Python循环：以惯用方式比较列表中的连续项 [英] Python looping: idiomatically comparing successive items in a list

问题描述

我需要遍历一个对象列表，比较它们：0对1,1对2,2对3等等（我正在使用pysvn来提取差异列表。）I最后只是循环索引，但我一直想知道是否有某种方法可以做到更接近惯用。这是Python;我不应该以某种聪明的方式使用迭代器吗？简单地循环索引似乎很清楚，但我想知道是否有一种更具表现力或简洁的方法。

I need to loop over a list of objects, comparing them like this: 0 vs. 1, 1 vs. 2, 2 vs. 3, etc. (I'm using pysvn to extract a list of diffs.) I wound up just looping over an index, but I keep wondering if there's some way to do it which is more closely idiomatic. It's Python; shouldn't I be using iterators in some clever way? Simply looping over the index seems pretty clear, but I wonder if there's a more expressive or concise way to do it.

for revindex in xrange(len(dm_revisions) - 1):
    summary = \
        svn.diff_summarize(svn_path,
                          revision1=dm_revisions[revindex],
                          revision2 = dm_revisions[revindex+1])

推荐答案

这称为滑动窗口。 文档。这是代码：

This is called a sliding window. There's an example in the itertools documentation that does it. Here's the code:

from itertools import islice

def window(seq, n=2):
    "Returns a sliding window (of width n) over data from the iterable"
    "   s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ...                   "
    it = iter(seq)
    result = tuple(islice(it, n))
    if len(result) == n:
        yield result    
    for elem in it:
        result = result[1:] + (elem,)
        yield result

你可以这样说：

for r1, r2 in window(dm_revisions):
    summary = svn.diff_summarize(svn_path, revision1=r1, revision2=r2)

当然你只关心n = 2的情况，所以你可以放弃更简单的东西：

Of course you only care about the case where n=2, so you can get away with something much simpler:

def adjacent_pairs(seq):
    it = iter(seq)
    a = it.next()
    for b in it:
        yield a, b
        a = b

for r1, r2 in adjacent_pairs(dm_revisions):
    summary = svn.diff_summarize(svn_path, revision1=r1, revision2=r2)

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