将列表更改为字典

Question

如何更改这样的列表：

[[0, 'Ealing Broadway', 103.89],
 [0, 'Notting Hill Gate', 103.89],
 [0, 'Mile End', 103.89],
 [1, 'Ealing Broadway', 59.089999999999996],
 [2, 'Notting Hill Gate', 40.279999999999994],
 [3, 'Mile End', 68.86999999999999]]

像字典一样

{0:{'length':103.89,'interchange':['Ealing Broadway','Notting Hill Gate','Mile End']},
1:{'length':59.089999999999996,'interchange':['Ealing Broadway']},
2:{'length':40.279999999999994,'interchange':['Notting Hill Gate']},
3:{'length':68.86999999999999,'interchange':['Mile End']}}

谢谢

I am trying to start with:

d2 = defaultdict(list)
for k, v in all_info:
    d2[k].append(v)

with_length=dict((k,list(v)) for k,v in d2.iteritems())
with_length

但它不起作用，我正在努力从哪里开始。

Answer 1

与 Majora 类似的答案，但首先使用 groupby。没有错误的查找，但可能需要事先进行排序。

from itertools import groupby

lst = [[0, 'Ealing Broadway', 103.89],
    [0, 'Notting Hill Gate', 103.89],
    [0, 'Mile End', 103.89],
    [1, 'Ealing Broadway', 59.089999999999996],
    [2, 'Notting Hill Gate', 40.279999999999994],
    [3, 'Mile End', 68.86999999999999]]

new_list = []
for key, group in groupby(lst, lambda x: x[0]):
    new_list.append(list(group))

main_dict = {}
for item in new_list:
    main_dict[item[0][0]] = {'length': item[0][2], 'interchange': [stn[1] for stn in item]}

Answer 2

这里有一个具体的例子来说明你将如何做到这一点：

l = [[0, 'Ealing Broadway', 103.89],
     [0, 'Notting Hill Gate', 103.89],
     [0, 'Mile End', 103.89],
     [1, 'Ealing Broadway', 59.089999999999996],
     [2, 'Notting Hill Gate', 40.279999999999994],
     [3, 'Mile End', 68.86999999999999]]

d = {}

for pair in l:
    if pair[0] not in d.keys():
        d[pair[0]] = { 'interchange': [] }

    d[pair[0]]['length'] = pair[2]
    d[pair[0]]['interchange'].append(pair[1])

这是假设您在向d[0] 添加元素时要覆盖d['length']。

Answer 3

b = {}
for i in a:
    if b.has_key(i[0]):
        b[i[0]]['interchange'].append(i[1])
    else:
        b[i[0]] = {'length': i[2], 'interchange': [i[1]]}

Answer 4

这是一种需要两次通过的方法。它的优点是易于理解。

import pprint

if __name__ == '__main__':
    rows = [
        [0, 'Ealing Broadway', 103.89],
        [0, 'Notting Hill Gate', 103.89],
        [0, 'Mile End', 103.89],
        [1, 'Ealing Broadway', 59.089999999999996],
        [2, 'Notting Hill Gate', 40.279999999999994],
        [3, 'Mile End', 68.86999999999999]]

    print('First Pass')
    d = {}
    for key, interchange, length in rows:
        inner_dict = d.setdefault((key, length), {})
        interchanges = inner_dict.setdefault('interchange', [])
        interchanges.append(interchange)

    pprint.pprint(d)

    print('=' * 72)
    print('Second Pass')
    d2 = {}
    for (key, length), v in d.items():
        v['length'] = length
        d2[key] = v

    pprint.pprint(d2)

输出

First Pass
{(0, 103.89): {'interchange': ['Ealing Broadway',
                               'Notting Hill Gate',
                               'Mile End']},
 (1, 59.089999999999996): {'interchange': ['Ealing Broadway']},
 (2, 40.279999999999994): {'interchange': ['Notting Hill Gate']},
 (3, 68.86999999999999): {'interchange': ['Mile End']}}
========================================================================
Second Pass
{0: {'interchange': ['Ealing Broadway', 'Notting Hill Gate', 'Mile End'],
     'length': 103.89},
 1: {'interchange': ['Ealing Broadway'], 'length': 59.089999999999996},
 2: {'interchange': ['Notting Hill Gate'], 'length': 40.279999999999994},
 3: {'interchange': ['Mile End'], 'length': 68.86999999999999}}

讨论

• 在第一遍中，我使用第一列和最后一列作为字典的键。这个字典的值是另一个字典 (inner_dict)
• 在第二遍中，我将键和值调整为最终形式。
• 此解决方案可能不是最有效或最优雅的，但我希望它易于理解

Answer 5

请将我的回答视为 Pandas（强大的 Python 数据分析工具包）模块方法的演示。

我很确定您是否想要快速处理大量数据 - pandas 是您的工具...

import pandas as pd

data = [[0, 'Ealing Broadway', 103.89],
        [0, 'Notting Hill Gate', 103.89],
        [0, 'Mile End', 103.89],
        [1, 'Ealing Broadway', 59.089999999999996],
        [2, 'Notting Hill Gate', 40.279999999999994],
        [3, 'Mile End', 68.86999999999999]
       ]

# create pandas DF
df = pd.DataFrame(data, columns=['route','interchange','length'])

原始 DF：

In [235]: df
Out[235]:
   route        interchange  length
0      0    Ealing Broadway  103.89
1      0  Notting Hill Gate  103.89
2      0           Mile End  103.89
3      1    Ealing Broadway   59.09
4      2  Notting Hill Gate   40.28
5      3           Mile End   68.87

让我们对数据进行分组：

In [239]: df.groupby(['route','length'])['interchange'].apply(lambda x: x.tolist()).reset_index()
Out[239]:
   route  length                                     interchange
0      0  103.89  [Ealing Broadway, Notting Hill Gate, Mile End]
1      1   59.09                               [Ealing Broadway]
2      2   40.28                             [Notting Hill Gate]
3      3   68.87                                      [Mile End]

我们也可以将其转换为字典列表：

In [240]: df.groupby(['route','length'])['interchange'].apply(lambda x: x.tolist()).reset_index().to_dict('record')
Out[240]:
[{'interchange': ['Ealing Broadway', 'Notting Hill Gate', 'Mile End'],
  'length': 103.89,
  'route': 0},
 {'interchange': ['Ealing Broadway'],
  'length': 59.089999999999996,
  'route': 1},
 {'interchange': ['Notting Hill Gate'],
  'length': 40.279999999999994,
  'route': 2},
 {'interchange': ['Mile End'], 'length': 68.86999999999999, 'route': 3}]

在我的家庭笔记本上为 600.000 行数据框计时：

设置：

In [245]: a = pd.concat([df] * 10**5)

合并aDF的形状：

In [246]: a.shape
Out[246]: (600000, 3)

时间：

In [251]: %timeit a.groupby(['route','length'])['interchange'].apply(lambda x: x.tolist()).reset_index()
10 loops, best of 3: 130 ms per loop

非向量化方法（用于循环/列表理解/等）：

In [262]: %paste
def roganjosh(lst):
    new_list = []
    for key, group in groupby(lst, lambda x: x[0]):
        new_list.append(list(group))

    main_dict = {}
    for item in new_list:
        main_dict[item[0][0]] = {'length': item[0][2], 'interchange': [stn[1] for stn in item]}

    return  main_dict
## -- End pasted text --

In [263]: lst = a.values.tolist()

In [264]: len(lst)
Out[264]: 600000

In [265]: %timeit roganjosh(lst)
1 loop, best of 3: 650 ms per loop