单个字典包含同一键下的多个列表。希望将同一键下的所有值组合到一个列表中

发布于 2025-02-11 00:11:16 字数 664 浏览 1 评论 0原文

我有一个单词，其中包含一个键下的多个列表 - 这是我的字典输出的一个示例：

print(dic)

输出：

{1: [[1, 2, 3, 4, 5, 6, 7],
  [5, 1, 4, 34, 3, 65, 0, 2],
  [2, 5, 4, 5, 7, 8, 4, 7]],
 2: [[1, 5, 4, 5, 0, 1, 4, 6],
  [2, 2, 3, 5, 6, 6, 3, 9]],
 3: [[12, 35, 42, 53, 70, 71, 74, 76],
  [6, 11, 16, 17, 38, 62, 66, 77]]}

我希望为每个键下的所有列表创建一个列表。因此，我要寻找的内容的输出示例将是这样的：

{1: [[1, 2, 3, 4, 5, 6, 7, 5, 1, 4, 34, 3, 65, 0, 2, 2, 5, 4, 5, 7, 8, 4, 7]],
 2: [[1, 5, 4, 5, 0, 1, 4, 6, 2, 2, 3, 5, 6, 6, 3, 9]],
 3: [[12, 35, 42, 53, 70, 71, 74, 76, 6, 11, 16, 17, 38, 62, 66, 77]]}

我尝试了几乎没有成功的理解，以及我在dic.items（）上迭代的循环。

原文

I have a single dictionary that contains multiple lists under a single key -
here is an example of the output of my dictionary:

print(dic)

Output:

{1: [[1, 2, 3, 4, 5, 6, 7],
  [5, 1, 4, 34, 3, 65, 0, 2],
  [2, 5, 4, 5, 7, 8, 4, 7]],
 2: [[1, 5, 4, 5, 0, 1, 4, 6],
  [2, 2, 3, 5, 6, 6, 3, 9]],
 3: [[12, 35, 42, 53, 70, 71, 74, 76],
  [6, 11, 16, 17, 38, 62, 66, 77]]}

I am looking to create a single list for all lists under each key.
So an output example of what i am looking for would look like this:

{1: [[1, 2, 3, 4, 5, 6, 7, 5, 1, 4, 34, 3, 65, 0, 2, 2, 5, 4, 5, 7, 8, 4, 7]],
 2: [[1, 5, 4, 5, 0, 1, 4, 6, 2, 2, 3, 5, 6, 6, 3, 9]],
 3: [[12, 35, 42, 53, 70, 71, 74, 76, 6, 11, 16, 17, 38, 62, 66, 77]]}

I have tried a dict comprehension to little success as well as a for loop where i iterate over dic.items().

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情仇皆在手 2025-02-18 00:11:16

您可以使用字典理解：

{key: [[item for sublist in value for item in sublist]] for key, value in data.items()}

此输出：

{
 1: [[1, 2, 3, 4, 5, 6, 7, 5, 1, 4, 34, 3, 65, 0, 2, 2, 5, 4, 5, 7, 8, 4, 7]],
 2: [[1, 5, 4, 5, 0, 1, 4, 6, 2, 2, 3, 5, 6, 6, 3, 9]],
 3: [[12, 35, 42, 53, 70, 71, 74, 76, 6, 11, 16, 17, 38, 62, 66, 77]]
}

You can use a dictionary comprehension:

{key: [[item for sublist in value for item in sublist]] for key, value in data.items()}

This outputs:

{
 1: [[1, 2, 3, 4, 5, 6, 7, 5, 1, 4, 34, 3, 65, 0, 2, 2, 5, 4, 5, 7, 8, 4, 7]],
 2: [[1, 5, 4, 5, 0, 1, 4, 6, 2, 2, 3, 5, 6, 6, 3, 9]],
 3: [[12, 35, 42, 53, 70, 71, 74, 76, 6, 11, 16, 17, 38, 62, 66, 77]]
}

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别闹i 2025-02-18 00:11:16

尝试使用 itertools.chain> itertools.chain.from_iterable/code>/code> ：

>>> import itertools

>>> {k : [list(itertools.chain.from_iterable(v))] for k,v in dct.items()}
{1: [[1, 2, 3, 4, 5, 6, 7, 5, 1, 4, 34, 3, 65, 0, 2, 2, 5, 4, 5, 7, 8, 4, 7]],
 2: [[1, 5, 4, 5, 0, 1, 4, 6, 2, 2, 3, 5, 6, 6, 3, 9]],
 3: [[12, 35, 42, 53, 70, 71, 74, 76, 6, 11, 16, 17, 38, 62, 66, 77]]}

Try with itertools.chain.from_iterable:

>>> import itertools

>>> {k : [list(itertools.chain.from_iterable(v))] for k,v in dct.items()}
{1: [[1, 2, 3, 4, 5, 6, 7, 5, 1, 4, 34, 3, 65, 0, 2, 2, 5, 4, 5, 7, 8, 4, 7]],
 2: [[1, 5, 4, 5, 0, 1, 4, 6, 2, 2, 3, 5, 6, 6, 3, 9]],
 3: [[12, 35, 42, 53, 70, 71, 74, 76, 6, 11, 16, 17, 38, 62, 66, 77]]}

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落叶缤纷 2025-02-18 00:11:16

使用 itertools.chain ：

from itertools import chain

out = {k: [list(chain.from_iterable(l))]
       for k,l in dic.items()}

输出：

{1: [[1, 2, 3, 4, 5, 6, 7, 5, 1, 4, 34, 3, 65, 0, 2, 2, 5, 4, 5, 7, 8, 4, 7]],
 2: [[1, 5, 4, 5, 0, 1, 4, 6, 2, 2, 3, 5, 6, 6, 3, 9]],
 3: [[12, 35, 42, 53, 70, 71, 74, 76, 6, 11, 16, 17, 38, 62, 66, 77]]}

Using itertools.chain:

from itertools import chain

out = {k: [list(chain.from_iterable(l))]
       for k,l in dic.items()}

Output:

{1: [[1, 2, 3, 4, 5, 6, 7, 5, 1, 4, 34, 3, 65, 0, 2, 2, 5, 4, 5, 7, 8, 4, 7]],
 2: [[1, 5, 4, 5, 0, 1, 4, 6, 2, 2, 3, 5, 6, 6, 3, 9]],
 3: [[12, 35, 42, 53, 70, 71, 74, 76, 6, 11, 16, 17, 38, 62, 66, 77]]}

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