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, 46], [1, 22, 53], [7, 8, 9]]
how do i sort c2 in such a way that for example:
for row in c2:
sort on row[2]
the result would be:
[[7,8,9],[14,25,46],[1,22,53]]
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]))
import operator; c2.sort(key=operator.itemgetter(2)) or c2.sort(key=operator.itemgetter(2, 1)) or c2.sort(key=operator.itemgetter(2, 1, 0)). It's apparently a fair bit faster than using a lambda expression.>>> 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]]
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])
key function which returns a single value to sort on, he's using a sort function which takes two values, compares then and returns an appropriate value. I'd argue that my option is easier though. :-)
key option, you should. It will in general be faster than the pure comparison-based sort.