如何编写代码来确定列表是否在Python中排序？

Question

How would I write a function that takes a list as an argument and verifies whether or not the list is sorted?

I want it to return true if sorted and false otherwise. For example:

>>> is_sorted([1, 5, 8, 10]) 
True

>>> is_sorted([4, 1, 7, 8])
False

However, I'm unsure of how to do this with any kind of list at all

Answer 1

The simplest version that gets O(n) performance:

def is_sorted(lst):
    return all(a <= b for a,b in zip(lst, lst[1:]))

Answer 2

An obvious way is to sort it and compare for equality with the original list:

#!/usr/bin/env python

def is_sorted(mylist):
    testlist = sorted(mylist[:])
    return testlist == mylist

print is_sorted([1, 4, 6, 7])
print is_sorted([1, 7, 6, 5])

outputs:

paul@horus:~/src/sandbox$ ./sorted.py
True
False
paul@horus:~/src/sandbox$ 

If you want something slightly more efficient (presuming you want to check if it's sorted in ascending order):

def is_sorted(mylist):
    last_item = mylist[0]

    for item in mylist:
        if item < last_item:
            return False
        last_item = item

    return True

Answer 3

This should be a O(n) solution by iterating through the list to check for values out of order. O(n) is worst case since it returns false the first element it finds out of order.

#!/usr/bin/env python

def is_sorted(mylist):
    for i, val in enumerate(mylist):
        if (i > 0 and mylist[i] < mylist[i-1]):
            return False
    return True

print is_sorted([1, 4, 6, 7])
print is_sorted([1, 7, 6, 5])

Output:

True
False

Answer 4

def is_sorted(mylist):
    return False not in [sorted(mylist)[idx] == val for idx, val in enumerate(mylist)]

>>> is_sorted([1, 2, 3, 4])
True

>>> is_sorted([1, 2, 4, 3])
False

Answer 5

Here is the most simplest way to do this:

def is_sorted(l):
    return True if l == sorted(l) else False