python - 如何使字典列表中的值唯一？

Question

我有一个 Python 字典列表，如下所示:

d = [{feature_a:1, feature_b:'Jul', feature_c:100}, {feature_a:2, feature_b:'Jul', feature_c:150}, {feature_a:1, feature_b:'Mar', feature_c:110}, ...]

我想要实现的是保持feature_a、_b 和_c 的唯一性。

例如，如果我们有 3 个具有相同 feature_a 和 _b 的条目，但有 3 个不同的 feature_c 100、100、150，那么运算后应该是100和150。

我怎样才能做到这一点？

============================================= =================更新:

好的，感谢 Anand 的出色回答，它工作得很好。不过，我还有一个问题。

假设我们有一个新的 feature_d 并且字典如下所示:

d = [{feature_a:1, feature_b:'Jul', feature_c:100, feature_d:'A'}, {feature_a:2, feature_b:'Jul', feature_c:150, feature_d: 'B'}, {feature_a:1, feature_b:'Mar', feature_c:110, feature_d:'F'}, ...]

我只想对 feature_a、_b 和 _c 进行重复数据删除，但将 feature_d 排除在外。我怎样才能做到这一点？

非常感谢。

Answer 1

如果初始d 列表的顺序不重要，您可以获取每个字典的.items() 并将其转换为frozenset()。 ，这是可哈希的，然后你可以将整个东西转换为 set() 或 frozenset() ，然后转换每个 frozenset() 返回词典。示例 -

uniq_d = list(map(dict, frozenset(frozenset(i.items()) for i in d)))

sets() 不允许重复元素。尽管您最终会失去列表的顺序。对于 Python 2.x，不需要 list(...)，因为 map() 返回一个列表。

---

示例/演示 -

>>> import pprint
>>> pprint.pprint(d)
[{'feature_a': 1, 'feature_b': 'Jul', 'feature_c': 100},
 {'feature_a': 2, 'feature_b': 'Jul', 'feature_c': 150},
 {'feature_a': 1, 'feature_b': 'Mar', 'feature_c': 110},
 {'feature_a': 1, 'feature_b': 'Jul', 'feature_c': 100},
 {'feature_a': 1, 'feature_b': 'Jul', 'feature_c': 150}]
>>> uniq_d = list(map(dict, frozenset(frozenset(i.items()) for i in d)))
>>> pprint.pprint(uniq_d)
[{'feature_a': 1, 'feature_b': 'Jul', 'feature_c': 100},
 {'feature_a': 1, 'feature_b': 'Jul', 'feature_c': 150},
 {'feature_a': 1, 'feature_b': 'Mar', 'feature_c': 110},
 {'feature_a': 2, 'feature_b': 'Jul', 'feature_c': 150}]

---

对于新需求——

However, what if that I have another feature_d but I only want to dedup feature_a, _b and _c

If two entries which have same feature_a, _b and _c, they are considered the same and duplicated, no matter what is in feature_d

执行此操作的一个简单方法是使用一个集合和一个新列表，仅将您需要的特征添加到集合中，并仅使用您需要的特征进行检查。示例 -

seen_set = set()
new_d = []
for i in d:
    if tuple([i['feature_a'],i['feature_b'],i['feature_c']]) not in seen_set:
        new_d.append(i)
        seen_set.add(tuple([i['feature_a'],i['feature_b'],i['feature_c']]))

示例/演示 -

>>> d = [{'feature_a':1, 'feature_b':'Jul', 'feature_c':100, 'feature_d':'A'},
...  {'feature_a':2, 'feature_b':'Jul', 'feature_c':150, 'feature_d': 'B'},
...  {'feature_a':1, 'feature_b':'Mar', 'feature_c':110, 'feature_d':'F'},
...  {'feature_a':1, 'feature_b':'Mar', 'feature_c':110, 'feature_d':'G'}]
>>> seen_set = set()
>>> new_d = []
>>> for i in d:
...     if tuple([i['feature_a'],i['feature_b'],i['feature_c']]) not in seen_set:
...         new_d.append(i)
...         seen_set.add(tuple([i['feature_a'],i['feature_b'],i['feature_c']]))
...
>>> pprint.pprint(new_d)
[{'feature_a': 1, 'feature_b': 'Jul', 'feature_c': 100, 'feature_d': 'A'},
 {'feature_a': 2, 'feature_b': 'Jul', 'feature_c': 150, 'feature_d': 'B'},
 {'feature_a': 1, 'feature_b': 'Mar', 'feature_c': 110, 'feature_d': 'F'}]