What is the runtime complexity of python list functions?

Question

I was writing a python function that looked something like this

def foo(some_list):
   for i in range(0, len(some_list)):
       bar(some_list[i], i)

so that it was called with

x = [0, 1, 2, 3, ... ]
foo(x)

I had assumed that index access of lists was O(1), but was surprised to find that for large lists this was significantly slower than I expected.

My question, then, is how are python lists are implemented, and what is the runtime complexity of the following

• Indexing: list[x]
• Popping from the end: list.pop()
• Popping from the beginning: list.pop(0)
• Extending the list: list.append(x)

For extra credit, splicing or arbitrary pops.

Answer 1

The answer is "undefined". The Python language doesn't define the underlying implementation. Here are some links to a mailing list thread you might be interested in.

• It is true that Python's lists have been implemented as contiguous vectors in the C implementations of Python so far.

• I'm not saying that the O() behaviours of these things should be kept a secret or anything. But you need to interpret them in the context of how Python works generally.

Also, the more Pythonic way of writing your loop would be this:

def foo(some_list):
   for item in some_list:
       bar(item)

Answer 2

Can't comment yet, so

if you need index and value then use enumerate:

for idx, item in enumerate(range(10, 100, 10)):
    print idx, item

Answer 3

there is a very detailed table on python wiki which answers your question.

However, in your particular example you should use enumerate to get an index of an iterable within a loop. like so:

for i, item in enumerate(some_seq):
    bar(item, i)

Answer 4

Lists are indeed O(1) to index - they are implemented as a vector with proportional overallocation, so perform much as you'd expect. The likely reason you were finding this code slower than you expected is the call to "range(0, len(some_list))".

range() creates a new list of the specified size, so if some_list has 1,000,000 items, you will create a new million item list up front. This behaviour changes in python3 (range is an iterator), to which the python2 equivalent is xrange, or even better for your case, enumerate