I have two lists like this:
a = [[1,2,3],[2,3,4],[5,6,7],[7,8,9]]
b = [1,2]
I would now like to filter list a, to keep only the items which contain the items in list b. So the desired output would look like this:
[[1,2,3],[2,3,4]]
I have tried some nested list comprehensions, which I could think of, but could not get the desired output. Any advice is appreciated.
you could try something like this :
print([i for i in a if any(map(i.__contains__,b))])
>>> [[1, 2, 3], [2, 3, 4]]
I would try something like this:
a = [[1,2,3],[2,3,4],[5,6,7],[7,8,9]]
b = [1,2]
result = [lst for lst in a if any(x in lst for x in b)]
A combination of list comprehension and sets would yield the wanted result. Note; I assume that repeated items and ordering is not of interest, if this is the case - a set won't work since it ignores ordering and only allows unique items.
A simple list comprehension would do, like below
filter_items = set(filter_items)
[sublist for sublist in original_list if not set(sublist).isdisjoint(filter_items)]
There's mainly one interesting part of this list comprehension, namely the if not set(sublist).isdisjoint(filter_items). Here you only keep the sublist if the set of sublist is not disjoint of the set filter_items i.e. none of the filter_items is in the sublist.
for your given example the provided answer would yield the following:
>>> a = [[1,2,3],[2,3,4],[5,6,7],[7,8,9]]
>>> b = set([1,2])
>>> [sublist for sublist in a if not set(sublist).isdisjoint(b)]
[[1, 2, 3], [2, 3, 4]]
set(sublist).intersection(b)? If the intersection is not empty then you want to keep the list... (set methods doesn't require a set as an argument)
b a set before comprehension to avoid repeated construction and use b.isdisjoint(sublist) (isdisjoint accepts any iterable, AFAIR).isdisjoint is more optimal here.isdisjoint performs short-circuit evaluation similar to any and will return False after first matching item, while intersection will check whole set anyway. It's like any(iterator) that won't exhaust iterator and will stop after first True. I really doubt that isdisjoint is implemented like not bool(intersection), it's inefficient.Using a set approach the in can be mimic with an intersection:
a = [[1,2,3],[2,3,4],[5,6,7],[7,8,9]]
b = [1,2]
b_as_set = set(b)
out = [l for l in a if b_as_set.intersection(l)]
# [[1, 2, 3], [2, 3, 4]]
I would try something like this.
a = [[1,2,3],[2,3,4],[5,6,7],[7,8,9]]
b = [1,2]
print([lst for lst in a if any([item in b for item in lst])])
any call
[i for i in a if i in b in i][i for i in a if i in i in b][[i for i in a if i in b] for ele in a][[i for i in b] for i in a]None of these produced the desired result.