从无序字典列表中删除重复项

Question

考虑以下内容：

[
[
{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'bob', 'score': 99}
],
[
{'name': 'frank', 'score': 100},
{'name': 'fred', 'score': 19},
{'name': 'bob', 'score': 99}
],
[
{'name': 'bob', 'score': 99},
{'name': 'frank', 'score': 100},
{'name': 'fred', 'score': 19}
],
[
{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'stu', 'score': 69}
]
]

忽略每个列表中字典的顺序，如何删除重复项，使输出仅为其中两个列表：一个带有bob，一个带有stu？

输出类似于：

[
[
{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'bob', 'score': 99}
],
[
{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'stu', 'score': 69}
]
]

Answer 1

您可以尝试类似的

dict_list =   [[{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'bob', 'score': 99}],
[{'name': 'frank', 'score': 100},
{'name': 'fred', 'score': 19},
{'name': 'bob', 'score': 99}],
[{'name': 'bob', 'score': 99},
{'name': 'frank', 'score': 100},
{'name': 'fred', 'score': 19}],
[{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'stu', 'score': 69}]]
# create list of names you've seen before
name_lists = []
# create lists of unique lists
unique_lists = []
# loop over each list you have
for L in dict_list:
# get list of names
names = [i['name'] for i in L]
# check if you've seen this set of names before
if set(names) not in [set(n) for n in name_lists]:
print(names)
# save these names
name_lists.append(names)
# add this list to your list of unique names
unique_lists.append(L)

输出：

['fred', 'frank', 'bob']
['fred', 'frank', 'stu']

unique_lists输出：

[[{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'bob', 'score': 99}],
[{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'stu', 'score': 69}]]

请注意，此方法将只保存第一组唯一名称的分数，并在重复该组名称时丢弃分数。如果预期相同的名称可能具有不同的分数，则可能需要保存每一组唯一的分数。在这种情况下，您可以按照PacketLoss下面给出的方法：

name_lists = []
unique_lists = []

for di, d in enumerate(dict_list):
# get list of name, score tuples
r = [(i['name'], i['score']) for i in d]
# sort tuples alphabetically by name
r.sort(key=lambda tup: tup[0])
# check if these names and scores have been seen before
if r not in name_lists:
name_lists.append(r)
unique_lists.append(dict_list[di])

Answer 2

由于排序被关闭，一个简单的==将不匹配，我们可以通过收集数据、将其排序为元组列表并检查以前是否见过匹配来解决此问题。

data = [[{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'bob', 'score': 99}],
[{'name': 'frank', 'score': 100},
{'name': 'fred', 'score': 19},
{'name': 'bob', 'score': 99}],
[{'name': 'bob', 'score': 99},
{'name': 'frank', 'score': 100},
{'name': 'fred', 'score': 19}],
[{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'stu', 'score': 69}]]
seen = list()
result = list()
for idx, d in enumerate(data):
r = [(i['name'], i['score']) for i in d]
r.sort(key=lambda tup: tup[0])
if r not in seen:
seen.append(r)
result.append(data[idx])

使用这种方法，我们检查分数和名称是否完全匹配，这意味着如果重复中的一个分数更改为98，它将不再被视为重复。

输出：

[[{'name': 'fred', 'score': 19}, {'name': 'frank', 'score': 100}, {'name': 'bob', 'score': 99}], [{'name': 'fred', 'score': 19}, {'name': 'frank', 'score': 100}, {'name': 'stu', 'score': 69}]]

修改数据中分数的输出：

data = [[{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'bob', 'score': 99}],
[{'name': 'frank', 'score': 100},
{'name': 'fred', 'score': 19},
{'name': 'bob', 'score': 99}],
[{'name': 'bob', 'score': 98},
{'name': 'frank', 'score': 100},
{'name': 'fred', 'score': 19}],
[{'name': 'fred', 'score': 19},
{'name': 'frank', 'score': 100},
{'name': 'stu', 'score': 69}]]
[[{'name': 'fred', 'score': 19}, {'name': 'frank', 'score': 100}, {'name': 'bob', 'score': 99}], [{'name': 'bob', 'score': 98}, {'name': 'frank', 'score': 100}, {'name': 'fred', 'score': 19}], [{'name': 'fred', 'score': 19}, {'name': 'frank', 'score': 100}, {'name': 'stu', 'score': 69}]]