使用Python联合查找实现

本文介绍了使用Python联合查找实现的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

这就是我想做的:我有一个包含几个等价关系的列表:

So here's what I want to do: I have a list that contains several equivalence relations:

l = [[1, 2], [2, 3], [4, 5], [6, 7], [1, 7]]

我想合并共享一个元素的集合.这是一个示例实现:

And I want to union the sets that share one element. Here is a sample implementation:

def union(lis):
  lis = [set(e) for e in lis]
  res = []
  while True:
    for i in range(len(lis)):
      a = lis[i]
      if res == []:
        res.append(a)
      else:
        pointer = 0
        while pointer < len(res):
          if a & res[pointer] != set([]) :
            res[pointer] = res[pointer].union(a)
            break
          pointer +=1
        if pointer == len(res):
          res.append(a)
     if res == lis:
      break
    lis,res = res,[]
  return res

它会打印

[set([1, 2, 3, 6, 7]), set([4, 5])]

这做对了，但是当等价关系太大时，它太慢了.我查看了有关联合查找算法的描述: http://en.wikipedia.org/wiki/不相交的set_data_structure 但是我仍然在编码Python实现时遇到问题.

This does the right thing but is way too slow when the equivalence relations is too large. I looked up the descriptions on union-find algorithm: http://en.wikipedia.org/wiki/Disjoint-set_data_structurebut I still having problem coding a Python implementation.

在`O(n)`时间运行的解决方案

Solution that runs in `O(n)` time

def indices_dict(lis):
    d = defaultdict(list)
    for i,(a,b) in enumerate(lis):
        d[a].append(i)
        d[b].append(i)
    return d

def disjoint_indices(lis):
    d = indices_dict(lis)
    sets = []
    while len(d):
        que = set(d.popitem()[1])
        ind = set()
        while len(que):
            ind |= que
            que = set([y for i in que
                         for x in lis[i]
                         for y in d.pop(x, [])]) - ind
        sets += [ind]
    return sets

def disjoint_sets(lis):
    return [set([x for i in s for x in lis[i]]) for s in disjoint_indices(lis)]

工作方式:

>>> lis = [(1,2),(2,3),(4,5),(6,7),(1,7)]
>>> indices_dict(lis)
>>> {1: [0, 4], 2: [0, 1], 3: [1], 4: [2], 5: [2], 6: [3], 7: [3, 4]})

indices_dict给出从等价#到lis中的索引的映射.例如. 1映射到lis中的索引0和4.

indices_dict gives a map from an equivalence # to an index in lis. E.g. 1 is mapped to index 0 and 4 in lis.

>>> disjoint_indices(lis)
>>> [set([0,1,3,4], set([2])]

disjoint_indices给出了不相交的索引集的列表.每个集合对应于等价的索引.例如. lis[0]和lis[3]是相同的，但不是lis[2].

disjoint_indices gives a list of disjoint sets of indices. Each set corresponds to indices in an equivalence. E.g. lis[0] and lis[3] are in the same equivalence but not lis[2].

>>> disjoint_set(lis)
>>> [set([1, 2, 3, 6, 7]), set([4, 5])]

disjoint_set将不相交的索引转换为它们的适当等值.

disjoint_set converts disjoint indices into into their proper equivalences.

很难看到 O(n) 的时间复杂度，但我会尽力解释.在这里，我将使用n = len(lis).

indices_dict当然可以在O(n)时间内运行，因为只有1个循环

indices_dict certainly runs in O(n) time because only 1 for-loop

disjoint_indices是最难看到的.它肯定在O(len(d))时间运行，因为当d为空时外循环停止，并且内循环在每次迭代中都删除d的元素.现在，len(d) <= 2n因为d是从等价#到索引的映射，并且lis中最多有2n个等价编号.因此，该函数在O(n)中运行.

disjoint_indices is the hardest to see. It certainly runs in O(len(d)) time since the outer loop stops when d is empty and the inner loop removes an element of d each iteration. now, the len(d) <= 2n since d is a map from equivalence #'s to indices and there are at most 2n different equivalence #'s in lis. Therefore, the function runs in O(n).

disjoint_sets.但是，您会注意到，i最多可以在lis中的所有n索引上运行，而x可以在2元组上运行，因此总复杂度为2n = O(n)

disjoint_sets is difficult to see because of the 3 combined for-loops. However, you'll notice that at most i can run over all n indices in lis and x runs over the 2-tuple, so the total complexity is 2n = O(n)

这篇关于使用Python联合查找实现的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

lis