteration's e is not the first e in the list.

Edit

I should note that this isn't a good way to do it, performance-wise. This is just a way that achieves it using only a list compression.

Answer 12

As a bonus, Counter is a simple way to get both the unique values and the count for each value:

from collections import Counter
l = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
c = Counter(l)

Answer 13

By using basic property of Python Dictionary:

inp=[u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
d={i for i in inp}
print d

Output will be:

set([u'nowplaying', u'job', u'debate', u'PBS', u'thenandnow'])

Answer 14

First thing, the example you gave is not a valid list.

example_list = [u'nowplaying',u'PBS', u'PBS', u'nowplaying', u'job', u'debate',u'thenandnow']

Suppose if above is the example list. Then you can use the following recipe as give the itertools example doc that can return the unique values and preserving the order as you seem to require. The iterable here is the example_list

from itertools import ifilterfalse

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 ifilterfalse(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

Answer 15

def get_distinct(original_list):
    distinct_list = []
    for each in original_list:
        if each not in distinct_list:
            distinct_list.append(each)
    return distinct_list

Answer 16

set can help you filter out the elements from the list that are duplicates. It will work well for str, int or tuple elements, but if your list contains dict or other list elements, then you will end up with TypeError exceptions.

Here is a general order-preserving solution to handle some (not all) non-hashable types:

def unique_elements(iterable):
    seen = set()
    result = []
    for element in iterable:
        hashed = element
        if isinstance(element, dict):
            hashed = tuple(sorted(element.iteritems()))
        elif isinstance(element, list):
            hashed = tuple(element)
        if hashed not in seen:
            result.append(element)
            seen.add(hashed)
    return result

Answer 17

If you want to get unique elements from a list and keep their original order, then you may employ OrderedDict data structure from Python's standard library:

from collections import OrderedDict

def keep_unique(elements):
    return list(OrderedDict.fromkeys(elements).keys())

elements = [2, 1, 4, 2, 1, 1, 5, 3, 1, 1]
required_output = [2, 1, 4, 5, 3]

assert keep_unique(elements) == required_output

In fact, if you are using Python ≥ 3.6, you can use plain dict for that:

def keep_unique(elements):
    return list(dict.fromkeys(elements).keys())

It's become possible after the introduction of "compact" representation of dicts. Check it out here. Though this "considered an implementation detail and should not be relied upon".

Answer 18

def setlist(lst=[]):
   return list(set(lst))

Answer 19

To get unique values from your list use code below:

trends = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
output = set(trends)
output = list(output)

IMPORTANT: Approach above won't work if any of items in a list is not hashable which is case for mutable types, for instance list or dict.

trends = [{'super':u'nowplaying'}, u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']
output = set(trends)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  TypeError: unhashable type: 'dict'

That means that you have to be sure that trends list would always contains only hashable items otherwise you have to use more sophisticated code:

from copy import deepcopy

try:
    trends = [{'super':u'nowplaying'}, [u'PBS',], [u'PBS',], u'nowplaying', u'job', u'debate', u'thenandnow', {'super':u'nowplaying'}]
    output = set(trends)
    output = list(output)
except TypeError:
    trends_copy = deepcopy(trends)
    while trends_copy:
        trend = trends_copy.pop()
        if trends_copy.count(trend) == 0:
            output.append(trend)
print output

Answer 20

I am surprised that nobody so far has given a direct order-preserving answer:

def unique(sequence):
    """Generate unique items from sequence in the order of first occurrence."""
    seen = set()
    for value in sequence:
        if value in seen:
            continue

        seen.add(value)

        yield value

It will generate the values so it works with more than just lists, e.g. unique(range(10)). To get a list, just call list(unique(sequence)), like this:

>>> list(unique([u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']))
[u'nowplaying', u'PBS', u'job', u'debate', u'thenandnow']

It has the requirement that each item is hashable and not just comparable, but most stuff in Python is and it is O(n) and not O(n^2), so will work just fine with a long list.

