如何在python中统一dict列表

Question

我有一个清单:

d = [{'x':1, 'y':2}, {'x':3, 'y':4}, {'x':1, 'y':2}]

{'x':1, 'y':2} 不止一次我想从列表中删除它.我的结果应该是:

 d = [{'x':1, 'y':2}, {'x':3, 'y':4} ]

注意: list(set(d))这里没有工作抛出错误.

Answer 1

如果你的价值是可以清洗的,这将有效:

>>> [dict(y) for y in set(tuple(x.items()) for x in d)]
[{'y': 4, 'x': 3}, {'y': 2, 'x': 1}]

编辑:

我尝试了它没有重复,它似乎工作正常

>>> d = [{'x':1, 'y':2}, {'x':3, 'y':4}]
>>> [dict(y) for y in set(tuple(x.items()) for x in d)]
[{'y': 4, 'x': 3}, {'y': 2, 'x': 1}]

和

>>> d = [{'x':1,'y':2}]
>>> [dict(y) for y in set(tuple(x.items()) for x in d)]
[{'y': 2, 'x': 1}]

Answer 2

Dicts不可清洗,所以你不能把它们放在一套.一种相对有效的方法是将对(key, value)转换为元组并对这些元组进行散列(随意消除中间变量):

tuples = tuple(set(d.iteritems()) for d in dicts)
unique = set(tuples)
return [dict(pairs) for pairs in unique]

如果值并不总是可以使用,则根本不可能使用集合,并且您必须使用in每元素检查的O(n ^ 2)方法.

Answer 3

避免这整个问题,而是使用namedtuples

from collections import namedtuple

Point = namedtuple('Point','x y'.split())
better_d = [Point(1,2), Point(3,4), Point(1,2)]
print set(better_d)

Answer 4

一个简单的循环：

tmp=[]

for i in d:
    if i not in tmp:
        tmp.append(i)        
tmp
[{'x': 1, 'y': 2}, {'x': 3, 'y': 4}]