Check item membership in set in Python -


hello i've been coding couple of months , know basics, i'm having set membership problem can't find solution.

i have list of lists of pairs of integers, , want remove list have "a" integer in them. thought using sets easiest way. bellow code:

## item test against.  = set([3])  ## list test.       groups = [[3, 2], [3, 4], [1, 2], [5, 4], [4, 3]]       ## list contain lists present ## in groups not contain "a" groups_no_a = []          group in groups:     group = set(group)     if in group:         groups_no_a.append(group)     ## thought problem had     ## clearing variable put in,        ## no remedy.      group.clear()     print groups_no_a

i had tried using s.issubset(t) until realized tested if every element in s in t.

thank you!

you want test if there no intersection:

if not & group:

or

if not a.intersection(group):

or, inversely, sets disjoint:

if a.isdisjoint(group):

the method forms take any iterable, don't have turn group set that. following one-liner work too:

groups_no_a = [group group in groups if a.isdisjoint(group)]

demo:

>>> = set([3])  >>> groups = [[3, 2], [3, 4], [1, 2], [5, 4], [4, 3]]      >>> [group group in groups if a.isdisjoint(group)] [[1, 2], [5, 4]]

if testing one element, creating sets going cost more in performance gain in testing membership, , doing:

3 not in group

where group short list.

you can use timeit module compare pieces of python code see works best specific typical list sizes.


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