Pythonic way to create union of all values contained in multiple lists
Question:
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?
Answers:
Unions are not supported by lists, which are ordered, but are supported by sets. Check out set.union.
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.
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 ]
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']
desired_result = [x for y in lists for x in y]
in comprehension way:
[*{ j for i in lists for j in i }]
or
[*functools.reduce(lambda x,y: {*x, *y}, 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?
Unions are not supported by lists, which are ordered, but are supported by sets. Check out set.union.
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.
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 ]
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']
desired_result = [x for y in lists for x in y]
in comprehension way:
[*{ j for i in lists for j in i }]
or
[*functools.reduce(lambda x,y: {*x, *y}, lists)]