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 ASC or DESC

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

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

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

Answer 1

If you want the fastest way for numpy arrays, use numba, which if you use conda should be already installed

The code will be fast because it will be compiled by numba

import numba
@numba.jit
def issorted(vec, ascending=True):
    if len(vec) < 2:
        return True
    if ascending:
        for i in range(1, len(vec)):
            if vec[i-1] > vec[i]:
                return False
        return True
    else:
        for i in range(1, len(vec)):
            if vec[i-1] < vec[i]:
                return False
        return True

and then:

>>> issorted(array([4,9,100]))
>>> True

Answer 2

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.

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))

For Python 3:

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

Answer 4

This is in fact the shortest way to do it using recursion:

if it's Sorted will print True else will print out False

 def is_Sorted(lst):
    if len(lst) == 1:
       return True
    return lst[0] <= lst[1] and is_Sorted(lst[1:])

 any_list = [1,2,3,4]
 print is_Sorted(any_list)

Answer 5

Not very Pythonic at all, but we need at least one reduce() answer, right?

def is_sorted(iterable):
    prev_or_inf = lambda prev, i: i if prev <= i else float('inf')
    return reduce(prev_or_inf, iterable, float('-inf')) < float('inf')

The accumulator variable simply stores that last-checked value, and if any value is smaller than the previous value, the accumulator is set to infinity (and thus will still be infinity at the end, since the 'previous value' will always be bigger than the current one).

Answer 6

A beautiful way to implement this is to use the imap function from itertools:

from itertools import imap, tee
import operator

def is_sorted(iterable, compare=operator.le):
  a, b = tee(iterable)
  next(b, None)
  return all(imap(compare, a, b))

This implementation is fast and works on any iterables.