Sorting a list of lists in Python

Question

c2=[]
row1=[1,22,53]
row2=[14,25,46]
row3=[7,8,9]

c2.append(row2)
c2.append(row1)
c2.append(row3)

c2 is now:

[[14, 25         


        

Answer 1

Well, your desired example seems to indicate that you want to sort by the last index in the list, which could be done with this:

sorted_c2 = sorted(c2, lambda l1, l2: l1[-1] - l2[-1])

Answer 2

>>> import operator
>>> c2 = [[14, 25, 46], [1, 22, 53], [7, 8, 9]]
>>> c2.sort(key=itemgetter(2))
>>> c2
[[7, 8, 9], [14, 25, 46], [1, 22, 53]]

Answer 3

The key argument to sort specifies a function of one argument that is used to extract a comparison key from each list element. So we can create a simple lambda that returns the last element from each row to be used in the sort:

c2.sort(key = lambda row: row[2])

A lambda is a simple anonymous function. It's handy when you want to create a simple single use function like this. The equivalent code not using a lambda would be:

def sort_key(row):
    return row[2]

c2.sort(key = sort_key)

If you want to sort on more entries, just make the key function return a tuple containing the values you wish to sort on in order of importance. For example:

c2.sort(key = lambda row: (row[2],row[1]))

or:

c2.sort(key = lambda row: (row[2],row[1],row[0]))