根据重叠项将Python列表分组到组中

2023年10月25日 178次阅读

我有一个列表列表,我试图根据他们的项目对它们进行分组或聚类.如果上一组中没有元素,嵌套列表将启动一个新组.

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

预期产出：

[
 [
  ['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']]

预期产出组：

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

最佳答案 您正在基于集合进行分组,因此使用集合来检测新组：

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

这给出了：

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

或者您可以再次将其强制转换为列表：

output = list(grouper(paths))

这假设这些组是连续的.如果您有不相交的组,则需要处理整个列表并循环遍历为每个项目构建的所有组：

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]
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