Why can't I sort this list?

Question

statlist = [('abc',5,1), ('bzs',66,1), ... ]
sorted(statlist, key=lambda x: int(x[1]))

I want to sort it by the integer largest to smallest. In this case, 5 and 66. But it doesn't seem to be working.

Answer 1

The sorted function returns a new list so you will need to assign the results of the function like this:

new_list = sorted(statlist, key=lambda x: int(x[1])) 

Answer 2

Use the .sort method for in place sorting:

statlist = [('abc',5,1), ('bzs',66,1), ... ]
statlist.sort(key=lambda x: int(x[1]))

If you do want to use sorted, then reassign the variable:

statlist = [('abc',5,1), ('bzs',66,1), ... ]
statlist = sorted(statlist, key=lambda x: int(x[1]))

For descending sort, use reverse:

statlist = [('abc',5,1), ('bzs',66,1), ... ]
statlist = sorted(statlist, key=lambda x: int(x[1]), reverse=True)

Then, you'd better use itemgetter instead of a lambda :

import operator
statlist = [('abc',5,1), ('bzs',66,1), ... ]
statlist = sorted(statlist, key=operator.itemgetter(1), reverse=True)

Answer 3

You can pass, key, and reverse to .sort function

>>> x.sort(key=lambda x:x[1],reverse=True)
>>> x
[('bzs', 66, 1), ('abc', 5, 1)]
>>>

Answer 4

for inplace sorting use

statlist.sort(key=lambda x: x[1])

for creating other list, with sorted data use

otherlist = sorted( statlist, key=lambda x: x[1] )

Answer 5

from operator import itemgetter
statlist = [('abc',5,1), ('bzs',66,1), ... ]

# statlist.sort modifiest the statlist, sorted returns a new one
# reverse puts the largest items to the front
statlist.sort(key=itemgetter(1), reverse=True)

Answer 6

In response to alex's comment that he thought that sorted() worked "like the sort function":

If it worked "like the sort function", it is unlikely to have been put in the library.

In any case, there is no sort function ... you refer to the sort method of list objects.

Simple demonstration using the interactive interpreter:

>>> alist = [3, 2, 1]; x = alist.sort(); print x; print alist
None
[1, 2, 3]
>>> alist = [3, 2, 1]; x = sorted(alist); print x; print alist
[1, 2, 3]
[3, 2, 1]

Here's a tip: look for patterns and similarities, but always verify your intuitive extrapolations. You might like to apply those ideas to reverse and reversed.

Answer 7

>>> s = [('xyz', 8, 1), ('abc',5,1), ('bzs',66,1) ]
>>> s = sorted(s, key=lambda x: int(x[1]))
>>> s.reverse()
>>> print s
[('bzs', 66, 1), ('xyz', 8, 1), ('abc', 5, 1)]

Answer 8

Hey when ever I am saving something to an array I don't tend to worry about order and then at the end I use sorted() for example like this statlist = sorted(statlist) and if you wanted it largest to smallest statlist = sorted(statlist, reverse = True) That's the simple way of getting largest to smallest!

Example code where I've used this (just an extract)

    while i <= math.sqrt(intnum):
        if (intnum % i) == 0:
            numbers.insert(0,i)
            numbers.insert(0,int(intnum/i))
            print(i,":", int(intnum/i))
        i += 1
    numbers = sorted(numbers, reverse = True)