Question

所以我需要一种用于在python中创建字典列表的最佳方法。

所以我有一个看起来像这样的列表：

'key'=>env(APP_KEY,'base64key')

所以我需要这样的输出：

[
   {'name': 'John', 'hobbies': ['Reading', 'Swimming']},
   {'name': 'Gina', 'hobbies': ['Skating', 'Cooking']},
   {'name': 'John', 'hobbies': ['Gardening', 'Swimming']}
]

如您所见，我需要为每个名称创建一组爱好，并且确实创建一个唯一的字典列表。

这是我尝试过的：

[
   {'name': 'John', 'hobbies': ['Reading', 'Swimming', 'Gardening']},
   {'name': 'Gina', 'hobbies': ['Skating', 'Cooking']},
]

但是并不需要创建一个集

有人可以以最佳方式帮助我吗？

谢谢。

Answer 1

如果您同意将输出的结构从名称更改为爱好集，而只是将其更改为词典，则可以在线性时间内完成（忽略边缘情况，即很多哈希冲突）：

from collections import defaultdict

data = [
    {'name': 'John', 'hobbies': ['Reading', 'Swimming']},
    {'name': 'Gina', 'hobbies': ['Skating', 'Cooking']},
    {'name': 'John', 'hobbies': ['Gardening', 'Swimming']}
]

output = defaultdict(set)

for d in data:
    output[d['name']].update(d['hobbies'])

print(output)
# defaultdict(<class 'set'>, {'John': {'Reading', 'Swimming', 'Gardening'},
#                             'Gina': {'Cooking', 'Skating'}})

如果您坚持使用字典列表，您仍然可以实现几乎线性时间（列表查找仍为O（n）），但具有将索引映射到名称的逻辑：

data = [
        {'name': 'John', 'hobbies': ['Reading', 'Swimming']},
        {'name': 'Gina', 'hobbies': ['Skating', 'Cooking']},
        {'name': 'John', 'hobbies': ['Gardening', 'Swimming']}
    ]

output = []
names_to_indices = {}
for d in data:
    if d['name'] not in names_to_indices:
        output.append({'name': d['name'], 'hobbies': d['hobbies']})
        names_to_indices[d['name']] = len(output) - 1
    else:
        index = names_to_indices[d['name']]
        for hobbie in d['hobbies']:
            if hobbie not in output[index]['hobbies']:
                output[index]['hobbies'].append(hobbie)
print(output)
# [{'name': 'John', 'hobbies': ['Reading', 'Swimming', 'Gardening']},
#  {'name': 'Gina', 'hobbies': ['Skating', 'Cooking']}]

如果您同意将业余爱好定为一组，则可以将其设为真正的线性时间（同样，如果我们忽略过多的哈希冲突的可能性），

data = [
        {'name': 'John', 'hobbies': ['Reading', 'Swimming']},
        {'name': 'Gina', 'hobbies': ['Skating', 'Cooking']},
        {'name': 'John', 'hobbies': ['Gardening', 'Swimming']}
    ]

output = []
names_to_indices = {}
for d in data:
    if d['name'] not in names_to_indices:
        output.append({'name': d['name'], 'hobbies': set(d['hobbies'])})
        names_to_indices[d['name']] = len(output) - 1
    else:
        index = names_to_indices[d['name']]
        output[index]['hobbies'].update(d['hobbies'])
print(output)
# [{'name': 'John', 'hobbies': {'Gardening', 'Swimming', 'Reading'}},
#  {'name': 'Gina', 'hobbies': {'Skating', 'Cooking'}}]

Answer 2

只需构造一个中间默认字典，即可使您在线性时间内完成此操作。最后转换回所需的结构。

inp = [
   {'name': 'John', 'hobbies': ['Reading', 'Swimming']},
   {'name': 'Gina', 'hobbies': ['Skating', 'Cooking']},
   {'name': 'John', 'hobbies': ['Gardening', 'Swimming']}
]

from collections import defaultdict
temp = defaultdict(set)
for d in inp:
    temp[d['name']].update(d['hobbies'])

result = [{'name':k, 'hobbies': list(v)} for k, v in temp.items()]

输出：

[{'name': 'John', 'hobbies': ['Gardening', 'Reading', 'Swimming']},
 {'name': 'Gina', 'hobbies': ['Cooking', 'Skating']}]

创建唯一词典列表的最佳方法

2 个答案: