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zealinux

求教 Python 合并元组算法

  •  
  •   zealinux Dec 12, 2018 3939 views
    This topic created in 2692 days ago, the information mentioned may be changed or developed.

    问个技术问题:

    比如: [("a", "b"), ("b", "c"), ("e", "f")]

    合并成

    [set("a", "b", "c"), set("e, "f")]

    (即与有一个元素有交集的,就合并进来)

    ============= 原列表中有一千个元组,半天没出结果

    我写的代码:

    # l = [("a", "b"), ("b", "c"), ("e", "f")] def merger(l): out_list = [] for item in l: temp_set = set(item) if len(out_list) == 0: out_list.append(temp_set) else: for item_set in out_list: if len(temp_set & item_set) > 0: item_set.update(temp_set) else: out_list.append(temp_set) 
    13 replies    2018-12-13 11:34:05 +08:00
    doraemon1293
        1
    doraemon1293  
       Dec 12, 2018
    set(itertools.chain(*arr))
    doraemon1293
        2
    doraemon1293  
       Dec 12, 2018
    没仔细看题,,,,忽略我写的吧..
    sikariba
        3
    sikariba  
       Dec 12, 2018
    不确定你程序半天没出结果的原因,但是你的 inner loop 里面不应该再从头遍历 out_list 了,因为前面的元素已经遍历过了,你把 l 里的元素换个顺序,改成[ ("e", "f"), ("a", "b"), ("b", "c")]再运行试试,肯定有重复的
    doraemon1293
        4
    doraemon1293  
       Dec 12, 2018
    union find
    ```
    from collections import defaultdict


    class DSU:
    def __init__(self):
    self.weights = {}
    self.parents = {}

    def find(self, x):
    if x not in self.parents:
    self.parents[x] = x
    self.weights[x] = 1
    return x
    else:
    path = [x]
    while self.parents[path[-1]] != path[-1]:
    path.append(self.parents[path[-1]])
    root = path[-1]
    for node in path:
    self.parents[node] = root
    return root

    def union(self, elements):
    roots = [self.find(e) for e in elements]
    heaviest_root = max([(self.weights[root], root) for root in roots])[1]
    for rot in roots:
    if root != heaviest_root:
    self.weights[heaviest_root] += self.weights[root]
    self.parents[root] = heaviest_root


    def merger(A):
    """
    :type A: List[int]
    :rtype: int
    """
    dsu = DSU()
    for a in A:
    dsu.union(a)
    d=defaultdict(set)
    for k,x in dsu.parents.items():
    d[x].add(k)
    return list(d.values())
    ```
    Jex
        5
    Jex  
       Dec 12, 2018   1
    如果是 [("a", "b"), ("b", "c"), ("c", "d")] 呢?全合并成一个?
    1nakaELYBbsXbZxY
        6
    1nakaELYBbsXbZxY  
       Dec 12, 2018
    建立一个无向图,然后输出 里面所有没有连接在一起的子图就可以了,非常简单的。
    sikariba
        7
    sikariba  
       Dec 12, 2018
    你程序死循环的原因应该是这里
    ```
    for item_set in out_list:
    if len(temp_set & item_set) > 0:
    item_set.update(temp_set)
    else:
    out_list.append(temp_set)
    ```
    你不断迭代 out_list,然后又不断对它 append,就永远都遍历不完。常见错误了。
    rabbbit
        8
    rabbbit  
       Dec 12, 2018
    ihciah
        9
    ihciah  
       Dec 12, 2018   1
    这不是并查集嘛
    mskf
        11
    mskf  
       Dec 12, 2018
    并查集。。。
    hustlibraco
        12
    hustlibraco  
       Dec 12, 2018
    # coding=utf-8
    # input = [('a', 'b'), ('b', 'c'), ('e', 'f')]
    # output = [{'a', 'b', 'c'}, {'e', 'f'}]


    def union_set(items):
    ret = []
    mark = {}
    for item in items:
    for n in item:
    if n in mark:
    for m in item:
    ret[mark[n]].add(m)
    mark[m] = mark[n]
    break
    else:
    index = len(ret)
    s = set()
    for n in item:
    s.add(n)
    mark[n] = index
    ret.append(s)

    return ret


    if __name__ == "__main__":
    inputs = [('a', 'b'), ('b', 'c'), ('a', 'e', 'f')]
    print(union_set(inputs))
    zealinux
        13
    zealinux  
    OP
       Dec 13, 2018
    @aijam great,代码比较优雅

    新学到两个技能:

    `for-else`
    `集合可用作判断,空集合可当 False`
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