列表理解，检查项目是否唯一

Question

我正在尝试编写一个列表理解语句，它只会添加当前不包含在列表中的项目。有没有办法检查当前正在构建的列表中的当前项目？这是一个简短的例子：

输入

{
    "Stefan" : ["running", "engineering", "dancing"],
    "Bob" : ["dancing", "art", "theatre"],
    "Julia" : ["running", "music", "art"]
}

输出

["running", "engineering", "dancing", "art", "theatre", "music"]

不使用列表理解的代码

output = []
for name, hobbies in input.items():
    for hobby in hobbies:
        if hobby not in output:
            output.append(hobby)

我的尝试

[hobby for name, hobbies in input.items() for hobby in hobbies if hobby not in ???]

Answer 1

使用一套：

dict = {
    "Stefan" : ["running", "engineering", "dancing"],
    "Bob" : ["dancing", "art", "theatre"],
    "Julia" : ["running", "music", "art"]
}

myset = set()
for _, value in dict.items():
    for item in value:
        myset.add(item)

print(myset)

Answer 2

这个怎么样：

set(dict['Bob']+dict['Stefan']+dict['Julia'])
>>> set(['art', 'theatre', 'dancing', 'engineering', 'running', 'music'])

或者更好：

dict = {
    "Stefan" : ["running", "engineering", "dancing"],
    "Bob" : ["dancing", "art", "theatre"],
    "Julia" : ["running", "music", "art"]
}

list_ = []
for y in dict.keys():
    list_ = list_ + dict[y]
list_ = set(list_)
>>> list_
set(['art', 'theatre', 'dancing', 'engineering', 'running', 'music'])

您可以将 list 函数应用于 list_，就像 list(list_) 应用于 return 列表而不是集合。

Answer 3

您可以使用 set 并设置理解：

{hobby for name, hobbies in input.items() for hobby in hobbies}

作为，我们这里不用name，所以可以用item.values()代替：

{hobby for hobbies in input.values() for hobby in hobbies}

如果你真的需要一个列表作为结果，你可以这样做（但请注意，通常你可以毫无问题地使用集合）：

list({hobby for hobbies in input.values() for hobby in hobbies})

Answer 4

正如this answer建议的那样：您可以使用唯一性过滤器：

def f7(seq):
    seen = set()
    seen_add = seen.add
    return [x for x in seq if not (x in seen or seen_add(x))]

并调用：

>>> f7(hobby for name, hobbies in input.items() for hobby in hobbies)
['running', 'engineering', 'dancing', 'art', 'theatre', 'music']

我会单独实施 唯一性过滤器，因为设计规则说 "different things should be handled by different classes/methods/components/whatever"。此外，如有必要，您可以简单地重复使用此方法。

另一个优点是 - 如 linked answer 中所写 - 项目的顺序得以保留。对于某些应用程序，这可能是必要的。

Answer 5

集和词典是你的朋友：

from collections import OrderedDict
from itertools import chain # 'flattens' collection of iterables

data = {
    "Stefan" : ["running", "engineering", "dancing"],
    "Bob" : ["dancing", "art", "theatre"],
    "Julia" : ["running", "music", "art"]
}

# using set is the easiest way, but sets are unordered:
print {hobby for hobby in chain.from_iterable(data.values())}
# output:
# set(['art', 'theatre', 'dancing', 'engineering', 'running', 'music'])


# or use OrderedDict if you care about ordering:
print OrderedDict(
        (hobby, None) for hobby in chain.from_iterable(data.values())
    ).keys()
# output:
# ['dancing', 'art', 'theatre', 'running', 'engineering', 'music']

Answer 6

如果你真的想要一个 listcomp 并且只需要一个 list-comp，你可以这样做

>>> s = []
>>> [s.append(j)  for i in d.values() for j in i if j not in s]
[None, None, None, None, None, None]
>>> s
['dancing', 'art', 'theatre', 'running', 'engineering', 'music']

此处，s 是副作用的结果，d 是您的原始词典。这里的独特优势是 您可以保留顺序，这与此处的大多数其他答案不同。

Note: This a bad way as it exploits the list-comp and the result is a side effect. Don't do it as a practice, This answer is just to show you that you can achieve it using a list comp alone

Answer 7

列表理解不太适合解决这个问题。我认为集合理解会更好，但由于已经在另一个答案中显示了这一点，我将展示一种使用紧凑的单行代码解决此问题的方法：

list(set(sum(hobbies_dict.values(), [])))

另一个有趣的解决方案是使用按位或运算符作为集合的并集运算符：

from operator import or_
from functools import reduce # Allowed, but unnecessary in Python 2.x
list(reduce(or_, map(set, hobbies_dict.values())))

Or（无意的双关语，我发誓），而不是使用按位或运算符，只需使用 set.union 并将您的值的解压缩集合映射传递给它。无需导入 or_ 和 reduce！这个想法的灵感来自 .

list(set.union(*map(set, hobbies_dict.values())))

Answer 8

还有另一种写法，它更能描述您实际正在做的事情，并且不需要嵌套（双 for）理解：

output = set.union(*[set(hobbies) for hobbies in input_.values()])

当您表示输入在概念上更合理时，这会变得更好，即使用 set 表示每个人的爱好（因为那里也不应该重复）：

input_ = {
    "Stefan" : {"running", "engineering", "dancing"},
    "Bob" : {"dancing", "art", "theatre"}, 
    "Julia" : {"running", "music", "art"}
}

output = set.union(*input_.values())

列表理解，检查项目是否唯一

List comprehension, check if item is unique

python

dictionary

list-comprehension

list