Python sort question

Question

I need to sort the following list of tuples in Python:

ListOfTuples = [('10', '2010 Jan 1;', 'Rapoport AM', 'Role of antiepileptic drugs as preventive agents for migraine', '20030417'), ('21', '2009 Nov;', 'Johannessen SI', 'Antiepilepticdrugs in epilepsy and other disorders--a population-based study of prescriptions', '19679449'),...]

My goal is to order his Descending year (listOfTuples [2]) and the Rising Author (listOfTuples [2]):

sorted(result, key = lambda item: (item[1], item[2]))

But that will not work. How can I get sorting stability?

Answer 1

def descyear_ascauth(atup):
  datestr = atup[1]
  authstr = atup[2]
  year = int(datestr.split(None, 1)[0])
  return -year, authstr

... sorted(result, key=descyear_ascauth) ...

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Answer 2

- . .

, :

import operator
ListOfTuples.sort(key=operator.itemgetter(2))
ListOfTuples.sort(key=lambda x: x[1][:4], reverse=True)

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Answer 3

, , , , :

data = [ ('a', 'a'), ('a', 'b'), ('b','a') ]

def sort_func( a, b ):
    # compare tuples with the 2nd entry switched
    # this inverts the sorting on the 2nd entry
    return cmp( (a[0], b[1]), (b[0], a[1]) ) 

print sorted( data )                    # [('a', 'a'), ('a', 'b'), ('b', 'a')]
print sorted( data, cmp=sort_func )     # [('a', 'b'), ('a', 'a'), ('b', 'a')]

Answer 4

, ( ):

import time
import operator

def sortkey(seq):
    strdate, author = seq[1], seq[2]
    spdate = strdate[:-1].split()
    month = time.strptime(spdate[1], "%b").tm_mon
    date = [int(spdate[0]), month] + map(int, spdate[2:])
    return map(operator.neg, date), author  

print sorted(result, key=sortkey)

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Answer 5

Here is the lambda version of Alex's answer. I think it looks more compact than Duncan's answer, but obviously most of Alex's answer's readability has been lost.

sorted(ListOfTuples, key=lambda atup: (-int(atup[1].split(None, 1)[0]), atup[2]))

Typically, compactness and readability and efficiency are required.