将Python列表列表根据重叠项分组

小编典典

将Python列表列表根据重叠项分组

algorithm

我有一个列表列表，我试图根据它们的项目对它们进行分组或聚类。如果上一个组中没有元素，则嵌套列表将开始一个新组。

输入：

paths = [  
        ['D', 'B', 'A', 'H'],  
        ['D', 'B', 'A', 'C'],  
        ['H', 'A', 'C'],  
        ['E', 'G', 'I'],  
        ['F', 'G', 'I']]

我失败的代码：

paths = [
    ['D', 'B', 'A', 'H'],
    ['D', 'B', 'A', 'C'],
    ['H', 'A', 'C'],
    ['E', 'G', 'I'],
    ['F', 'G', 'I']
]
groups = []
paths_clone = paths
for path in paths:
    for node in path:
        for path_clone in paths_clone:
            if node in path_clone:
                if not path == path_clone:
                    groups.append([path, path_clone])
                else:
                    groups.append(path)
print groups

预期产量：

[
 [
  ['D', 'B', 'A', 'H'],
  ['D', 'B', 'A', 'C'],
  ['H', 'A', 'C']
 ],
 [
  ['E', 'G', 'I'],
  ['F', 'G', 'I']
 ]
]

另一个例子：

paths = [['shifter', 'barrel', 'barrel shifter'],
         ['ARM', 'barrel', 'barrel shifter'],
         ['IP power', 'IP', 'power'],
         ['ARM', 'barrel', 'shifter']]

预期的输出组：

output = [
         [['shifter', 'barrel', 'barrel shifter'],
         ['ARM', 'barrel', 'barrel shifter'],
         ['ARM', 'barrel', 'shifter']],
         [['IP power', 'IP', 'power']],
         ]

阅读 372

2020-07-28

共1个答案

小编典典

您是根据集合进行分组，因此请使用集合来检测新的分组：

def grouper(sequence):
    group, members = [], set()

    for item in sequence:
        if group and members.isdisjoint(item):
            # new group, yield and start new
            yield group
            group, members = [], set()
        group.append(item)
        members.update(item)

    yield group

这给出：

>>> for group in grouper(paths):
...     print group
... 
[['D', 'B', 'A', 'H'], ['D', 'B', 'A', 'C'], ['H', 'A', 'C']]
[['E', 'G', 'I'], ['F', 'G', 'I']]

或者您可以将其再次投射到列表中：

output = list(grouper(paths))

这假定组是连续的。如果您有不相交的组，则需要处理整个列表并遍历到目前为止为每个项目构造的所有组：

def grouper(sequence):
    result = []  # will hold (members, group) tuples

    for item in sequence:
        for members, group in result:
            if members.intersection(item):  # overlap
                members.update(item)
                group.append(item)
                break
        else:  # no group found, add new
            result.append((set(item), [item]))

    return [group for members, group in result]

2020-07-28