[ACCEPTED]-Get difference from two lists in Python-python
Short answer, yes:
list(set(l1) - set(l2)), though this will not 7 keep order.
Long answer, no, since internally 6 the CPU will always iterate. Though if you 5 use
set() that iteration will be done highly 4 optimized and will be much faster then your 3 list comprehension (not to mention that 2 checking for membership
value in list is much faster with 1 sets then lists).
You can't do it without iteration. Even 6 if you call a single method, internally 5 that will iterate.
Your approach is fine 4 for a small list, but you could use this 3 approach instead for larger lists:
s2 = set(l2) result = [i for i in l1 if not i in s2 ]
This will 2 be fast and will also preserve the original 1 order of the elements in l1.
If you don't care about the order of the 1 elements, you can use sets:
l1 = set([2, 3, 4, 5]) l2 = set([0, 1, 2, 3]) print l1 - l2
The conversion to sets is great when your 5 list elements can be converted to sets. Otherwise 4 you'll need something like Mark Byers' solution. If you have 3 large lists to compare you might not want 2 to pay the memory allocation overhead and 1 simplify his line to:
[l1.remove(m) for m in l1 if m in l2]
You can use use set_1.difference_update(set_2) for 1 in place difference:
>>sl1 = set([2, 3, 4, 5]) >>sl2 = set([0, 1, 2, 3]) >>sl1.difference_update(sl2) >>sl1 set([4, 5])
You can do this simply as follows:
list( set(l1) - set(l2) )
This should 1 do the trick.
Convert them to sets and use the difference 1 operator:
l1=[2,3,4,5] l2=[0,1,2,3] answer = set(l1) - set(l2)
Using the built-in module set
>>> a = set([1,2,3,4,5]) >>> b = set([1,3,5]) >>> a.difference(b) set([2, 4])
>>> a = set([1,2,3,4,5]) >>> b = [1,3,5] >>> a.difference(b) set([2, 4])
Simply as it is programming, a simple task 3 can be done in a variety of ways. We can 2 use list comprehension methods like this 1 for exactly the same problem
fruits = ["apple", "banana", "cherry", "kiwi", "mango"] ddd = ["apple", "banana", "mango"] newlist = [x for x in fruits if x not in ddd] print(newlist)
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