A function that checks if the list is sorted

[micseydel]

The OP's solution is superior to the one-liner in two ways:
The big-O of the OP's solution is O(n), whereas the version using sorted() is O(n**2) (or whatever Tim-sort is; definitely not linear)
The OP's code can return immediately upon finding that the list isn't sorted, it doesn't even have to do n work.

As others have mentioned enumerate(), I tend to use it instead of range(len()) just to communicate to the reader that I really do need the index. That said, I would use itertools' islice (and izip if Python 2) and avoid indexes entirely, the only time I find myself actually needing indexes is for some interview questions.

[ThomasL]

The OP's solution is superior to the one-liner in two ways

The OP´s code is no solution as it doesn´t work.
But i understand what you mean :-)

[micseydel]

The original poster did write a correct solution, unless I'm mistaken. That's what I'm referring to.

[ThomasL]

Typical misunderstanding, i thought you referred to the OP´s first code.

[MIRA]

Hi,

@ppoyan89: iterating over an iterable with for x in range(len(iterable)) is still a bad anti-pattern. Don't you have any motivation to use what you learned before?

Except this, this is solvable without iteration but by the build-in function sorted:

def is_sorted(iterable):
    return iterable == sorted(iterable)

Regards, noisefloor

hi,
your code is only when you need to sort a list, and the question is basically asking you to see if the list is sorted return true otherwise false which doesn't work with your code

[Pedroski55]

Of course you must compare things which are comparable.

from string import ascii_lowercase
from random import choice, randint

mylist1 = [choice(ascii_lowercase) for i in range(5)]
mylist2 = [i for i in ascii_lowercase]
mylist3 = [randint(1,10) for i in range(10)]
mylist4 = [i for i in range(10)]

def is_sorted(alist):
    for i in range(1, len(alist)-1):
        # exit as soon as the condition is not true
        if not alist[i-1] <= alist[i] <= alist[i+1]:
            return False
    return True

[DeaD_EyE]

The useful function itertools.pairwise was added to Python 3.10.

https://docs.python.org/3/library/iterto...s.pairwise

This isn't Pythonic: range(len(something))

from itertools import pairwise
from string import ascii_lowercase
from random import choice, randint


test_data = [
    [choice(ascii_lowercase) for i in range(5)],
    [i for i in ascii_lowercase],
    [randint(1, 10) for i in range(10)],
    [i for i in range(10)],
]


def is_sorted(iterable):
    for last, current in pairwise(iterable):
        if last > current:
            return False
    return True


# test
for data in test_data:
    print(f"'{'-'.join(map(str, data))}' is {'' if is_sorted(data) else 'not'} sorted")