Python 如何组合两个元组列表以创建嵌套字典?
我有两个元组列表,如:Python 如何组合两个元组列表以创建嵌套字典?,python,Python,我有两个元组列表,如: tin = [ ('a1', 'meow'), ('b1', 'woof'), ('c1', 'mooo'), ('d1', 'oink'), ] rop = [ ('b1', 'forest'), ('a1', 'home'), ('d1', 'shed'), ] 如何将它们合并到一个字典中,使结果如下所示: full = [
tin = [
('a1', 'meow'),
('b1', 'woof'),
('c1', 'mooo'),
('d1', 'oink'),
]
rop = [
('b1', 'forest'),
('a1', 'home'),
('d1', 'shed'),
]
如何将它们合并到一个字典中,使结果如下所示:
full = [
{'a1' : {'sound': 'meow',
'place': 'home'}
{'b1' : {'sound': 'woof',
'place': 'forest'}
{'c1' : {'sound': 'mooo',
'place': None}
{'d1' : {'sound': 'oink',
'place': 'shed'}
]
我让它这样工作:
my_dict = {}
for k, v in tin :
if not my_dict.get(k):
my_dict[k] = {}
my_dict[k]['sound'] = v
else:
my_dict[k]['sound'] = v
for k, v in rop:
if not my_dict.get(k):
my_dict[k] = {}
my_dict[k]['place'] = v
else:
my_dict[k]['place'] = v
{
'a1': {'place': 'home', 'sound': 'meow'},
'b1': {'place': 'forest', 'sound': 'woof'},
'c1': {'place': None, 'sound': 'mooo'},
'd1': {'place': 'shed', 'sound': 'oink'}
}
但是它非常冗长,我认为应该有更具python风格的东西。使用一个简单的迭代 演示: 输出:
{'a1': {'sound': 'meow', 'place': 'home'}, 'b1': {'sound': 'woof', 'place': 'forest'}, 'c1': {'sound': 'mooo', 'place': None}, 'd1': {'sound': 'oink', 'place': 'shed'}}
您可以使用itertools.groupby:
输出:
{'a1': {'sound': 'meow', 'place': 'home'}, 'b1': {'sound': 'woof', 'place': 'forest'}, 'c1': {'sound': 'mooo', 'place': None}, 'd1': {'sound': 'oink', 'place': 'shed'}}
您确定要将全部内容作为列表吗?如果你说它应该是一个dict,那么它应该是这样的:
my_dict = {}
for k, v in tin :
if not my_dict.get(k):
my_dict[k] = {}
my_dict[k]['sound'] = v
else:
my_dict[k]['sound'] = v
for k, v in rop:
if not my_dict.get(k):
my_dict[k] = {}
my_dict[k]['place'] = v
else:
my_dict[k]['place'] = v
{
'a1': {'place': 'home', 'sound': 'meow'},
'b1': {'place': 'forest', 'sound': 'woof'},
'c1': {'place': None, 'sound': 'mooo'},
'd1': {'place': 'shed', 'sound': 'oink'}
}
由以下代码生成:
# note: converting tin and rop to dict:
tin = dict(tin)
rop = dict(rop)
full = {}
for k in set(tin.keys()) | set(rop.keys()):
full[k] = {'sound': tin.get(k, None), 'place': rop.get(k, None)}
顺便说一下,如果你真的想要一个目录列表,请使用以下命令:
full = []
for k in set(tin.keys()) | set(rop.keys()):
full.append({k: {'sound': tin.get(k, None), 'place': rop.get(k, None)}})