[英]How to merge multiple dictionaries from separate lists if they share any key-value pairs?
如果它们共享一个公用的键值对,如何合并多个列表中的字典?
例如,这是三个字典列表:
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'}]
所需结果:
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此函数仅将列表作为参数而不是键作为参数,因为我只想插入任意数量的字典列表,而不必担心哪个键名是重叠的(在这种情况下,它们将所有三个键名组合在一起的键名是“ category”和“ type”)。
我应该注意,索引无关紧要,因为它仅应基于公共元素。
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
对于此问题,将dict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字典,因此我们可以将每个元组视为图中的一个节点,将元组对视为边。 有了图形后,问题就会减少到找到图形的连接组件。
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,则可以使用例如pillmuncher的实现 :
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。
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