Pythonic way to check if a list is sorted or not


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.

Asked By: anijhaw



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.


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.

Answered By: aaronasterling

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 😉

Answered By: Wai Yip Tung

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 <=.

Answered By: Wai Yip Tung

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

# python2 only
if str is bytes:
    from itertools import izip as zip

def is_sorted(l):
    return all(a <= b for a, b in zip(l, l[1:]))
Answered By: PaulMcG

I ran a benchmark and sorted(lst, reverse=True) == lst was the fastest for long lists, and all(l[i] >= l[i+1] for i in xrange(len(l)-1)) was the fastest for short lists. These benchmarks were run on a MacBook Pro 2010 13" (Core2 Duo 2.66GHz, 4GB 1067MHz DDR3 RAM, Mac OS X 10.6.5).

UPDATE: I revised the script so that you can run it directly on your own system. The previous version had bugs. Also, I have added both sorted and unsorted inputs.

  • Best for short sorted lists: all(l[i] >= l[i+1] for i in xrange(len(l)-1))
  • Best for long sorted lists: sorted(l, reverse=True) == l
  • Best for short unsorted lists: all(l[i] >= l[i+1] for i in xrange(len(l)-1))
  • Best for long unsorted lists: all(l[i] >= l[i+1] for i in xrange(len(l)-1))

So in most cases there is a clear winner.

UPDATE: aaronasterling’s answers (#6 and #7) are actually the fastest in all cases. #7 is the fastest because it doesn’t have a layer of indirection to lookup the key.

#!/usr/bin/env python

import itertools
import time

def benchmark(f, *args):
    t1 = time.time()
    for i in xrange(1000000):
    t2 = time.time()
    return t2-t1

L1 = range(4, 0, -1)
L2 = range(100, 0, -1)
L3 = range(0, 4)
L4 = range(0, 100)

# 1.
def isNonIncreasing(l, key=lambda x,y: x >= y): 
    return all(key(l[i],l[i+1]) for i in xrange(len(l)-1))
print benchmark(isNonIncreasing, L1) # 2.47253704071
print benchmark(isNonIncreasing, L2) # 34.5398209095
print benchmark(isNonIncreasing, L3) # 2.1916718483
print benchmark(isNonIncreasing, L4) # 2.19576501846

# 2.
def isNonIncreasing(l):
    return all(l[i] >= l[i+1] for i in xrange(len(l)-1))
print benchmark(isNonIncreasing, L1) # 1.86919999123
print benchmark(isNonIncreasing, L2) # 21.8603689671
print benchmark(isNonIncreasing, L3) # 1.95684289932
print benchmark(isNonIncreasing, L4) # 1.95272517204

# 3.
def isNonIncreasing(l, key=lambda x,y: x >= y): 
    return all(key(a,b) for (a,b) in itertools.izip(l[:-1],l[1:]))
print benchmark(isNonIncreasing, L1) # 2.65468883514
print benchmark(isNonIncreasing, L2) # 29.7504849434
print benchmark(isNonIncreasing, L3) # 2.78062295914
print benchmark(isNonIncreasing, L4) # 3.73436689377

# 4.
def isNonIncreasing(l):
    return all(a >= b for (a,b) in itertools.izip(l[:-1],l[1:]))
print benchmark(isNonIncreasing, L1) # 2.06947803497
print benchmark(isNonIncreasing, L2) # 15.6351969242
print benchmark(isNonIncreasing, L3) # 2.45671010017
print benchmark(isNonIncreasing, L4) # 3.48461818695

# 5.
def isNonIncreasing(l):
    return sorted(l, reverse=True) == l
print benchmark(isNonIncreasing, L1) # 2.01579380035
print benchmark(isNonIncreasing, L2) # 5.44593787193
print benchmark(isNonIncreasing, L3) # 2.01813793182
print benchmark(isNonIncreasing, L4) # 4.97615599632

