Question

我的数据格式如下：

Group  Item-1  Item-2
0       7       13
0      10        4
1       2        8
1       3        1
1       4        3
1       6       28
1       8        6
...

我需要Python 2.7为每个组输出Item-1和Item-2之间的增长链/树/连接。因此对于除0-13和10-4之外的第0组，没有链/连接，但对于第1组，输出将类似于：

1  (2, 8, 6, 28), (4, 3, 1)

Answer 1

这似乎是union-find或disjoint-set算法的一种情况。这是我用来保存在工具箱中的一个实现：

from collections import defaultdict

class UnionFind:
    def __init__(self):
        self.leaders = defaultdict(lambda: None)

    def find(self, x):
        l = self.leaders[x]
        if l is not None:
            l = self.find(l)
            self.leaders[x] = l
            return l
        return x

    def union(self, x, y):
        lx, ly = self.find(x), self.find(y)
        if lx != ly:
            self.leaders[lx] = ly

    def get_groups(self):
        groups = defaultdict(set)
        for x in self.leaders:
            groups[self.find(x)].add(x)
        return groups

以下是如何将其应用于您的数据：

# parse data
data = """Group  Item-1  Item-2
0       7       13
0      10        4
1       2        8
1       3        1
1       4        3
1       6       28
1       8        6"""
data = [[int(x) for x in line.split()] for line in data.splitlines()[1:]]

# get mapping {group_number: [list of pairs]}
groups = defaultdict(list)
for g, x, y in data:
    groups[g].append((x, y))

# for each group, add pairs to union find structure and get groups
for group, links in groups.items():
    union = UnionFind()
    for x, y in links:
        union.union(x, y)
    print group, union.get_groups().values()

输出是：

0 [set([10, 4]), set([13, 7])]
1 [set([1, 3, 4]), set([8, 2, 28, 6])]

用于创建树的Python列表比较

1 个答案: