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)

让我加上我的。

对目标字典进行排序，使{'a': 1， 'b': 2}和{'b': 2， 'a': 1}不会被区别对待 将其设置为json格式 通过set重复数据删除(因为set不适用于字典) 同样，通过json.loads将其转换为dict

import json

[json.loads(i) for i in set([json.dumps(i) for i in [dict(sorted(i.items())) for i in target_dict]])]

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)

扩展John La Rooy (Python -唯一字典列表)的答案，使其更加灵活:

def dedup_dict_list(list_of_dicts: list, columns: list) -> list:
    return list({''.join(row[column] for column in columns): row
                for row in list_of_dicts}.values())

调用函数:

sorted_list_of_dicts = dedup_dict_list(
    unsorted_list_of_dicts, ['id', 'name'])

在python 3中，简单的技巧，但基于唯一字段(id):

data = [ {'id': 1}, {'id': 1}]

list({ item['id'] : item for item in data}.values())