Answer 21

In addition to the previous answers, which say you can convert your list to set, you can do that in this way too

mylist = [u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenadnow']
mylist = [i for i in set(mylist)]

output will be

[u'nowplaying', u'job', u'debate', u'PBS', u'thenadnow']

though order will not be preserved.

Another simpler answer could be (without using sets)

>>> t = [v for i,v in enumerate(mylist) if mylist.index(v) == i]
[u'nowplaying', u'PBS', u'job', u'debate', u'thenadnow']

Answer 22

1. At the begin of your code just declare your output list as empty: output=[]
2. Instead of your code you may use this code trends=list(set(trends))

Answer 23

Set is a collection of un-ordered and unique elements. So, you can use set as below to get a unique list:

unique_list = list(set([u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job', u'debate', u'thenandnow']))

Answer 24

You can use sets. Just to be clear, I am explaining what is the difference between a list and a set. sets are unordered collection of unique elements.Lists are ordered collection of elements. So,

    unicode_list=[u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job',u'debate', u'thenandnow']
    list_unique=list(set(unicode_list))
    print list_unique
[u'nowplaying', u'job', u'debate', u'PBS', u'thenandnow']

But: Do not use list/set in naming the variables. It will cause error: EX: Instead of use list instead of unicode_list in the above one.

list=[u'nowplaying', u'PBS', u'PBS', u'nowplaying', u'job',u'debate', u'thenandnow']
        list_unique=list(set(list))
        print list_unique
    list_unique=list(set(list))
TypeError: 'list' object is not callable

Answer 25

use set to de-duplicate a list, return as list

def get_unique_list(lst):
        if isinstance(lst,list):
            return list(set(lst))

Answer 26

My solution to check contents for uniqueness but preserve the original order:

def getUnique(self):
    notunique = self.readLines()
    unique = []
    for line in notunique: # Loop over content
        append = True # Will be set to false if line matches existing line
        for existing in unique:
            if line == existing: # Line exists ? do not append and go to the next line
                append = False
                break # Already know file is unique, break loop
        if append: unique.append(line) # Line not found? add to list
    return unique

Edit: Probably can be more efficient by using dictionary keys to check for existence instead of doing a whole file loop for each line, I wouldn't use my solution for large sets.

Answer 27

I know this is an old question, but here's my unique solution: class inheritance!:

class UniqueList(list):
    def appendunique(self,item):
        if item not in self:
            self.append(item)
            return True
        return False

Then, if you want to uniquely append items to a list you just call appendunique on a UniqueList. Because it inherits from a list, it basically acts like a list, so you can use functions like index() etc. And because it returns true or false, you can find out if appending succeeded (unique item) or failed (already in the list).

To get a unique list of items from a list, use a for loop appending items to a UniqueList (then copy over to the list).

Example usage code:

unique = UniqueList()

for each in [1,2,2,3,3,4]:
    if unique.appendunique(each):
        print 'Uniquely appended ' + str(each)
    else:
        print 'Already contains ' + str(each)

Prints:

Uniquely appended 1
Uniquely appended 2
Already contains 2
Uniquely appended 3
Already contains 3
Uniquely appended 4

Copying to list:

unique = UniqueList()

for each in [1,2,2,3,3,4]:
    unique.appendunique(each)

newlist = unique[:]
print newlist

Prints:

[1, 2, 3, 4]

Answer 28

For long arrays

s = np.empty(len(var))

s[:] = np.nan

for  x in  set(var):

    x_positions = np.where(var==x)

    s[x_positions[0][0]]=x


sorted_var=s[~np.isnan(s)]

Answer 29

Try this function, it's similar to your code but it's a dynamic range.

def unique(a):

    k=0
    while k < len(a):
        if a[k] in a[k+1:]:
            a.pop(k)
        else:
            k=k+1



    return a

Answer 30

Use the following function:

def uniquefy_list(input_list):
"""
This function  takes a list as input and return a list containing only unique elements from the input list

"""
output_list=[]
for elm123 in input_list:
    in_both_lists=0
    for elm234 in output_list:
        if elm123 == elm234:
            in_both_lists=1
            break
    if in_both_lists == 0:
        output_list.append(elm123)

return output_list