Modify Python sorted() build-in func to return a list with values from lambda

user_8927462 0
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Given a python list:

nums = [9,4,5,7,-1,2,-3,0]

The task is to write a function that will return a new list with squares of numbers of original list in ascending order. Expected output:

[0, 1, 4, 9, 16, 25, 49, 81]

The simplest way to do this is just create a new list and then apply Python build-in sorted() function:

def my_sort(nums):
    new_list = [i**2 for i in nums]
    return sorted(new_list)

print(my_sort(nums))

output as expected:

[0, 1, 4, 9, 16, 25, 49, 81]

However there is another option is to use "key" argument and pass lambda function inside sorted():

new_list = sorted(nums, key=lambda x: x**2)
print(new_list)

output:

[0, -1, 2, -3, 4, 5, 7, 9]

Here we have the same sorting behavior with difference that sorted() function return sorted list with original values.

My question - is there any way to override python build-in sorted() function that it will return a sored list with values from lambda?

UPD:

Apparently i have to explain a little where this question came from, for greater clarity. I had a python interview where this question was asked. When i implemented my basic solution, the interviewer said that in terms of big O() complexity, this solution is not entirely effective, arguing that fisrt we create a new array\list at the beginning and only then apply sorting algorithm over it and suggested writing sorting "on the fly". I was confused because believed that there was no point in writing a custom sorting algorithm and no doubt it will perform slower than the built-in solution, i woud never write my own "Timesort" algorithm (which as i kwon implemented in sorted()). Later, thinking about this question, there was a suggestion that a solution with sorting via a key would perform faster than sorting with the preliminary creation of a new list. Therefore, this question arose. However, further tests showed that soution when we create a new list works faster than sorting through the key=lambda function. But of course, this is beyond the scope of this question.

python
4 Answers

is there any way to override python build-in sorted() function that it will return a sored list with values from lambda?

Sorry, but this is the WrongWayToIt™. It fights how the tools were designed to be used. Instead, the right way is to convert the data first and then sort it:

>>> nums = [9,4,5,7,-1,2,-3,0]
>>> sorted(map(lambda x: x**2, nums))
[0, 1, 4, 9, 16, 25, 49, 81]

If for some reason you still have to rewrite sorted(), follow essentially same pattern in the new function:

>>> def new_sorted(data, key):
...     return sorted(map(key, data))
...

>>> new_sorted(nums, key=lambda x: x*x)
[0, 1, 4, 9, 16, 25, 49, 81]
user_4244990

You have got it all wrong. That's not how the key parameter is supposed to work. The value of the key parameter should be a function (or other callable) that takes a single argument and returns a key to use for sorting purposes.

That's why you are getting the original values.

See the docs: https://docs.python.org/3.12/howto/sorting.html

user_157431180

You can do it by importing builtins module to use original sorted method and write your own in current scope

import builtins
def sorted(l,/, *, key=None, reverse=False, f=None):
    bsorted = builtins.sorted
    if f is None:
        return bsorted(l, key=key, reverse=reverse)
    else:
        l = [f(x) for x in l]
    return bsorted(l, key=key, reverse=reverse)

However you are overwriting a well-known builtin method to add new functionality, which could cause confusion when other people read your code. You should better use another name instead of doing it.

user_91577870

You can replace the original sorted function globally by setting builtins.sorted. my_own_sorted is your own implementation.

import builtins
builtins.sorted = my_own_sorted
user_47895610
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