# 6.
def isNonIncreasing(l, key=lambda x, y: x >= y): 
    for i, el in enumerate(l[1:]):
        if key(el, l[i-1]):
            return False
    return True
print benchmark(isNonIncreasing, L1) # 1.06842684746
print benchmark(isNonIncreasing, L2) # 1.67291283607
print benchmark(isNonIncreasing, L3) # 1.39491200447
print benchmark(isNonIncreasing, L4) # 1.80557894707

# 7.
def isNonIncreasing(l):
    for i, el in enumerate(l[1:]):
        if el >= l[i-1]:
            return False
    return True
print benchmark(isNonIncreasing, L1) # 0.883186101913
print benchmark(isNonIncreasing, L2) # 1.42852401733
print benchmark(isNonIncreasing, L3) # 1.09229516983
print benchmark(isNonIncreasing, L4) # 1.59502696991
Answered By: Nathan Farrington

As noted by @aaronsterling the following solution is the shortest and seems fastest when the array is sorted and not too small:
def is_sorted(lst):
return (sorted(lst) == lst)

If most of the time the array is not sorted, it would be desirable to use a solution that does not scan the entire array and returns False as soon as an unsorted prefix is discovered. Following is the fastest solution I could find, it is not particularly elegant:

def is_sorted(lst):
    it = iter(lst)
        prev = next(it)
    except StopIteration:
        return True
    for x in it:
        if prev > x:  # For reverse, use <
            return False
        prev = x
    return True

Using Nathan Farrington’s benchmark, this achieves better runtime than using sorted(lst) in all cases except when running on the large sorted list.

Here are the benchmark results on my computer.

sorted(lst)==lst solution

  • L1: 1.23838591576
  • L2: 4.19063091278
  • L3: 1.17996287346
  • L4: 4.68399500847

Second solution:

  • L1: 0.81095790863
  • L2: 0.802397012711
  • L3: 1.06135106087
  • L4: 8.82761001587
Answered By: Amit Moscovich

Although I don’t think there is a guarantee for that the sorted built-in calls its cmp function with i+1, i, it does seem to do so for CPython.

So you could do something like:

def my_cmp(x, y):
   cmpval = cmp(x, y)
   if cmpval < 0:
      raise ValueError
   return cmpval

def is_sorted(lst):
      sorted(lst, cmp=my_cmp)
      return True
   except ValueError:
      return False

print is_sorted([1,2,3,5,6,7])
print is_sorted([1,2,5,3,6,7])

Or this way (without if statements -> EAFP gone wrong? 😉 ):

def my_cmp(x, y):
   assert(x >= y)
   return -1

def is_sorted(lst):
      sorted(lst, cmp=my_cmp)
      return True
   except AssertionError:
      return False
Answered By: Anthon

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).

Answered By: orlade

I’d do this (stealing from a lot of answers here [Aaron Sterling, Wai Yip Tung, sorta from Paul McGuire] and mostly Armin Ronacher):

from itertools import tee, izip

def pairwise(iterable):
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

def is_sorted(iterable, key=lambda a, b: a <= b):
    return all(key(a, b) for a, b in pairwise(iterable))

One nice thing: you don’t have to realize the second iterable for the series (unlike with a list slice).

Answered By: hughdbrown

I use this one-liner based on numpy.diff():

def issorted(x):
    """Check if x is sorted"""
    return (numpy.diff(x) >= 0).all() # is diff between all consecutive entries >= 0?

I haven’t really timed it against any other method, but I assume it’s faster than any pure Python method, especially for large n, since the loop in numpy.diff (probably) runs directly in C (n-1 subtractions followed by n-1 comparisons).

However, you need to be careful if x is an unsigned int, which might cause silent integer underflow in numpy.diff(), resulting in a false positive. Here’s a modified version:

def issorted(x):
    """Check if x is sorted"""
        if x.dtype.kind == 'u':
            # x is unsigned int array, risk of int underflow in np.diff
            x = numpy.int64(x)
    except AttributeError:
        pass # no dtype, not an array
    return (numpy.diff(x) >= 0).all()
Answered By: Martin Spacek

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.

