如果初始
d列表的顺序不重要，则可以将
.items()每个词典的转换为
frozenset()，然后将其转换为可哈希的，然后将整个内容转换为
set()或
frozenset()，然后再将其转换
frozenset()回词典。范例-
uniq_d = list(map(dict, frozenset(frozenset(i.items()) for i in d)))
sets()不允许重复的元素。虽然您最终会失去列表的顺序。对于Python 2.x，
list(...)不需要，因为会
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}]对于新要求-
但是，如果我还有另一个feature_d但我只想删除feature_a，_b和_c怎么办？
如果两个具有相同feature_a，_b和_c的条目，则无论feature_d中的内容如何，都将它们视为相同且重复的条目
一种简单的方法是使用集合和新列表，仅将所需的功能添加到集合中，然后仅使用所需的功能进行检查。范例-
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'}]
