获取最后一个列表项索引的最pythonic方法

Question

Given a list:

l1 = [0, 211, 576, 941, 1307, 1672, 2037]

What is the most pythonic way of getting the index of the last element of the list. Given that Python lists are zero-indexed, is it:

len(l1) - 1

Or, is it the following which uses Python's list operations:

l1.index(l1[-1])

Both return the same value, that is 6.

Answer 1

Only the first is correct:

>>> lst = [1, 2, 3, 4, 1]
>>> len(lst) - 1
4
>>> lst.index(lst[-1])
0

However it depends on what do you mean by "the index of the last element".

Note that index must traverse the whole list in order to provide an answer:

In [1]: %%timeit lst = list(range(100000))
   ...: lst.index(lst[-1])
   ...: 
1000 loops, best of 3: 1.82 ms per loop

In [2]: %%timeit lst = list(range(100000))
len(lst)-1
   ...: 
The slowest run took 80.20 times longer than the fastest. This could mean that an intermediate result is being cached.
10000000 loops, best of 3: 109 ns per loop

Note that the second timing is in nanoseconds versus milliseconds for the first one.

Answer 2

You should use the first. Why?

>>> l1 = [1,2,3,4,3]
>>> l1.index(l1[-1])
2

Answer 3

Bakuriu's answer is great!

In addition, it should be mentioned that you rarely need this value. There are usually other and better ways to do what you want to do. Consider this answer as a sidenote :)

As you mention, getting the last element can be done this way:

lst = [1,2,4,2,3]
print lst[-1]  # 3

If you need to iterate over a list, you should do this:

for element in lst:
    # do something with element

If you still need the index, this is the preferred method:

for i, element in enumerate(lst):
    # i is the index, element is the actual list element