Definitely works in Python 3 and above for integers or strings:

def tail(t):
    return t[:]

letters = ['a', 'b', 'c', 'd', 'e']
rest = tail(letters)
if letters == rest:
    print ('Given list is SORTED.')
    print ('List NOT Sorted.')


Another way of finding if the given list is sorted or not

trees1 = list ([1, 4, 5, 3, 2])
trees2 = list (trees1)
if trees1 == trees2:
    print ('trees1 is SORTED')
    print ('Not sorted')
Answered By: cooldeal


from itertools import tee

def is_sorted(l):
    l1, l2 = tee(l)
    next(l2, None)
    return all(a <= b for a, b in zip(l1, l2))
Answered By: Sergey11g

This uses recursion:

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

 some_list = [1,2,3,4]

Note that this will raise RuntimeError: maximum recursion depth exceeded for long sequences.

Answered By: Baraa

This is similar to the top answer, but I like it better because it avoids explicit indexing. Assuming your list has the name lst, you can generate
(item, next_item) tuples from your list with zip:

all(x <= y for x,y in zip(lst, lst[1:]))

In Python 3, zip already returns a generator, in Python 2 you can use itertools.izip for better memory efficiency.

Small demo:

>>> lst = [1, 2, 3, 4]
>>> zip(lst, lst[1:])
[(1, 2), (2, 3), (3, 4)]
>>> all(x <= y for x,y in zip(lst, lst[1:]))
>>> lst = [1, 2, 3, 2]
>>> zip(lst, lst[1:])
[(1, 2), (2, 3), (3, 2)]
>>> all(x <= y for x,y in zip(lst, lst[1:]))

The last one fails when the tuple (3, 2) is evaluated.

Bonus: checking finite (!) generators which cannot be indexed:

>>> def gen1():
...     yield 1
...     yield 2
...     yield 3
...     yield 4
>>> def gen2():
...     yield 1
...     yield 2
...     yield 4
...     yield 3
>>> g1_1 = gen1()
>>> g1_2 = gen1()
>>> next(g1_2)
>>> all(x <= y for x,y in zip(g1_1, g1_2))
>>> g2_1 = gen2()
>>> g2_2 = gen2()
>>> next(g2_2)
>>> all(x <= y for x,y in zip(g2_1, g2_2))

Make sure to use itertools.izip here if you are using Python 2, otherwise you would defeat the purpose of not having to create lists from the generators.

Answered By: timgeb

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
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
        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
Answered By: tal

Just to add another way (even if it requires an additional module): iteration_utilities.all_monotone:

>>> from iteration_utilities import all_monotone
>>> listtimestamps = [1, 2, 3, 5, 6, 7]
>>> all_monotone(listtimestamps)

>>> all_monotone([1,2,1])

To check for DESC order:

>>> all_monotone(listtimestamps, decreasing=True)

>>> all_monotone([3,2,1], decreasing=True)

There is also a strict parameter if you need to check for strictly (if successive elements should not be equal) monotonic sequences.

It’s not a problem in your case but if your sequences contains nan values then some methods will fail, for example with sorted:

def is_sorted_using_sorted(iterable):
    return sorted(iterable) == iterable

>>> is_sorted_using_sorted([3, float('nan'), 1])  # definitely False, right?

>>> all_monotone([3, float('nan'), 1])

Note that iteration_utilities.all_monotone performs faster compared to the other solutions mentioned here especially for unsorted inputs (see benchmark).

