CODEWITHSUNDEEP
pythonlistset-union
AJ.
AJ.

Reputation: 28174

Pythonic way to create union of all values contained in multiple lists

I have a list of lists:

lists = [[1,4,3,2,4], [4,5]]

I want to flatten this list and remove all duplicates; or, in other words, apply a set union operation:

desired_result = [1, 2, 3, 4, 5]

What's the easiest way to do this?

Upvotes: 122

Views: 76980

Answers (7)

Alex Martelli
Alex Martelli

Reputation: 881675

Since you seem to be using Python 2.5 (it would be nice to mention in your Q if you need an A for versions != 2.6, the current production one, by the way;-) and want a list rather than a set as the result, I recommend:

import itertools

...

return list(set(itertools.chain(*result_list)))

itertools is generally a great way to work with iterators (and so with many kinds of sequences or collections) and I heartily recommend you become familiar with it. itertools.chain, in particular, is documented here.

Upvotes: 19

Randy
Randy

Reputation: 171

in comprehension way:

[*{ j for i in lists for j in i }]

or

[*functools.reduce(lambda x,y: {*x, *y}, lists)]

Upvotes: 3

Mike G.
Mike G.

Reputation: 1

desired_result = [x for y in lists for x in y]

Upvotes: -2

sth
sth

Reputation: 229603

set.union does what you want:

>>> results_list = [[1,2,3], [1,2,4]]
>>> results_union = set().union(*results_list)
>>> print(results_union)
set([1, 2, 3, 4])

You can also do this with more than two lists.

Upvotes: 211

Justin R.
Justin R.

Reputation: 24031

Unions are not supported by lists, which are ordered, but are supported by sets. Check out set.union.

Upvotes: 2

GrvTyagi
GrvTyagi

Reputation: 4487

You can also follow this style 

In [12]: a = ['Orange and Banana', 'Orange Banana']
In [13]: b = ['Grapes', 'Orange Banana']
In [14]: c = ['Foobanana', 'Orange and Banana']

In [20]: list(set(a) | set(b) | set(c))
Out[20]: ['Orange and Banana', 'Foobanana', 'Orange Banana', 'Grapes']

In [21]: list(set(a) & set(b) | set(c))
Out[21]: ['Orange and Banana', 'Foobanana', 'Orange Banana']

Upvotes: 3

Bear
Bear

Reputation: 1127

I used the following to do intersections, which avoids the need for sets.

a, b= [[1,2,3], [1,2]]
s = filter( lambda x: x in b, a)

or,

s = [ x for x in b if x in a ]

Upvotes: 0

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