Pythonic way to check if a list is sorted or not

Here is a one liner:

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

If using Python 2, use xrange instead of range.

For reverse=True, use >= instead of <=.

I would just use

if sorted(lst) == lst:
    # code here

unless it's a very big list in which case you might want to create a custom function.

if you are just going to sort it if it's not sorted, then forget the check and sort it.

lst.sort()

and don't think about it too much.

if you want a custom function, you can do something like

def is_sorted(lst, key=lambda x: x):
    for i, el in enumerate(lst[1:]):
        if key(el) < key(lst[i]): # i is the index of the previous element
            return False
    return True

This will be O(n) if the list is already sorted though (and O(n) in a for loop at that!) so, unless you expect it to be not sorted (and fairly random) most of the time, I would, again, just sort the list.