Get unique values from a list in python [duplicate] – Dev

The best answers to the question “Get unique values from a list in python [duplicate]” in the category Dev.

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:

To be consistent with the type I would use:

mylist = list(set(mylist))

ANSWER:

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:

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:

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.from_keys 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.from_keys 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