sorting a list of tuples in Python

Question

While working on a problem from Google Python class, I formulated following result by using 2-3 examples from Stack overflow-

def sort_last(tuples):
    return [b for a,b in sorted((tup[1], tup) for tup in tuples)]

print sort_last([(1, 3), (3, 2), (2, 1)])

I learned List comprehension yesterday, so know a little about list comprehension but I am confused how this solution is working overall. Please help me to understand this (2nd line in function).

Answer 1

That pattern is called decorate-sort-undecorate.

1. You turn each (1, 3) into (3, (1, 3)), wrapping each tuple in a new tuple, with the item you want to sort by first.
2. You sort, with the outer tuple ensuring that the second item in the original tuple is sorted on first.
3. You go back from (3, (1, 3)) to (1, 3) while maintaining the order of the list.

In Python, explicitly decorating is almost always unnecessary. Instead, use the key argument of sorted:

sorted(list_of_tuples, key=lambda tup: tup[1]) # or key=operator.itemgetter(1)

Or, if you want to sort on the reversed version of the tuple, no matter its length:

sorted(list_of_tuples, key=lambda tup: tup[::-1]) 
                              # or key=operator.itemgetter(slice(None, None, -1))

Answer 2

Lets break it down:

In : [(tup[1],tup) for tup in tuples]

Out: [(3, (1, 3)), (2, (3, 2)), (1, (2, 1))]

So we just created new tuple where its first value is the last value of inner tuple - this way it is sorted by the 2nd value of each tuple in 'tuples'.

Now we sort the returned list:

In: sorted([(3, (1, 3)), (2, (3, 2)), (1, (2, 1))])

Out: [(1, (2, 1)), (2, (3, 2)), (3, (1, 3))]

So we now have our list sorted by its 2nd value of each tuple. All that is remain is to extract the original tuple, and this is done by taking only b from the for loop.

The list comprehension iterates the given list (sorted([...] in this case) and returns the extracted values by order.