Python sorting by key, breaking ties

Question

original = ["aga", "aaa", "aba"]

dict = {
    "aba": 1,
    "aaa": 0,
    "aga": 1
}

I need to sort by dict values and tie breakers need to keep original order, how would I do that? Example very simplified.

I tried:

final = sorted(sorted(original, key=lambda x: (dict[x]), key=original.index))

I was under the impression that the built-in sort method _does_ preserve original order in the case of ties. `sorted(original, key=lambda x: (dict[x]))` gives `['aaa', 'aga', 'aba']`. Is that not what you want? If not, can you give the exact desired contents of `final`? — Kevin, Apr 03 '17 at 16:29
Possible duplicate of [Python sorting list of dictionaries by multiple keys](http://stackoverflow.com/questions/1143671/python-sorting-list-of-dictionaries-by-multiple-keys) — Prune, Apr 03 '17 at 16:40
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Moses Koledoye · Accepted Answer · 2018-10-01T17:12:57.973

6

Just sort them based on their value in the dictionary. The original order will be preserved for ties:

final = sorted(original, key=dct.get)
print final
# ['aaa', 'aga', 'aba']

Timsort - Python's standard sorting algorithm - is stable; items that compare equal retain their relative order.

On another note, do not use names like dict or list to avoid making builtins unusable later on in your code.

edited Oct 01 '18 at 17:12

answered Apr 03 '17 at 16:31

Moses Koledoye

score 0 · Answer 2 · answered Apr 03 '17 at 16:40

0

Dictionaries in Python don't store original order. However, you could use an OrderedDict instead to maintain the order you wish:

import collections
d = collections.OrderedDict()
d['aba'] = 1
d['aaa'] = 0
d['aga'] = 1
print(d)

answered Apr 03 '17 at 16:40

Neil

2 Answers2