How do operator.itemgetter() and sort() work?

Answer for Python beginners

In simpler words:

1. The key= parameter of sort requires a key function (to be applied to be objects to be sorted) rather than a single key value and
2. that is just what operator.itemgetter(1) will give you: A function that grabs the first item from a list-like object.

(More precisely those are callables, not functions, but that is a difference that can often be ignored.)

Looks like you're a little bit confused about all that stuff.

operator is a built-in module providing a set of convenient operators. In two words operator.itemgetter(n) constructs a callable that assumes an iterable object (e.g. list, tuple, set) as input, and fetches the n-th element out of it.

So, you can't use key=a[x][1] there, because python has no idea what x is. Instead, you could use a lambda function (elem is just a variable name, no magic there):

a.sort(key=lambda elem: elem[1])

Or just an ordinary function:

def get_second_elem(iterable):
    return iterable[1]

a.sort(key=get_second_elem)

So, here's an important note: in python functions are first-class citizens, so you can pass them to other functions as a parameter.

Other questions:

1. Yes, you can reverse sort, just add reverse=True: a.sort(key=..., reverse=True)
2. To sort by more than one column you can use itemgetter with multiple indices: operator.itemgetter(1,2), or with lambda: lambda elem: (elem[1], elem[2]). This way, iterables are constructed on the fly for each item in list, which are than compared against each other in lexicographic(?) order (first elements compared, if equal - second elements compared, etc)
3. You can fetch value at [3,2] using a[2,1] (indices are zero-based). Using operator... It's possible, but not as clean as just indexing.

Refer to the documentation for details:

1. operator.itemgetter explained
2. Sorting list by custom key in Python