Search results
From Python version 2.6 on you can use multiple arguments to set.intersection(), like. u = set.intersection(s1, s2, s3) If the sets are in a list, this translates to: u = set.intersection(*setlist) where *a_list is list expansion. Note that set.intersection is not a static method, but this uses the functional notation to apply intersection of ...
Here's some Python 2 / Python 3 code that generates timing information for both list-based and set-based methods of finding the intersection of two lists. The pure list comprehension algorithms are O(n^2), since in on a list is a linear search. The set-based algorithms are O(n), since set search is O(1), and set creation is O(n) (and converting ...
Apr 14, 2017 · set.intersection(*map(set,d3)) Will actually work, though because d3 already contains sets you can just do: set.intersection(*d3) And, in fact, only the first one needs to be a set - the others can be any iterable, and intersection will setify them by itself. The problem you're having doesn't seem to be in this code - rather,
Apr 29, 2015 · The output is the equivalent of (a | b) - (a & b), the union of both sets minus the intersection of both sets. Producing the output takes O (M+N) time for sets of length M and N, respectively; M steps to copy set a then N steps to alter that set based on each value in b: def symmetric_difference(a, b): result = set(a)
Oct 4, 2010 · sets = iter(map(set, d)) result = sets.next() for s in sets: result = result.intersection(s) return result. for newer versions of python: the intersection method takes an arbitrary amount of arguments. result = set(d[0]).intersection(*d[1:]) alternatively, you can intersect the first set with itself to avoid slicing the list and making a copy:
The "worst case" assumes data that is inappropriate for use in the hash table used by dict and set.The data would have to be something such that every datum had exactly the same hash value -- this would force the hash table to do something akin to a linear search to do the __contains__ check.
else: # no intersection. If you really need True or False. bool(set(L) & set(M)) After running some timings, this seems to be a good option to try too. m_set=set(M) any(x in m_set for x in L) If the items in M or L are not hashable you have to use a less efficient approach like this. any(x in M for x in L)
Then here is your solution for Python 2: c3 = [filter(lambda x: x in c1, sublist) for sublist in c2] In Python 3 filter returns an iterable instead of list, so you need to wrap filter calls with list(): c3 = [list(filter(lambda x: x in c1, sublist)) for sublist in c2] Explanation:
Aug 6, 2013 · 113. Place both series in Python's set container then use the set intersection method: s1.intersection(s2) and then transform back to list if needed. Just noticed pandas in the tag. Can translate back to that: pd.Series(list(set(s1).intersection(set(s2)))) From comments I have changed this to a more Pythonic expression, which is shorter and ...
May 7, 2014 · If the order depends on a third list then you just use Peter DeGlopper's answer but replace set_2 with a combination of the two lists. set_2=set(l1) & set(l2) then intersection = [x for x in list_3 if x in set_2] Still no need to do linear searches. – Duncan. Apr 4, 2017 at 8:30.
