python - 如何在 python 中仅展平列表列表的第二级(稍后将其变成字典)？

Question

我有一个列表列表，稍后我想把它变成一本字典。问题是列表的列表非常不规则:对于一个键，存在 1 到 4 个列表，其中的数字属于该键。原始数据集存储在 json 中。这是一个示例数据集:

data = [36146779,
[17628,35633, 2847385, 71393, 41814],[51068348,49722,3255134,66598],[103475099, 1337536, 1136863360,257],
22971125,
[230806,116805,118456,9031, 3573662],[719279707,299836,40722,35134,668],[1337536, 1136863360,257],
111125168,
[719279707,299836,40722,35138],[17628,35633, 2847385],
71280747,
[806,116805,11845],[17628,35633, 2847385]]

我能够完全展平列表列表，但我无法将这个平面列表变成具有给定键的字典

ex_eco = ["36146779","22971125","111125168","71280747"]

(我确信也可以直接将第一个嵌套列表转成字典，但我找不到解决方案，所以我尝试了这种方式)

def flatten(l):
  out = []
  for item in l:
    if isinstance(item, (list, tuple)):
      out.extend(flatten(item))
    else:
      out.append(item)
  return out

flattened_eco = flatten(data)

print(flattened_eco[0:100])

我得到的是一个平面列表:

[36146779, 17628, 35633, 2847385, 71393, 41814, 51068348, 49722, 3255134, 66598, 103475099, 1337536, 1136863360, 257, 22971125, 230806, 116805, 118456, 9031, 3573662, 719279707, 299836, 40722, 35134, 668, 1337536, 1136863360, 257, 111125168, 719279707, 299836, 40722, 35138, 17628, 35633, 2847385, 71280747, 806, 116805, 11845, 17628, 35633, 2847385]

我需要的是这样的:

[36146779,
[17628,35633, 2847385, 71393,41814,51068348,49722,3255134,66598,103475099,1337536, 1136863360,257],
22971125,
[230806,116805,118456,9031,573662,719279707,299836,40722,35134,668,1337536, 1136863360,257],
111125168,
[719279707,299836,40722,35138,17628,35633, 2847385],
71280747,
[806,116805,11845,17628,35633, 2847385]
]

Answer 1

import itertools

def flatten(data):
    flattened = []
    for key, value in itertools.groupby(data, type):
        if key == int:
            flattened.append(next(value))
        else:
            flattened.append(list(itertools.chain.from_iterable(value)))
    return flattened

例子

>>> data = [36146779,
            [17628,35633, 2847385, 71393, 41814],[51068348,49722,3255134,66598],[103475099, 1337536, 1136863360,257],
            22971125,
            [230806,116805,118456,9031, 3573662],[719279707,299836,40722,35134,668],[1337536, 1136863360,257],
            111125168,
            [719279707,299836,40722,35138],[17628,35633, 2847385],
            71280747,
            [806,116805,11845],[17628,35633, 2847385]]

>>> flatten(data)
[36146779, 
 [17628, 35633, 2847385, 71393, 41814, 51068348, 49722, 3255134, 66598, 103475099, 1337536, 1136863360, 257],
 22971125,
 [230806, 116805, 118456, 9031, 3573662, 719279707, 299836, 40722, 35134, 668, 1337536, 1136863360, 257],
 111125168,
 [719279707, 299836, 40722, 35138, 17628, 35633, 2847385],
 71280747,
 [806, 116805, 11845, 17628, 35633, 2847385]]

要更进一步并创建字典，只需将函数的最后一行更改为 dict comprehension

def dictify(data):
    flattened = []
    for key, value in itertools.groupby(data, type):
        if key == int:
            flattened.append(next(value))
        else:
            flattened.append(list(itertools.chain.from_iterable(value)))
    return {key:value for key,value in zip(flattened[::2], flattened[1::2])}

>>> dictify(data)
{36146779: [17628, 35633, 2847385, 71393, 41814, 51068348, 49722, 3255134, 66598, 103475099, 1337536, 1136863360, 257],
 22971125: [230806, 116805, 118456, 9031, 3573662, 719279707, 299836, 40722, 35134, 668, 1337536, 1136863360, 257],
 111125168: [719279707, 299836, 40722, 35138, 17628, 35633, 2847385],
 71280747: [806, 116805, 11845, 17628, 35633, 2847385]}