How key in python sorted method can use a list

The key argument to sorted means "Pretend the value is the result of this function instead of the actual value." So when you sort 'abc' with the lookup table you gave, it does this:

                # [1st, 2nd, 3rd] sort order
lookup.get('a') # [ -3,   0,   0]
lookup.get('b') # [  0,  -1,  -2]
lookup.get('c') # [  0,  -2,  -1]

Then it will figure out the sorted order of the above values. Lists are sorted lexicographically meaning the first element is compared first, just like in a dictionary ("aardvark" comes before "beaver" and also before "ant").

After looking at the first elements (-3, 0, 0) we know 'a' has the smallest value, but we don't know which of 'b' and 'c' is smaller. But as soon as we see the second elements (0, -1, -2), we know that 'c' is smaller, so the final order is 'acb' without ever consulting the third elements (0, -2, -1).

so from your example, imagine you have the following

lookup = defaultdict(list)
lookup['a'] = [-3, 0, 0]
lookup['b'] = [0, -1, -2]
lookup['c'] = [0, -2, -1]
inputs = ['abc', 'acb', 'acb'] # a list of strings
# note that the key params of sort usually takes a function
result = ''.join(sorted(
            inputs[0], # this is the first value 'abc' of the input list 
            key=lookup.get # passing in lookup.get()
         ))

the sort function passing in each value of the string 'abc'

lookup.get(a) # first 
lookup.get(b) # next 
lookup.get(c) # next 

To understand the logic of comparison, it's internal for most data structure, you can implement yours for a custom class , __lt__ less than . __gt__ greater than

class my_int(int):
   def __lt__(a,b):
       return (a % b) % 2 != 0
   def __gt__(a,b):        
       return (a % b) % 2 == 0

Suppose you have a list of animals:

>>> animals=['aarvark','zebra','giraffe','bear','dog','cat','badger','ant']

Sorted lexicographically, or in alphabetical order, aardvark is sorted before ant and both before zebra:

>>> sorted(animals)
['aarvark', 'ant', 'badger', 'bear', 'cat', 'dog', 'giraffe', 'zebra']

Now suppose your 10 year old tells you I want all animals that start with 'b' sorted first, then 'z' then alphabetically.

With a key function, this is trivial to accomplish:

>>> lookup=['b','z']
>>> key_func=lambda s: (lookup.index(s[0]),s) if s[0] in lookup else (len(lookup),s)
>>> sorted(animals, key=key_func)
['badger', 'bear', 'zebra', 'aarvark', 'ant', 'cat', 'dog', 'giraffe']

Before the key function was added to Python sorting routines, the common approach to a problem like this was called Decorate, Sort, Undecorate and can be seen here:

>>> ts=sorted([(lookup.index(s[0]),s) if s[0] in lookup else (len(lookup), s) for s in animals])
>>> ts
[(0, 'badger'), (0, 'bear'), (1, 'zebra'), (2, 'aarvark'), (2, 'ant'), (2, 'cat'), (2, 'dog'), (2, 'giraffe')]
>>> [t[1] for t in ts]
['badger', 'bear', 'zebra', 'aarvark', 'ant', 'cat', 'dog', 'giraffe']

(BTW: This example is way easier and faster if you use a dict instead of a list:

>>> lookup={'b':0, 'z':1}
>>> sorted(animals, key=lambda s: (lookup.get(s[0], len(lookup)),s))
['badger', 'bear', 'zebra', 'aarvark', 'ant', 'cat', 'dog', 'giraffe']

That is the right way but your question involved list lookup...)

Key functions allow you to modify how the sort order is interpreted. For another example, consider if you wanted to sort by integers found in the sort strings and then alphabetically.

Here is the list:

>>> nl=['zebra65','ant101','bear5','no num', '2 num first', 's with 1 and 2']

If you just use the default, it comes out ASCIIbetically:

>>> sorted(nl)
['2 num first', 'ant101', 'bear5', 'no num', 's with 1 and 2', 'zebra65']

With a simple regex and key function, you can find all the numbers and form a tuple for sorting by number then the string:

import re

def find_n(s):
    ml=re.findall(r'(\d+)', s)
    if ml:
        return tuple(map(int, ml))+(s,)
    return (0,s)

>>> sorted(nl, key=find_n)
['no num', 's with 1 and 2', '2 num first', 'bear5', 'zebra65', 'ant101']