Question

这是我当前的词典列表：

list1 = [{"one" : 1, "too": 2, "three": 3}, {"one" : 10, "too": 20, "three": 30} , {"one" : 1, "too": 26, "three": 35}, {"one" : 1, "too": 236, "three": 365}]

我想基于键“ one”从列表中删除所有重复字典。这意味着在键“一个”中该列表的另一个字典中的值是相同的，将其删除。我希望输出为：

list1 = [{"one" : 1, "too": 2, "three": 3}, {"one" : 10, "too": 20, "three": 30}]

Answer 1

您应该保留set键的可见值"one"。然后，您可以使用如下所示的简单循环或棘手的理解（利用or运算符的作用）：

list1 = [{"one" : 1, "too": 2, "three": 3}, {"one" : 10, "too": 20, "three": 30} , {"one" : 1, "too": 26, "three": 35}, {"one" : 1, "too": 236, "three": 365}]

seen = set()
list1[:] = [seen.add(d["one"]) or d for d in list1 if d["one"] not in seen]
# [{'one': 1, 'too': 2, 'three': 3}, {'one': 10, 'too': 20, 'three': 30}]

切片分配只是确保这是原始列表上的一个突变，例如如果要使用它来修改函数参数。无论如何，您都应该从头开始构建列表，因为从list重复删除是反模式的表现很差。

Answer 2

尝试：

list1 = [{"one" : 1, "too": 2, "three": 3}, {"one" : 10, "too": 20, "three": 30} , {"one" : 1, "too": 26, "three": 35}, {"one" : 1, "too": 236, "three": 365}]

l = [d for index,d in enumerate(list1) if d["one"] not in set(map(lambda x:x["one"],list1[:index]))]

结果：

[{'one': 1, 'too': 2, 'three': 3}, {'one': 10, 'too': 20, 'three': 30}]

Answer 3

这将为您提供唯一列表以及列表中的唯一键。

代码：

list1 = [{"one" : 1, "too": 2, "three": 3}, {"one" : 10, "too": 20, "three": 30} , {"one" : 1, "too": 26, "three": 35}, {"one" : 1, "too": 236, "three": 365}]
list2=[]
unikey=[]

for i in range(0,len(list1)):
    if list1[i]['one']:

        if list1[i]['one'] not in unikey:
            unikey.append(list1[i]['one'])
            list2.append(list1[i])

print('Unique List -> ',list2)
print('Unique Keys -> ', unikey)

结果：

Unique List ->  [{'one': 1, 'too': 2, 'three': 3}, {'one': 10, 'too': 20, 'three': 30}]
Unique Keys ->  [1, 10]

根据特定的键和值从列表中删除重复字典

3 个答案: