通过添加元组的第二个和第三个元素对元组列表进行排序

Question

i have a list of tuples which is :

[('g', 10, 2), ('o', 6, 11), ('v', 2, 4), ('t', 1, 15), ('x', 40, 3), ('m', 4, 4), ('k', 10, 2), ('f', 14, 1), ('p', 70, 90), ('l', 21, 7), ('n', 1, 27), ('a', 39, 70), ('d', 11, 10), ('h', 21, 10), ('c', 10, 19), ('b', 8, 1), ('e', 30, 39), ('i', 23, 29), ('r', 8, 7), ('q', 2, 2), ('s', 18, 86)]

and I'm struggling about how to sort them so they can be placed like this:

[('p', 70, 90), ('a', 39, 70), ('s', 18, 84), ('è', 27, 45), ('e', 30, 39), ('i', 23, 29), ('x', 40, 3), ('h', 21, 10), ('c', 10, 19), ('l', 20, 7), ('d', 11, 10), ('o', 6, 11), ('t', 1, 15), ('f', 14, 1), ('r', 8, 7), ('g', 10, 2), ('k', 10, 2), ('n', 1, 11), ('b', 8, 1), ('m', 4, 4), ('v', 2, 4), ('q', 2, 2)]

It should be the addition of the second and third element of the tuple , and when they are the same ex. (a,10,9) and (b,9,10) they should be sorted alphabetically. It's written in python 3.5 and I can't call any library

Answer 1

You can set a tuple as the key to sort your list:

sorted(lst, key = lambda x: (sum(x[1:]), x[0]))

In this way, it will firstly sort by the sum of the last two elements of the tuple and then by the first element of the tuple.

And if you like the result in descending order, as @Moses commented, you can specify the the reverse parameter to be True :

sorted(lst, key = lambda x: (sum(x[1:]), x[0]), reverse = True)

Update : To handle descending, ascending order differently, since the sum here is numeric, you can negate the sum as well. In this way, it will be sorted in descending order for the sum but alphabetically for the first letter.

sorted(lst, key = lambda x: (-sum(x[1:]), x[0]))

Answer 2

sorted(a, key=lambda x: (sum(x[1:3]), x[0]))

Where a is your list. If you need reversed:

sorted(a, key=lambda x: (sum(x[1:3]), x[0]), reverse=True)

Answer 3

a = [('g', 10, 2), ('o', 6, 11), ('v', 2, 4), ('t', 1, 15),
     ('x', 40, 3), ('m', 4, 4), ('k', 10, 2), ('f', 14, 1),
     ('p', 70, 90), ('l', 21, 7), ('n', 1, 27), ('a', 39, 70),
     ('d', 11, 10), ('h', 21, 10), ('c', 10, 19), ('b', 8, 1),
     ('e', 30, 39), ('i', 23, 29), ('r', 8, 7), ('q', 2, 2),
     ('s', 18, 86)]

Create a function that will return the sum of the items and use that function a the sort key. I like to use operator.itemgetter() when extracting items from iterables.

import operator
second_third_first_items = operator.itemgetter(1, 2, 0)
def key(thing):
    *two_three, one = second_third_first_items(thing)
    return (sum(two_three), one)
a.sort(key = key, reverse = True)

---

key without operator.itemgetter

def key(thing):
    one, *two_three = thing[:3]
    return (sum(two_three), one)