Get The Items Not Repeated In A List

Take two lists, second with same items than first plus some more: a = [1,2,3] b = [1,2,3,4,5] I want to get a third one, containing only the new items (the ones not repeated): c =

Solution 1:

at the very least use a list comprehension:

[x for x in a + b if(a + b).count(x)== 1]

otherwise use the set class:

list(set(a).symmetric_difference(set(b)))

there is also a more compact form:

list(set(a) ^ set(b))

Solution 2:

If the order is not important and you can ignore repetitions within a and b, I would simply use sets:

>>> set(b) - set(a)
set([4, 5])

Sets are iterable, so most of the times you do not need to explicitly convert them back to list. If you have to, this does it:

>>> list(set(b) - set(a))
[4, 5]

Solution 3:

I'd say go for the set variant, where

set(b) ^ set(a)   (set.symmetric_difference())

only applies if you can be certain that a is always a subset of b, but in that case has the advantage of being commutative, ie. you don't have to worry about calculating set(b) ^ set(a) or set(a) ^ set(b); or

set(b) - set(a)    (set.difference())

which matches your description more closely, allows a to have extra elements not in b which will not be in the result set, but you have to mind the order (set(a) - set(b) will give you a different result).

Solution 4:

Here are some different possibilities with the sets

>>> a = [1, 2, 3, 4, 5, 1, 2]
>>> b = [1, 2, 5, 6]
>>> print list(set(a)^set(b))
[3, 4, 6]
>>> print list(set(a)-set(b))
[3, 4]
>>> print list(set(b)-set(a))
[6]
>>> print list(set(a)-set(b))+list(set(b)-set(a))
[3, 4, 6]
>>>

Solution 5:

Another solution using only lists:

a = [1, 2, 3]
b = [1, 2, 3, 4, 5]
c = [n for n in a + b if n not in a or n not in b]