I can perform
a = [1,2,3] b = [4,5,6] a.extend(b) # a is now [1,2,3,4,5,6] Is there way to perform an action for extending list and adding new items to the beginning of the list?
Like this
a = [1,2,3] b = [4,5,6] a.someaction(b) # a is now [4,5,6,1,2,3] I use version 2.7.5, if it is important.
You can assign to a slice:
a[:0] = b Demo:
>>> a = [1,2,3] >>> b = [4,5,6] >>> a[:0] = b >>> a [4, 5, 6, 1, 2, 3] Essentially, list.extend() is an assignment to the list[len(list):] slice.
You can 'insert' another list at any position, just address the empty slice at that location:
>>> a = [1,2,3] >>> b = [4,5,6] >>> a[1:1] = b >>> a [1, 4, 5, 6, 2, 3] 
-1 is not the end of the list, it is the last element. a[-1:] is symmetrical with a[:1] at the start of the list, both contain 1 element. This is why I explicitly mention len(list) in my answer.This is what you need ;-)
a = b + a 
a, try this: >>> a is b+a (b+a is neither b nor a, but a completely new object) ;)
You could use collections.deque:
import collections a = collections.deque([1, 2, 3]) b = [4, 5, 6] a.extendleft(b[::-1]) 
If you need fast operations and you need to be able to access arbitrary elements, try a treap or red-black tree.
>>> import treap as treap_mod >>> treap = treap_mod.treap() >>> for i in range(100000): ... treap[i] = i ... >>> treap[treap.find_min() - 1] = -1 >>> treap[100] 100 Most operations on treaps and red-black trees can be done in O(log(n)). Treaps are purportedly faster on average, but red-black trees give a lower variance in operation times.
b.extend(a)?ais calledvery_importantandbis calledaux. You may want to keep the former and forget about the latter.