Sorting tuples in python and keeping the relative order

Question

Input = [("M", 19), ("H", 19), ("A", 25)]

Output =[("A", 25), ("M" ,19), ("H", 19)]

It should sort alphabetically but when the second value is equal then it should remain in place without changing their respective places. Here M and H both have value as 19 so it is already sorted.

Answer 1

You could group the items by the second value of each tuple using itertools.groupby, sort groupings by the first item in each group, then flatten the result with itertools.chain.from_iterable:

from operator import itemgetter
from itertools import groupby, chain

def relative_sort(Input):
    return list(
        chain.from_iterable(
            sorted(
                (
                    tuple(g)
                    for _, g in groupby(
                        sorted(Input, key=itemgetter(1)), key=itemgetter(1)
                    )
                ),
                key=itemgetter(0),
            )
        )
    )

Output:

>>> relative_sort([("M", 19), ("H", 19), ("A", 25)])
[('A', 25), ('M', 19), ('H', 19)]
>>> relative_sort([("B", 19), ("B", 25), ("M", 19), ("H", 19), ("A", 25)])
[('B', 19), ('M', 19), ('H', 19), ('B', 25), ('A', 25)]
>>> relative_sort([("A", 19), ("B", 25), ("M", 19), ("J", 30), ("H", 19)])
[('A', 19), ('M', 19), ('H', 19), ('B', 25), ('J', 30)]

Answer 2

IIUC, you can group tuples by the second element. First gather them together using sorted based on the tuples' second element, then throw them in a list with groupby based on the second element. This sort will preserve the order you have them in already (this sort may also be unnecessary depending on your data).

import itertools

Input = [('M', 19), ('H', 19), ('A', 25)]
sort1 = sorted(Input, key=lambda x: x[1])
grouped = []
for _, g in itertools.groupby(sort1, lambda x: x[1]):
    grouped.append(list(g))

Then sort these grouped lists based on the first letter and finally "unlist" them.

sort2 = sorted(grouped, key=lambda x: x[0][0])
Output = [tup for sublist in sort2 for tup in sublist]