返回的组本身就是一个共享基础对象的迭代器 可通过groupby()迭代.因为源是共享的，所以当 groupby()对象是高级的，上一个组不再 可见. The returned group is itself an iterator that shares the underlying iterable with groupby(). Because the source is shared, when the groupby() object is advanced, the previous group is no longer visible.让我们将其分解为几个阶段.Let's break it down into stages.from itertools import groupbya = list(groupby("cccccaaaaatttttsssssss"))print(a)b = a[0][1]print(b)print('So far, so good')print(list(b))print('What?!') 输出[('c', <itertools._grouper object at 0xb715104c>), ('a', <itertools._grouper object at 0xb715108c>), ('t', <itertools._grouper object at 0xb71510cc>), ('s', <itertools._grouper object at 0xb715110c>)]<itertools._grouper object at 0xb715104c>So far, so good[]What?!我们的itertools._grouper object at 0xb715104c是空的，因为它与groupby返回的父"迭代器共享其内容，而这些项现在都消失了，因为第一个list调用在父对象上进行了迭代.Our itertools._grouper object at 0xb715104c is empty because it shares its contents with the "parent" iterator returned by groupby, and those items are now gone because that first list call iterated over the parent.如果您尝试在任何迭代器上进行两次迭代(例如，简单的生成器表达式)，则实际上没有什么不同.It's really no different to what happens if you try to iterate twice over any iterator, eg a simple generator expression.g = (c for c in 'python')print(list(g))print(list(g)) 输出['p', 'y', 't', 'h', 'o', 'n'][]顺便说一句，如果您实际上不需要groupby组的内容，这是另一种获取长度的方法.比建立一个列表来查找长度要便宜一些(并且使用更少的RAM).BTW, here's another way to get the length of a groupby group if you don't actually need its contents; it's a little cheaper (and uses less RAM) than building a list just to find its length.from itertools import groupbyfor k, g in groupby("cccccaaaaatttttsssssss"): print(k, sum(1 for _ in g)) 输出c 5a 5t 5s 7 这篇关于如何将itertools的“石斑鱼"变成将对象放入列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！
