python统计列表中的项目并保持其发生顺序(python count items in list and keep their order of occurrance)

python统计列表中的项目并保持其发生顺序(python count items in list and keep their order of occurrance)

鉴于：列表，例如l = [4,4,4,4,5,5,5,6,7,7,7]待办事项：获取元素的数量并保持其发生顺序，例如：[（ 4,4），（5,3），（6,1），（7,3）]

我可以这样做：

tmpL = [(i,l.count(i)) for i in l] tmpS = set() cntList = [x for x in tmpL if x not in tmpS and not tmpS.add(x)]

但是有更好的方法吗？ 我在这里看到了这个链接，但它将计数排序并因此打乱了顺序。

编辑：性能不是解决方案的问题，最好是内置的东西。

Given: a list, such as l=[4,4,4,4,5,5,5,6,7,7,7] Todo: get the count of an element and keep their occurrence order, e.g.: [(4,4),(5,3),(6,1),(7,3)]

I could do it with:

tmpL = [(i,l.count(i)) for i in l] tmpS = set() cntList = [x for x in tmpL if x not in tmpS and not tmpS.add(x)]

But is there a better way? I have seen the link here, but it sorts the counts and hence breaks the order.

Edit: performance is not an issue for the solution, preferable something built-in.

最满意答案

使用groupby ：

>>> l = [4,4,4,4,5,5,5,6,7,7,7,2,2] >>> from itertools import groupby >>> [(i, l.count(i)) for i,_ in groupby(l)] [(4, 4), (5, 3), (6, 1), (7, 3), (2, 2)]

Use groupby:

>>> l = [4,4,4,4,5,5,5,6,7,7,7,2,2] >>> from itertools import groupby >>> [(i, l.count(i)) for i,_ in groupby(l)] [(4, 4), (5, 3), (6, 1), (7, 3), (2, 2)]