Ascending Order in Python

Question

I'm trying to get this code to associate True with the variable is_ascending if the list numbers is in ascending order and False if otherwise

for i in range(2, len(numbers)) :
if numbers[i] < numbers[i-1] :
    is_ascending = False
    break
else :
    is_ascending = True

But the program im doing, this is for a class, says this won't work. Any help would be appreciated

Your indentation isn't quite correct. Have you tested the code on a test list of `numbers`? — Matt Messersmith, Sep 21 '18 at 15:04
You should use `range(1, len(numbers))` because 1 means the second element and 0 index would refer to the first element. That being said, there are indentation errors in your code as well — Sheldore, Sep 21 '18 at 15:06

score 2 · Answer 1 · answered Sep 21 '18 at 15:07

You should use range(1, len(numbers)) as 1 means the second element and 0 index would refer to the first element. Besides that, your if-else statements should be inside the for loop (indentation issue). Your code works as below for a unsorted list and also for a sorted list. Try running it with numbers = [1,2,3,4,5] and it will print True

numbers = [1,2,4,3,5]

for i in range(1, len(numbers)):
    if numbers[i] < numbers[i-1]:
        is_ascending = False
        break
    else:
        is_ascending = True

print (is_ascending)        
> False

Domino · Answer 2 · 2018-09-21T19:25:46.007

Bazingaa's answer is good and more readable to people not as familiar with functional programming. There's still two things you could improve on.

You could iterate over range(len(numbers) - 1) and compare each element with the next one instead of the previous. It's a personal preference but I find it more intuitive than the two-parameter notation of range.
You could use the all function as in this answer: https://stackoverflow.com/a/3755251/2813263

It would give you something like this:

is_ascending = all(numbers[i] <= numbers[i+1] for i in range(len(numbers) - 1))

score 1 · Accepted Answer · answered Sep 21 '18 at 15:05

1

if sorted(numbers) == numbers:
    is_ascending = True

That'll do it.

answered Sep 21 '18 at 15:05

ltd9938

1,444
1
15
29

1

I added an else statement but besides for that, it worked. Thank you! – silver_music Sep 21 '18 at 15:08
2

you could even do a one liner : `is_ascending = sorted(numbers) == numbers`. – Diane M Sep 21 '18 at 15:11
This does work and looks nice, but is much less efficient as you don't break when an element is reached. It is at best O(n) instead of O(1) – Domino Sep 21 '18 at 15:20

score 1 · Answer 4 · answered Sep 21 '18 at 15:26

1

If you build a list of the outcome of the paired comparisons:

numbers = [10,20,30,40,20,50]
outcomes = [ numbers[i] > numbers[i-1] for i in range(1,len(numbers))]
print(outcomes)

This outputs:

[True, True, True, False, True]

The False is for the 40 & 20 comparison.

Then you can use all() to determine if the list is in ascending order

numbers = [10,20,30,40,50,60]
is_ascending = all([ numbers[i] > numbers[i-1] for i in range(1,len(numbers)) ])
print(is_ascending) # prints True
numbers = [10,5,20,30,40,50,60]
is_ascending = all([ numbers[i] > numbers[i-1] for i in range(1,len(numbers)) ])
print(is_ascending) # prints False

answered Sep 21 '18 at 15:26

T Burgis

1,395
7
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We must have posted at the same time. Note that you can drop the square brackets, which will turn the list comprehension into a generator. Instead of creating a list and then checking all elements, it will iterate over the loop and stop as soon as an incorrect value is found. – Domino Sep 21 '18 at 15:40
Yes, we must have been typing at the same time. Thanks for pointing out about dropping the []. A good tip. – T Burgis Sep 21 '18 at 15:55

Ascending Order in Python

4 Answers4