How to build a sorting algorithm with 2 keys in Python?

Question

I’m aware that Python has a very nice function sorted() for this but I’d like to implement my own function to understand the logic.
I have a list of strings and I’d like to sort them first by length, then alphabetically.
For example,

Input:

['daring','adequate','bold','bait','cold','beautiful']

Output:

['bait','bold','cold','daring','adequate','beautiful']

How do I build a function that does this?

I started with building a simple quicksort function for sorting only alphabetically, but now I can’t think of a efficient way of progressing from it to include 2 keys at once.
Here is the code: (assumes strings are all in one case)

def quick_sort(sequence):
    length = len(sequence)
    if length <=1:
        return sequence
    else:
        pivot = sequence.pop()

    items_greater = []
    items_lower = []

    for item in sequence:
        if item >pivot:
            items_greater.append(item)

        else:
            items_lower.append(item)

    return quick_sort(items_lower) + [pivot] + quick_sort(items_greater)
print(quick_sort(['daring','adequate','bold','bait','cold','beautiful']))

TLDR: How to turn this into a Length/Alphebetical sort?
Thanks

Asked By: Joydurn Yup

||

Source

Answer 1

You can do this by adding a ‘key function’ that maps an item into a tuple (item_length, item) prior with pivot in your algorithm. For example:

def quick_sort(sequence, key_fn):
    length = len(sequence)
    if length <=1:
        return sequence
    else:
        pivot = sequence.pop()

    items_greater = []
    items_lower = []

    for item in sequence:
        if key_fn(item) > key_fn(pivot):
            items_greater.append(item)

        else:
            items_lower.append(item)


    return quick_sort(items_lower, key_fn) + [pivot] + quick_sort(items_greater, key_fn)


print(quick_sort(['daring','adequate','bold','bait','cold','beautiful'], lambda e: (len(e),e)))

will return

['bait', 'bold', 'cold', 'daring', 'adequate', 'beautiful']

Answered By: Alberto Garcia

Answer 2

You can implement a compare function that compares any two values in the way you define it.

def compare(a, b):
    if len(a) != len(b):
        return len(a) > len(b)
    else:
        return a > b


def quick_sort(sequence):
    length = len(sequence)
    if length <= 1:
        return sequence
    else:
        pivot = sequence.pop()

    items_greater = []
    items_lower = []

    for item in sequence:
        if compare(item, pivot):
            items_greater.append(item)

        else:
            items_lower.append(item)

    return quick_sort(items_lower) + [pivot] + quick_sort(items_greater)


print(quick_sort(['daring', 'adequate', 'bold', 'bait', 'cold', 'beautiful']))

this will print

['bait', 'bold', 'cold', 'daring', 'adequate', 'beautiful']

Answered By: seermer

Answer 3

This is how you can write it into your quicksort function assuming that you only care about the alphabetical order based on the first letter of each word.

def quick_sort(sequence):
    length = len(sequence)
    if length <=1:
        return sequence
    else:
        pivot = sequence.pop()

    items_greater = []
    items_lower = []

    for item in sequence:
        if item[0] > pivot[0]:
            if item[0] == pivot[0]:
                if len(item) > pivot:
                    items_greater.append(item)
                else:
                    items_lower.append(item)
            items_greater.append(item)

        else:
            items_lower.append(item)

    return quick_sort(items_lower) + [pivot] + quick_sort(items_greater)
print(quick_sort(['daring','adequate','bold','bait','cold','beautiful']))

returns

['adequate', 'bold', 'bait', 'beautiful', 'cold', 'daring']

Answered By: Rihhard

How to build a sorting algorithm with 2 keys in Python?

Question:

Input:

Output:

Answers: