如果共享任何键值对,如何合并来自单独列表的多个词典?

时间:2015-06-23 23:30:01

标签: python list dictionary pattern-matching itertools

如果共享一个共同的键值对,如何组合多个列表中的词典?

例如,以下是三个词典列表:

l1 = [{'fruit':'banana','category':'B'},{'fruit':'apple','category':'A'}]
l2 = [{'type':'new','category':'A'},{'type':'old','category':'B'}]
l3 = [{'order':'2','type':'old'},{'order':'1','type':'new'}]

期望的结果:

l = [{'fruit':'apple','category':'A','order':'1','type':'new'},{'fruit':'banana','category':'B','order':'2','type':'old'}]

棘手的部分是我希望这个函数只将列表作为参数而不是键,因为我只想插入任意数量的字典列表而不关心哪些键名是重叠的(在这种情况下,将所有三个组合在一起的键名称为'category'和'type')。

我应该注意索引应该无关紧要,因为它只应该基于共同的元素。

这是我的尝试:

def combine_lists(*args):
    base_list = args[0]
    L = []
    for sublist in args[1:]:
        L.extend(sublist)
    for D in base_list:
        for Dict in L:
            if any([tup in Dict.items() for tup in D.items()]): 
                D.update(Dict)
    return base_list

1 个答案:

答案 0 :(得分:1)

对于这个问题,将dicts视为元组列表很方便:

In [4]: {'fruit':'apple','category':'A'}.items()
Out[4]: [('category', 'A'), ('fruit', 'apple')]

由于我们希望连接共享键值对的dicts,我们可以考虑每个 元组作为图形中的节点,元组对作为边缘。一旦你有了图表 问题被简化为找到图形的连通组件。

使用networkx

import itertools as IT
import networkx as nx

l1 = [{'fruit':'apple','category':'A'},{'fruit':'banana','category':'B'}]
l2 = [{'type':'new','category':'A'},{'type':'old','category':'B'}]
l3 = [{'order':'1','type':'new'},{'order':'2','type':'old'}]

data = [l1, l2, l3]
G = nx.Graph()
for dct in IT.chain.from_iterable(data):
    items = list(dct.items())
    node1 = node1[0]
    for node2 in items:
        G.add_edge(node1, node22)

for cc in nx.connected_component_subgraphs(G):
    print(dict(IT.chain.from_iterable(cc.edges())))

产量

{'category': 'A', 'fruit': 'apple', 'type': 'new', 'order': '1'}
{'category': 'B', 'fruit': 'banana', 'type': 'old', 'order': '2'}

如果您希望删除networkx依赖关系,可以使用,例如pillmuncher's implementation

import itertools as IT

def connected_components(neighbors):
    """
    https://stackoverflow.com/a/13837045/190597 (pillmuncher)
    """
    seen = set()
    def component(node):
        nodes = set([node])
        while nodes:
            node = nodes.pop()
            seen.add(node)
            nodes |= neighbors[node] - seen
            yield node
    for node in neighbors:
        if node not in seen:
            yield component(node)

l1 = [{'fruit':'apple','category':'A'},{'fruit':'banana','category':'B'}]
l2 = [{'type':'new','category':'A'},{'type':'old','category':'B'}]
l3 = [{'order':'1','type':'new'},{'order':'2','type':'old'}]

data = [l1, l2, l3]
G = {}
for dct in IT.chain.from_iterable(data):
    items = dct.items()
    node1 = items[0]
    for node2 in items[1:]:
        G.setdefault(node1, set()).add(node2)
        G.setdefault(node2, set()).add(node1)

for cc in connected_components(G):
    print(dict(cc))

打印与上面相同的结果。