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"]

（我确定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法）

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

示例

>>> 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]]

要更进一步创建字典，只需将函数的最后一行更改为 dict 理解

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]}

Answer 2

从您的输入来看，您似乎不需要递归函数，只需使用迭代：

result = {}
keys = []
for d in data:
  if type(d) == int:
    result[d] = []
    keys.append(d)
  else:
    result[keys[-1]] += d

使用提供的数据输出：

{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]}

Answer 3

extend()- 用于合并两个列表。

例如

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]]

new_dict = {}
temp=None
for x in data:
    if not isinstance(x, list):
        new_dict[x] = []
        temp = x
    else:
        new_dict[temp].extend(x)

print(new_dict)

O/P:

{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]}

Answer 4

一个版本使用 itertools.groupby:

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]]

from itertools import groupby

def generate(d):
    for v, g in groupby(d, lambda k: isinstance(k, (tuple, list))):
        if not v:
            yield from g
        else:
            yield sum(g, [])

from pprint import pprint
pprint([*generate(data)], width=180)

打印：

[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]]

如何在 python 中仅展平列表列表的第二级（稍后将其变成字典）？

How to only flatten the second level of a list of lists (to later turn it into a dictionary) in python?

python

dictionary

list

flatten