开发者

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

开发者 https://www.devze.com 2022-12-18 12:03 出处:网络
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:

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 开发者_如何学编程do this?


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.


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.


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']    


in comprehension way:

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

or

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


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


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 ]


desired_result = [x for y in lists for x in y]
0

精彩评论

暂无评论...
验证码 换一张
取 消

关注公众号