比较列表中所有相邻元素的Pythonic方法(Pythonic way of comparing all adjacent elements in a list)

比较列表中所有相邻元素的Pythonic方法(Pythonic way of comparing all adjacent elements in a list)

我想知道是否有更多的Pythonic方法来做到以下几点：

A = some list i = 0 j = 1 for _ in range(1, len(A)): #some operation between A[i] and A[j] i += 1 j += 1

我觉得这应该/可以做得不同。 想法？

编辑：因为有些人要求要求。 我想要一个通用的答案。 也许要检查A [i]，A [j]是否在一定范围内，或者它们是否相等。 或者，也许我想要做一些元素的“涓滴”。 越一般，越好。

I want to know if there's a more Pythonic way of doing the following:

A = some list i = 0 j = 1 for _ in range(1, len(A)): #some operation between A[i] and A[j] i += 1 j += 1

I feel like this should/could be done differently. Ideas?

EDIT: Since some are asking for requirements. I wanted a general-purpose answer. Maybe to check if A[i], A[j] are between a certain range, or if they're equal. Or maybe I wanted to do a "trickle-up" of elements. The more general, the better.

最满意答案

zip可以让你组合多个迭代器：

for i,j in zip(range(0,len(A)-1), range(1,len(A))): #some operation between A[i] and A[j]

你也可以在范围对象上使用enumerate ：

for i,j in enumerate(range(1,len(A)): #some operation between A[i] and A[j]

请注意，与其他答案不同的是，您可以访问A的索引而不仅仅是项目，如果您想要对A[i]或A[j]使用任何分配 ，则这是必要的，例如，这里是非常基本的气泡排序：

A = list(range(10)) found1=True while found1: found1=False for i,j in enumerate(range(1,len(A))): if A[i] < A[j]: A[i],A[j] = A[j],A[i] found1=True print(A)

这只有在迭代A的索引时才有可能。

zip lets you combine multiple iterators:

for i,j in zip(range(0,len(A)-1), range(1,len(A))): #some operation between A[i] and A[j]

you can also use enumerate on a range object:

for i,j in enumerate(range(1,len(A)): #some operation between A[i] and A[j]

Note that unlike the other answers this gives you access to the indices of A not just the items, this is necessary if you want to use any assignment to A[i] or A[j], for example here is a very basic bubble sort:

A = list(range(10)) found1=True while found1: found1=False for i,j in enumerate(range(1,len(A))): if A[i] < A[j]: A[i],A[j] = A[j],A[i] found1=True print(A)

this is only possible when you iterate over the indices of A.