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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?

2019/10/09
1
823
10/9/2019 4:06:00 AM


To be consistent with the type I would use:

mylist = list(set(mylist))
2016/05/27

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 - 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.

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)
0.4188511371612549

timeit.timeit('[x for x in mylist if x not in used and (used.append(x) or True)]', setup='used = [];'+setup)
0.6157128810882568

timeit.timeit('reduce(lambda l, x: l.append(x) or l if x not in l else l, mylist, [])', setup=setup)
1.8778090476989746

timeit.timeit('reduce(lambda l, x: l+[x] if x not in l else l, mylist, [])', setup=setup)
2.13108491897583

timeit.timeit('reduce(lambda l, x: l if x in l else l+[x], mylist, [])', setup=setup)
2.207760810852051

timeit.timeit('[x for i, x in enumerate(mylist) if i == mylist.index(x)]', setup=setup)
2.3621110916137695

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
2019/03/11

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'}
2019/10/09

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.

2019/10/09

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))
2018/02/07