假设我有一个字典列表:
[
{'id': 1, 'name': 'john', 'age': 34},
{'id': 1, 'name': 'john', 'age': 34},
{'id': 2, 'name': 'hanna', 'age': 30},
]
如何获得唯一字典的列表(删除重复项)?
[
{'id': 1, 'name': 'john', 'age': 34},
{'id': 2, 'name': 'hanna', 'age': 30},
]
对象可以放入集合中。您可以使用对象而不是字典,如果需要,在所有set插入后转换回字典列表。例子
class Person:
def __init__(self, id, age, name):
self.id = id
self.age = age
self.name = name
my_set = {Person(id=2, age=3, name='Jhon')}
my_set.add(Person(id=3, age=34, name='Guy'))
my_set.add({Person(id=2, age=3, name='Jhon')})
# if needed convert to list of dicts
list_of_dict = [{'id': obj.id,
'name': obj.name,
'age': obj.age} for obj in my_set]
a = [
{'id':1,'name':'john', 'age':34},
{'id':1,'name':'john', 'age':34},
{'id':2,'name':'hanna', 'age':30},
]
b = {x['id']:x for x in a}.values()
print(b)
输出:
[{“年龄”:34岁“id”:1、“名称”:“约翰”},{“id”:“年龄”:30日2时,“名字”:“汉娜”}]
我总结了我最喜欢的尝试:
https://repl.it/@SmaMa/Python-List-of-unique-dictionaries
# ----------------------------------------------
# Setup
# ----------------------------------------------
myList = [
{"id":"1", "lala": "value_1"},
{"id": "2", "lala": "value_2"},
{"id": "2", "lala": "value_2"},
{"id": "3", "lala": "value_3"}
]
print("myList:", myList)
# -----------------------------------------------
# Option 1 if objects has an unique identifier
# -----------------------------------------------
myUniqueList = list({myObject['id']:myObject for myObject in myList}.values())
print("myUniqueList:", myUniqueList)
# -----------------------------------------------
# Option 2 if uniquely identified by whole object
# -----------------------------------------------
myUniqueSet = [dict(s) for s in set(frozenset(myObject.items()) for myObject in myList)]
print("myUniqueSet:", myUniqueSet)
# -----------------------------------------------
# Option 3 for hashable objects (not dicts)
# -----------------------------------------------
myHashableObjects = list(set(["1", "2", "2", "3"]))
print("myHashAbleList:", myHashableObjects)
这里有一个内存开销很小的实现,代价是不像其他实现那样紧凑。
values = [ {'id':2,'name':'hanna', 'age':30},
{'id':1,'name':'john', 'age':34},
{'id':1,'name':'john', 'age':34},
{'id':2,'name':'hanna', 'age':30},
{'id':1,'name':'john', 'age':34},]
count = {}
index = 0
while index < len(values):
if values[index]['id'] in count:
del values[index]
else:
count[values[index]['id']] = 1
index += 1
输出:
[{'age': 30, 'id': 2, 'name': 'hanna'}, {'age': 34, 'id': 1, 'name': 'john'}]