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]

我不知道你是否只希望列表中dicts的id是唯一的，但如果目标是有一组dict，其中所有键的值都是唯一的。你应该像这样使用元组键:

>>> L=[
...     {'id':1,'name':'john', 'age':34},
...    {'id':1,'name':'john', 'age':34}, 
...    {'id':2,'name':'hanna', 'age':30},
...    {'id':2,'name':'hanna', 'age':50}
...    ]
>>> len(L)
4
>>> L=list({(v['id'], v['age'], v['name']):v for v in L}.values())
>>>L
[{'id': 1, 'name': 'john', 'age': 34}, {'id': 2, 'name': 'hanna', 'age': 30}, {'id': 2, 'name': 'hanna', 'age': 50}]
>>>len(L)
3

希望它能帮助你或其他有顾虑的人....

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)

非常简单的选择:

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


D = dict()
for l in L: D[l['id']] = l
output = list(D.values())
print output

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]