I would like to collect all the items that exist in every sublist. Let's say the list of lists looks like this:
list[0] = ['A', 'B', 'C', 'D']
list[1] = ['X', 'B', 'A']
list[2] = ['R', 'C', 'A', 'B', 'X']
'A' and 'B' exist in every sublist and my goal is to save these to another list:
list2 = ['A', 'B']
I've been trying to use list comprehension but I can't figure out how to get it to work the way I want. Any help is greatly appreciated!
Use set intersections, and then convert the result back to a list.
your_list = [
['A', 'B', 'C', 'D'],
['X', 'B', 'A'],
['R', 'C', 'A', 'B', 'X']
]
print(set.intersection(*map(set,your_list)))
If you know the values per list are unique in the first place, and you don't care about order, then you can just use a list of sets in your original code, so this simplifies to:
your_list = [
set(['A', 'B', 'C', 'D']),
set(['X', 'B', 'A']),
set(['R', 'C', 'A', 'B', 'X'])
]
print(set.intersection(*your_list))
Note don't call your own variables list, as it collides with the built-in list type and will get really confusing.
If order doesn't matter, you can use set.intersection:
l = [['A', 'B', 'C', 'D'],
['X', 'B', 'A'],
['R', 'C', 'A', 'B', 'X']]
list2 = list(set.intersection(*(set(sl) for sl in l)))
print(list2)
Output:
['A', 'B']
# or ['B', 'A']
Try this :
>>> list1 = [['A', 'B', 'C', 'D'], ['X', 'B', 'A'], ['R', 'C', 'A', 'B', 'X']]
>>> list2 = list(set(list1[0]).intersection(list1[1]).intersection(list1[2]))
>>> list2
['A', 'B']
