将列表更改为字典

Question

我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]]

一本字典里

{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

但是它不起作用，我在挣扎着从哪里开始。

Answer 1

与Majora类似的答案，但首先使用groupby。没有虚假的查找，但可能需要事先排序。

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

下面是一个具体的例子，说明您将如何做到这一点：

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])

这是假设在向d['length']添加元素时要覆盖d[0]。

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]]}