在列表中查找ContiguousCount項？

Question

給出一個清單：

>>> l = ['x', 'x', 'y', 'y', 'x']

我可以通過使用collections.Counter得到列表的計數：

>>> from collections import Counter
>>> Counter(l)
Counter({'x': 3, 'y': 2})

如何計算連續項而不是列表中元素的全局計數？ 例如

>>> l = ['x', 'x', 'y', 'y', 'x']
>>> ContiguousCounter(l)
[('x',2), ('y',2), ('x', 1)]

>>> l = ['x', 'x', 'y', 'y', 'x', 'x', 'x', 'y']
>>> ContiguousCounter(l)
[('x',2), ('y',2), ('x', 3), ('y', 1)]

Answer 1

你可以使用內置的itertools.groupby函數：

In [3]: from itertools import groupby

In [4]: l = ['x', 'x', 'y', 'y', 'x']

In [5]: list(groupby(l))
Out[5]: 
[('x', <itertools._grouper at 0x7fd94716f1d0>),
 ('y', <itertools._grouper at 0x7fd94716f208>),
 ('x', <itertools._grouper at 0x7fd94716f240>)]

In [6]: [(x, len(list(g))) for x, g in groupby(l)]
Out[6]: [('x', 2), ('y', 2), ('x', 1)]