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},
]

当前回答

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)

其他回答

可能有更优雅的解决方案,但我认为最好添加一个更详细的解决方案,使其更容易遵循。这里假设没有唯一键,你有一个简单的k,v结构,并且你使用的python版本保证了列表顺序。这适用于原来的职位。

data_set = [
    {'id': 1, 'name': 'john', 'age': 34},
    {'id': 1, 'name': 'john', 'age': 34},
    {'id': 2, 'name': 'hanna', 'age': 30},
]

# list of keys
keys = [k for k in data_set[0]]

# Create a List of Lists of the values from the data Set
data_set_list = [[v for v in v.values()] for v in data_set]

# Dedupe
new_data_set = []
for lst in data_set_list:
    # Check if list exists in new data set
    if lst in new_data_set:
        print(lst)
        continue
    # Add list to new data set
    new_data_set.append(lst)

# Create dicts
new_data_set = [dict(zip(keys,lst)) for lst in new_data_set]    

print(new_data_set)

这里提到的所有答案都很好,但在一些答案中,如果字典项有嵌套的列表或字典,就会面临错误,所以我建议简单的答案

a = [str(i) for i in a]
a = list(set(a))
a = [eval(i) for i in a]

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'}]

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)

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())