Pythonic way to check if a list is sorted or not

问题:

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.

回答1:

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))

回答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.

回答3:

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:]))

回答4:

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.

回答5:

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).

回答6:

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: aaronsterling\'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):
        f(*args)
    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

回答7:

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):
   try:
      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):
   try:
      sorted(lst, cmp=my_cmp)
      return True
   except AssertionError:
      return False

回答8:

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).

回答9:

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\"\"\"
    try:
        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()

回答10:

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:]))
True
>>> 
>>> 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:]))
False

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)
1
>>> all(x <= y for x,y in zip(g1_1, g1_2))
True
>>>
>>> g2_1 = gen2()
>>> g2_2 = gen2()
>>> next(g2_2)
1
>>> all(x <= y for x,y in zip(g2_1, g2_2))
False

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.

回答11:

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 ;)

回答12:

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)
    try:
        prev = it.next()
    except StopIteration:
        return True
    for x in it:
        if prev > x:
            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

回答13:

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

回答14:

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)
True

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

To check for DESC order:

>>> all_monotone(listtimestamps, decreasing=True)
False

>>> all_monotone([3,2,1], decreasing=True)
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])  # definetly False, right?
True

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

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

回答15:

Lazy

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))

回答16:

Simplest way:

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

回答17:

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)
rest.sort()
if letters == rest:
    print (\'Given list is SORTED.\')
else:
    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)
trees2.sort()
if trees1 == trees2:
    print (\'trees1 is SORTED\')
else:
    print (\'Not sorted\')

回答18:

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)

回答19:

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(al[i])
            print(al[i+1])
            print(\'Not sorted\')
            return -1

    else :
        print(\'sorted\')
        return  true