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]
这里有很多答案,所以让我再补充一个:
import json
from typing import List
def dedup_dicts(items: List[dict]):
dedupped = [ json.loads(i) for i in set(json.dumps(item, sort_keys=True) for item in items)]
return dedupped
items = [
{'id': 1, 'name': 'john', 'age': 34},
{'id': 1, 'name': 'john', 'age': 34},
{'id': 2, 'name': 'hanna', 'age': 30},
]
dedup_dicts(items)
这里有一个相当紧凑的解决方案,尽管我怀疑不是特别有效(委婉地说):
>>> ds = [{'id':1,'name':'john', 'age':34},
... {'id':1,'name':'john', 'age':34},
... {'id':2,'name':'hanna', 'age':30}
... ]
>>> map(dict, set(tuple(sorted(d.items())) for d in ds))
[{'age': 30, 'id': 2, 'name': 'hanna'}, {'age': 34, 'id': 1, 'name': 'john'}]
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)
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”}]
这是我找到的解决方案:
usedID = []
x = [
{'id':1,'name':'john', 'age':34},
{'id':1,'name':'john', 'age':34},
{'id':2,'name':'hanna', 'age':30},
]
for each in x:
if each['id'] in usedID:
x.remove(each)
else:
usedID.append(each['id'])
print x
基本上你检查ID是否存在于列表中,如果存在,删除字典,如果不存在,将ID追加到列表中