Answered By: MSeifert

How about this one ? Simple and straightforward.

def is_list_sorted(al):

    llength =len(al)

    for i in range (llength):
        if (al[i-1] > al[i]):
            print('Not sorted')
            return -1

    else :
        return  true
Answered By: Zuckerberg

Simplest way:

def isSorted(arr):
  i = 1
  while i < len(arr):
    if(result[i] < result[i - 1]):
      return False
    i += 1
  return True
Answered By: Aluis92

Python 3.6.8

from more_itertools import pairwise

class AssertionHelper:
    def is_ascending(cls, data: iter) -> bool:
        for a, b in pairwise(data):
            if a > b:
                return False
        return True

    def is_descending(cls, data: iter) -> bool:
        for a, b in pairwise(data):
            if a < b:
                return False
        return True

    def is_sorted(cls, data: iter) -> bool:
        return cls.is_ascending(data) or cls.is_descending(data)
>>> AssertionHelper.is_descending((1, 2, 3, 4))
>>> AssertionHelper.is_ascending((1, 2, 3, 4))
>>> AssertionHelper.is_sorted((1, 2, 3, 4))

Answered By: Chweng Mega
from functools import reduce

# myiterable can be of any iterable type (including list)
isSorted = reduce(lambda r, e: (r[0] and (r[1] or r[2] <= e), False, e), myiterable, (True, True, None))[0]

The derived reduction value is a 3-part tuple of (sortedSoFarFlag, firstTimeFlag, lastElementValue). It initially starts with (True, True, None), which is also used as the result for an empty list (regarded as sorted because there are no out-of-order elements). As it processes each element it calculates new values for the tuple (using previous tuple values with the next elementValue):

[0] (sortedSoFarFlag) evaluates true if: prev_0 is true and (prev_1 is true or prev_2 <= elementValue)
[1] (firstTimeFlag): False
[2] (lastElementValue): elementValue

The final result of the reduction is a tuple of:

[0]: True/False depending on whether the entire list was in sorted order
[1]: True/False depending on whether the list was empty
[2]: the last element value

The first value is the one we’re interested in, so we use [0] to grab that from the reduce result.

Answered By: Mr Weasel

As I don’t see this option above I will add it to all the answers.
Let denote the list by l, then:

import numpy as np

# Trasform the list to a numpy array
x = np.array(l)

# check if ascendent sorted:
all(x[:-1] <= x[1:])

# check if descendent sorted:
all(x[:-1] >= x[1:])
Answered By: FBruzzesi

A solution using assignment expressions (added in Python 3.8):

def is_sorted(seq):
    seq_iter = iter(seq)
    cur = next(seq_iter, None)
    return all((prev := cur) <= (cur := nxt) for nxt in seq_iter)

z = list(range(10))

import random

z = []


[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[1, 7, 5, 9, 4, 0, 8, 3, 2, 6]
Answered By: PaulMcG

Starting in Python 3.10, the new pairwise function provides a way to slide through pairs of consecutive elements, and thus find if all of these pairs satisfy the same predicate of ordering:

from itertools import pairwise

all(x <= y for x, y in pairwise([1, 2, 3, 5, 6, 7]))
# True

The intermediate result of pairwise:

pairwise([1, 2, 3, 5, 6, 7])
# [(1, 2), (2, 3), (3, 5), (5, 6), (6, 7)]
Answered By: Xavier Guihot

The third-party package method more_itertools.is_sorted hasn’t been mentioned yet:

import more_itertools

ls = [1, 4, 2]


ls2 = ["ab", "c", "def"]

print(more_itertools.is_sorted(ls2, key=len))
Answered By: Zane Dufour

This approach using Pandas is very slow, but it’s noted for completeness.

from typing import Sequence

import pandas as pd

def is_sorted(seq: Sequence, reverse: bool = False) -> bool:
    index = pd.Index(seq)
    if reverse:
        return index.is_monotonic_decreasing
    return index.is_monotonic_increasing
Answered By: Asclepius

Try this:

def is_sorted(arr) -> bool:
    for i in range(1, len(arr)):
        if arr[i] < arr[i - 1]:
            return False
    return True
Answered By: Rustam-Z
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