在python中排序列表

Question

I am trying to sort lists on a list : each list contains [seq1,seq2,score] , I want to sort the list L according to the score of (seq1,seq2) from the max score to the minimum score, then each list take a rank to each (seq1,seq2) according to the score

L=[ ['AA', 'CG', 0],['AA', 'AA', 4], ['CG', '--', -1]]

the sorted list must be:

L=[['AA', 'AA', 4], ['AA', 'CG', 0], ['CG', '--', -1]]

for the ranks:

['AA', 'AA', 4] has rank 1
['AA', 'CG', 0] has rank 2
['CG', '--', -1] has rank 3

how can I do it? I tried:

def getKey():/* to get the score from each list*/
    scorelist=score()/*this is the list L*/
    return scorelist[][2]



def sort_list():
   scorelist=score()
    p=sorted(s, key=getKey)
    return p

Answer 1

You can use itemgetter as a function to get the value to sort the list by. Then to get the rank, you can use enumerate .

from operator import itemgetter

score_list = score()
score_list.sort(key=itemgetter(2), reversed=True) # sort list descending by third element
for i, values in enumerate(score_list):
    print(values, i+1) # print values and rank

If several items can share the same score, you can use groupby to assign them the same rank:

from operator import itemgetter
from itertools import groupby

score_list = score()
score_list.sort(key=itemgetter(2), reversed=True)
for i, group in enumerate(groupby(score_list)):
    for item in group:
        print(item, i+1)

With this code, if you have 2 items ranked first, the next item will be ranked second. If you want to rank it third (as expected), you can increment the rank in the inner loop (and save it rank for display):

from operator import itemgetter
from itertools import groupby

score_list = score()
score_list.sort(key=itemgetter(2), reversed=True)
rank = 1 # counter for the rank
for group in groupby(score_list):
    current_rank = rank # save the rank
    for item in group:
        print(item, current_rank)
        rank += 1 # rank increment in inner loop

Answer 2

You're right on track with using the sorted function. The key can be an itemgetter which yields the third list element. In that case, you'll need to reverse the generated list to get the desired order:

>>> import operator
>>> sorted(L, key=operator.itemgetter(2), reverse=True)
[['AA', 'AA', 4], ['AA', 'CG', 0], ['CG', '--', -1]]

Answer 3

Try this:

for result in L:
   result.reverse()
L.sort()
for result in L:
   result.reverse()
L.reverse()

In this code you set the number of any list on the first place and sort it with the sort function. After you have a sorted list from the smallest to the highest number. So you have to use the reverse function again so that you can use it for your ranked system