python - 如何从列表项制作嵌套字典？

Question

我有一个列表列表，例如:

a = [
     ['A', 'America', 'LA', '1123', '2014-05-01', [('A', '211'), ('AD', '398')]],
     ['D', 'America', 'LA', '1135', '2014-05-01', [('A', '211'), ('AD', '398')]],
     ['I', 'America', 'SF', '1145', '2014-05-01', [('A', '211'), ('AD', '398')]],
     ['A', 'England', 'LND', '3564', '2014-05-01', [('A', '211'), ('AD', '398')]],
     ['D', 'Dubai', 'DUB', '9990', '2014-05-01', [('A', '211'), ('AD', '398')]],
     ['D', 'Dubai', 'DUX', '9670', '2014-05-01', [('A', '211'), ('AD', '398')]],
     ['I', 'Dubai', 'DUB', '9800', '2014-05-01', [('A', '211'), ('AD', '398')]],
    ]

我想要一个像这样的嵌套字典:

d = {
     'America': {
                 'LA': {
                        '1123'  : ['A', '2014-05-01', [('A', '211'), ('AD', '398')]],
                        '1135': ['D', '2014-05-01', [('A', '211'), ('AD', '398')]]
                       },
                 'SF': {
                        '1145': ['I', '2014-05-01', [('A', '211'), ('AD', '398')]]
                       }
                 },
     'England': {
                 'LND': {
                        '3564': ['A', '2014-05-01', [('A', '211'), ('AD', '398')]]
                        }

                 },
     'Dubai': {
                 'DUB': {
                        '9990': ['D', '2014-05-01', [('A', '211'), ('AD', '398')]],
                        '9800': ['I', '2014-05-01', [('A', '211'), ('AD', '398')]]
                        },
                 'DUX': {
                        '9670': ['D', '2014-05-01', [('A', '211'), ('AD', '398')]]
                       }
                 }
    }

我试过了还是不行，请检查我的代码并帮助我!

pd = defaultdict(dict)
for row in a:
    country = defaultdict(dict)
    for c in a:
        dest = defaultdict(list)
        for d in a:
            if c[2] == d[2]:
                dest[d[3]].append([d[1], d[0],d[4],d[5],d[6], d[7]])
            else:
                continue
        country[c[2]] = dest
    pd[row[1]] = country

Answer 1

from operator import itemgetter
from itertools import groupby
getter, d = itemgetter(0, 4, 5), {}
for key, grp in groupby(sorted(a, key=lambda x: x[3]), key=lambda x: x[3]):
    for item in grp:
        d.setdefault(item[1], {}).setdefault(
            item[2], {})[item[3]] = list(getter(item))
print d

输出

{'America': {'LA': {'1123': ['A',
                             '2014-05-01',
                             [('A', '211'), ('AD', '398')]],
                    '1135': ['D',
                             '2014-05-01',
                             [('A', '211'), ('AD', '398')]]},
             'SF': {'1145': ['I',
                             '2014-05-01',
                             [('A', '211'), ('AD', '398')]]}},
 'Dubai': {'DUB': {'9800': ['I',
                            '2014-05-01',
                            [('A', '211'), ('AD', '398')]],
                   '9990': ['D',
                            '2014-05-01',
                            [('A', '211'), ('AD', '398')]]},
           'DUX': {'9670': ['D',
                            '2014-05-01',
                            [('A', '211'), ('AD', '398')]]}},
 'England': {'LND': {'3564': ['A',
                              '2014-05-01',
                              [('A', '211'), ('AD', '398')]]}}}