通过嵌套字典键获取唯一列表项

Question

我将列表作为字典值嵌套在另一个名为 data 的字典中。我一直在尝试找到一种快速的方法来从特定的嵌套键中获取所有唯一列表项，例如key1 或key2。

我想出了以下功能，这似乎不是很有效。有什么想法可以加快速度并变得更加 Pythonic 吗？

Python 函数

def get_uniq_by_value(data, val_name):
    results = []
    for key, value in data.iteritems():
        for item in value[val_name]:
            if item not in results:
                results.append(item)
    return results

示例数据

data = {
"top1": {
    "key1": [
        "there is no spoon", "but dictionaries are hard",
    ],
    "key2": [
        "mad max fury road was so good",
    ]
},
"top2": {
    "key1": [
        "my item", "foo bar"
    ],
    "key2": [
        "blah", "more junk"
    ]
},

Answer 1

如果顺序无关紧要，您可以使用set / set comprehension 来获得所需的结果 -

def get_uniq_by_value(data, val_name):
    return {val for value in data.values() for val in value.get(val_name,[])}

如果您想要一个列表作为结果，您可以在集合推导上使用list() 在返回之前将结果集转换为列表。

演示 -

>>> def get_uniq_by_value(data, val_name):
...     return {val for value in data.values() for val in value.get(val_name,[])}
...
>>> data = {
... "top1": {
...     "key1": [
...         "there is no spoon", "but dictionaries are hard",
...     ],
...     "key2": [
...         "mad max fury road was so good",
...     ]
... },
... "top2": {
...     "key1": [
...         "my item", "foo bar"
...     ],
...     "key2": [
...         "blah", "more junk"
...     ]
... }}
>>> get_uniq_by_value(data,"key1")
{'but dictionaries are hard', 'my item', 'foo bar', 'there is no spoon'}

---

如下面的 cmets 所示，如果顺序很重要并且 data 已经是 OrderedDict 的 collections.OrderedDict，则可以使用新的 OrderedDict ，并将列表中的元素添加为键，@ 987654329@ 将避免任何重复并保留添加键的顺序。

您也可以使用OrderedDict.fomkeys 在一行中完成此操作，如 cmets 中所示。示例 -

from collections import OrderedDict
def get_uniq_by_value(data, val_name):
    return list(OrderedDict.fromkeys(val for value in data.values() for val in value.get(val_name,[])))

请注意，这仅适用于data 是嵌套的 OrderedDict，否则data 的元素将不会以任何特定的顺序开始。

演示 -

>>> from collections import OrderedDict
>>> data = OrderedDict([
... ("top1", OrderedDict([
...     ("key1", [
...         "there is no spoon", "but dictionaries are hard",
...     ]),
...     ("key2", [
...         "mad max fury road was so good",
...     ])
... ])),
... ("top2", OrderedDict([
...     ("key1", [
...         "my item", "foo bar"
...     ]),
...     ("key2", [
...         "blah", "more junk"
...     ])
... ]))])
>>>
>>> def get_uniq_by_value(data, val_name):
...     return list(OrderedDict.fromkeys(val for value in data.values() for val in value.get(val_name,[])))
...
>>> get_uniq_by_value(data,"key1")
['there is no spoon', 'but dictionaries are hard', 'my item', 'foo bar']