Compare list items in pythonic way

Question

对于list说l = [1, 2, 3, 4]如何以Python方式比较l[0] < l[1] < l[2] < l[3] ？

Answer 1

那这个呢：

l == sorted(l)

Answer 2

There's a general approach, that consists in zipping the list with itself, with an offset.

In this situation, you can wrap it up in a list comprehension:

[a < b for a, b in zip(l[:-1], l[1:])]
# [l[0] < l[1], l[1] < l[2], ...]

This will give you a list of booleans. To check if the list is globally sorted, you can use all :

all([a < b for a, b in zip(l[:-1], l[1:])])

But then, you don't need to build the actual list, a generator is enough:

all(a < b for a, b in zip(l[:-1], l[1:]))

Notice that the square brackets [] are dropped: the elements of the comprehensions will be evaluated lazily, which will avoid the creation of a useless list.

This solution has a O(n) complexity: you iterate twice over l with the zip , then once over the zipped values with all .

On the other hand, l == sorted(l) looks smart and is elegant, but you'll have to pay the O(n*log(n)) cost of sorting l . If your goal is to sort l if it is not sorted, just directly call l.sort() .

Answer 3

尝试这个：

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

Answer 4

你可以这样

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

Answer 5

另一种方法是使用.sort（）方法，在这种情况下，您必须完全返回一个新列表。