Remove the last N elements of a list

Question

Is there a a better way to remove the last N elements of a list.

for i in range(0,n):
    lst.pop( )

Answer 1

Should be using this:

a[len(a)-n:] = []

or this:

del a[len(a)-n:]

It's much faster, since it really removes items from existing array. The opposite (a = a[:len(a)-1]) creates new list object and less efficient.

>>> timeit.timeit("a = a[:len(a)-1]\na.append(1)", setup="a=range(100)", number=10000000)
6.833014965057373
>>> timeit.timeit("a[len(a)-1:] = []\na.append(1)", setup="a=range(100)", number=10000000)
2.0737061500549316
>>> timeit.timeit("a[-1:] = []\na.append(1)", setup="a=range(100)", number=10000000)
1.507638931274414
>>> timeit.timeit("del a[-1:]\na.append(1)", setup="a=range(100)", number=10000000)
1.2029790878295898

If 0 < n you can use a[-n:] = [] or del a[-n:] which is even faster.