如何判断列表是否按升序排列？(How do I tell if a list is in ascending order? Just needs to return True or False)

如何判断列表是否按升序排列？(How do I tell if a list is in ascending order? Just needs to return True or False)

我有它的方式它检查前两个数字是否上升后停止。

在检查完整个列表之前，如何让它继续运行？

def isAscending(xs): for n in range(len(xs) + 1): if xs[n] > xs[n+1]: return False else: if xs[n] < xs[n+1]: return True

The way I have it it stops after checking if the first two digits are ascending.

How do I make it keep running until it has checked the whole list?

def isAscending(xs): for n in range(len(xs) + 1): if xs[n] > xs[n+1]: return False else: if xs[n] < xs[n+1]: return True

最满意答案

仅在结尾处返回True ，即在检查每个元素之后。 使用最少的更改修复代码：

def isAscending(xs): for n in range(len(xs) - 1): if xs[n] > xs[n+1]: return False return True print isAscending([1,2,3,4]) # True print isAscending([1,2,4,3]) # False

简短解决方案

>>> lst = [1,2,3,4] >>> sorted(lst) == lst True >>> lst = [1,2,4,3] >>> sorted(lst) == lst False

在O（n）中运行的更好的简短解决方案（排序为O（n log n））：

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

为了使最后一个内存更有效，如果您使用的是Python 2，请使用itertools.izip代替zip 。

Only return True at the end, that is after every element has been checked. Fixing your code with minimal changes:

def isAscending(xs): for n in range(len(xs) - 1): if xs[n] > xs[n+1]: return False return True print isAscending([1,2,3,4]) # True print isAscending([1,2,4,3]) # False

Short solution:

>>> lst = [1,2,3,4] >>> sorted(lst) == lst True >>> lst = [1,2,4,3] >>> sorted(lst) == lst False

Better short solution that runs in O(n) (sorting is O(n log n)):

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

To make the last one more memory efficient, use itertools.izip in place of zip if you are using Python 2.