迭代列表中的每两个元素 [英] Iterating over every two elements in a list

问题描述

如何制作 for 循环或列表推导式，以便每次迭代都给我两个元素?

How do I make a for loop or a list comprehension so that every iteration gives me two elements?

l = [1,2,3,4,5,6]

for i,k in ???:
    print str(i), '+', str(k), '=', str(i+k)

输出:

1+2=3
3+4=7
5+6=11

推荐答案

你需要一个pairwise()(或grouped()代码>) 实现.

You need a pairwise() (or grouped()) implementation.

def pairwise(iterable):
    "s -> (s0, s1), (s2, s3), (s4, s5), ..."
    a = iter(iterable)
    return zip(a, a)

for x, y in pairwise(l):
   print("%d + %d = %d" % (x, y, x + y))

或者，更一般地说:

def grouped(iterable, n):
    "s -> (s0,s1,s2,...sn-1), (sn,sn+1,sn+2,...s2n-1), (s2n,s2n+1,s2n+2,...s3n-1), ..."
    return zip(*[iter(iterable)]*n)

for x, y in grouped(l, 2):
   print("%d + %d = %d" % (x, y, x + y))

在 Python 2 中，您应该导入 izip 作为 Python 3 内置 zip() 函数.

In Python 2, you should import izip as a replacement for Python 3's built-in zip() function.

martineau-a-python-dictionary-from-a-line-of-text/4356415#4356415">他对我的回答问题，我发现这非常有效，因为它只在列表上迭代一次，并且不会在此过程中创建任何不必要的列表.

All credit to martineau for his answer to my question, I have found this to be very efficient as it only iterates once over the list and does not create any unnecessary lists in the process.

注意:这不应与 pairwise recipe 在 Python 自己的 itertools 文档，产生 s ->(s0, s1), (s1, s2), (s2, s3), ...，正如 所指出的@lazyr 在评论中.

N.B: This should not be confused with the pairwise recipe in Python's own itertools documentation, which yields s -> (s0, s1), (s1, s2), (s2, s3), ..., as pointed out by @lazyr in the comments.

对于那些想在 Python 3 上使用 mypy 进行类型检查的人来说，补充一点:

Little addition for those who would like to do type checking with mypy on Python 3:

from typing import Iterable, Tuple, TypeVar

T = TypeVar("T")

def grouped(iterable: Iterable[T], n=2) -> Iterable[Tuple[T, ...]]:
    """s -> (s0,s1,s2,...sn-1), (sn,sn+1,sn+2,...s2n-1), ..."""
    return zip(*[iter(iterable)] * n)

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