It's O(n), also check out: http://wiki.python.org/moin/TimeComplexity

This page documents the time-complexity (aka "Big O" or "Big Oh") of various operations in current CPython. Other Python implementations (or older or still-under development versions of CPython) may have slightly different performance characteristics. However, it is generally safe to assume that they are not slower by more than a factor of O(log n)...

>>> ["foo", "bar", "baz"].index("bar")
1

See the documentation for the built-in .index() method of the list:

list.index(x[, start[, end]])

Return zero-based index in the list of the first item whose value is equal to x. Raises a ValueError if there is no such item.

The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.

Caveats

Linear time-complexity in list length

An index call checks every element of the list in order, until it finds a match. If the list is long, and if there is no guarantee that the value will be near the beginning, this can slow down the code.

This problem can only be completely avoided by using a different data structure. However, if the element is known to be within a certain part of the list, the start and end parameters can be used to narrow the search.

For example:

>>> import timeit
>>> timeit.timeit('l.index(999_999)', setup='l = list(range(0, 1_000_000))', number=1000)
9.356267921015387
>>> timeit.timeit('l.index(999_999, 999_990, 1_000_000)', setup='l = list(range(0, 1_000_000))', number=1000)
0.0004404920036904514

The second call is orders of magnitude faster, because it only has to search through 10 elements, rather than all 1 million.

Only the index of the first match is returned

A call to index searches through the list in order until it finds a match, and stops there. If there could be more than one occurrence of the value, and all indices are needed, index cannot solve the problem:

>>> [1, 1].index(1) # the `1` index is not found.
0

Instead, use a list comprehension or generator expression to do the search, with enumerate to get indices:

>>> # A list comprehension gives a list of indices directly:
>>> [i for i, e in enumerate([1, 2, 1]) if e == 1]
[0, 2]
>>> # A generator comprehension gives us an iterable object...
>>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1)
>>> # which can be used in a `for` loop, or manually iterated with `next`:
>>> next(g)
0
>>> next(g)
2

The list comprehension and generator expression techniques still work if there is only one match, and are more generalizable.

Raises an exception if there is no match

As noted in the documentation above, using .index will raise an exception if the searched-for value is not in the list:

>>> [1, 1].index(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: 2 is not in list

If this is a concern, either explicitly check first using item in my_list, or handle the exception with try/except as appropriate.

The explicit check is simple and readable, but it must iterate the list a second time. See What is the EAFP principle in Python? for more guidance on this choice.

Time complexity is O(n) . Have a look at the link

http://wiki.python.org/moin/TimeComplexity

Accessing any array element is in constant time, since it is a known by memory location (i.e. a pointer.)

The array does not need to go through prior elements in order to access the nth element (i.e. it is not like a linked list.) The location of all elements is known before hand and can simply be accessed directly.

Updated thanks to comments.

array[-x] is syntactic sugar for array[len(lst) - x]. So it's still a simple constant access to a pointer and takes no time.

You can see this answer for a bit more info. While it is about C, the concepts should be the same.

The problem you have in your current methods is that you make placing one element dependent on the entire opposing list. Instead, simply make a chart of where each element is in each list.

Use a dict. Go through list1, filling in key:value to be the element and its index. This gives you

{ 5: 0, 8: 1, 1: 2, 4: 3 }

Now, go through list 2. Access each element; if it exists (use get for the fail-safe functionality), print the stored index (list1) and the current index (list2). If it doesn't exist, then simply skip printing.

O(N) complexity.