I have a sorted list of values i need to filter such that the only values returned are not the same as the value to its left.
For example mylist=[1,2,2,3,3,3,3,6,7,9,9,11] would return [1,2,3,6,7,9,11].
I've done this task using for and if loops but wondering if there is not a more elegant solution using list comprehensions.
Thanks!
I don't think there's a way to do it with list comprehension, but a better way to do it would be:
mylist=[1,2,2,3,3,3,3,6,7,9,9,11]
uniqueList = list(set(mylist))
In a sorted list, that is by definition the same thing as list(set(mylist)), which just uniquifies your list.
But, code golfing a bit, since you asked for a list-comp, there's the charming:
seen = set()
[x for x in mylist if x not in seen and not seen.add(x)]
Out[3]: [1, 2, 3, 6, 7, 9, 11]
Or:
from itertools import zip_longest
[tup[0] for tup in zip_longest(mylist,mylist[1:]) if len(set(tup)) == 2]
Out[11]: [1, 2, 3, 6, 7, 9, 11]
The latter of which works with your requirement on non-sorted lists as well.
As you can see, both of these are less readable than list(set(mylist)), which should be preferred if you know that your input lists will always be sorted.
set.add does not return anything, so that part is equivalent to and not None which is always true. It also adds x to the set, so the second time around x in seen is True and it gets filtered out.
You can try something like this:
>>> m=set( mylist)
>>> m
set([1, 2, 3, 6, 7, 9, 11])
