CODEWITHSUNDEEP
python
EddyIT
EddyIT

Reputation: 155

How to test if a list is sorted in ascending order

This is the question of the exercise: write a function that checks if a list is sorted in ascending order.

def ascending(lst):
    for k in range(0,len(lst)):
        if lst[k] < lst[k+1]:
            print('Ok')
        else:
            print('NOk, the number ' + str(lst[k]) + ' is greater than his next ' + str(lst[k+1]))
    return 'Bye!'

lst = [1,3,2,4,5]
print(ascending(lst))

I expect the output: Ok, Ok, NOk the number 3 is greather than his next 2, Ok ... and I get it but, at the very end of the problem, the error message is obviously "IndexError: list index out of range". I understood that the problem is at the end of the if statement because for k = 4, k+1 = 5 (out of range) but I don't know how to solve it.

Upvotes: 2

Views: 433

Answers (3)

user2390182
user2390182

Reputation: 73470

The easiest:

def ascending(lst):
    lst == sorted(lst)

But this is log-linear and does not short-circuit. Better:

def ascending(lst):
    return all(a <= b for a, b in zip(lst, lst[1:]))

or in Python >= 3.10:

from itertools import pairwise

def ascending(lst):
    return all(a <= b for a, b in pairwise(lst))

Upvotes: 0

s3dev
s3dev

Reputation: 9711

A different approach:

If your task is as simple as 'Test if a list is sorted in ascending order'; then how about a simple function like:

def issorted(lst):
    return lst == sorted(lst)

print('Is sorted: ', issorted([5, 4, 1, 2]))
print('Is sorted: ', issorted([1, 2, 3, 4]))
print('Is sorted: ', issorted(['a', 'b', 'd', 'c']))
print('Is sorted: ', issorted(['w', 'x', 'y', 'z']))

Which outputs:

Is sorted:  False
Is sorted:  True
Is sorted:  False
Is sorted:  True

Upvotes: 0

Prune
Prune

Reputation: 77850

Your problem is here:

for k in range(0,len(lst)):
    if lst[k] < lst[k+1]:

When k=4 ( which is len(list) ), then k+1 is out of range. Make your loop statement

for k in range(0,len(lst) - 1):

Upvotes: 2

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