Pythonic way to check if a list is sorted or not

Question

Is there a pythonic way to check if a list is already sorted in AESC or DESC.

listtimestamps=[1,2,3,5,6,7]

something like listtimestamps.isSorted() that returns True or False.

EDIT: I want to input a list of timestamps for some messages and check if the the transactions appeared in the correct order.

and I can write a custom function too :) , Thanks

EDIT: Thanks for pointing out camel case issues :)

EDIT: Also I am sorry if I wasn't clear I don't need to sort it afterwards, I just need to check if the list is sorted, I am using it to solve a decision problem

Answer 1

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, y: x < y):
    for i, el in enumerate(lst[1:]):
        if key(el, lst[i-1]):
            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.

Answer 2

SapphireSun is quite right. You can just use lst.sort(). Python's sort implementation (TimSort) check if the list is already sorted. If so sort() will completed in linear time. Sounds like a Pythonic way to ensure a list is sorted ;)

Answer 3

Actually we are not giving the answer anijhaw is looking for. Here is the one liner:

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

Answer 4

This iterator form is 10-15% faster than using integer indexing:

from itertools import izip
isSorted = lambda l : all(a <= b for a,b in izip(l[:-1],l[1:]))

Answer 5

Here's an other way with eval and chained comparison from python :

a=[3,4,5,1]
isSorted = eval("<".join([str(v) for v in a]))

EDIT : Limitations :

This way is actually more a funny way to use chained comparison because there's a lot of problems :

• For a=[x] it returns x
• For big arrays, it raises a SystemError (com_backpatch: offset too large) (cf. here)