Question

我需要像这样转换列表列表：

[['S', 'NP', 'VP'], ['NP', 'Det', 'N'], ['NP', 'NP', 'PP'], ['VP', 'V', 'NP'], ['VP', 'VP', 'PP'], ['PP', 'P', 'NP'], ['Det', "'the'"], ['N', "'pirate'"], ['N', "'sailor'"], ['N', "'telescope'"], ['V', "'sees'"], ['P', "'with'"]]

像这样的字典：

{'S':['NP', 'VP'], 'NP': ['Det', 'N'], ['NP', 'PP'], 'VP': ['V', 'NP'], ['VP', 'PP'], 'PP': ['P', 'NP'], 'Det': ["'the'"], 'N': ["'pirate'"], ["'sailor'"], ["'telescope'"], 'V': ["'sees'"], 'P': ["'with'"]}

我尝试通过from collections import default dict使用此方法：

g = defaultdict(dict)
for i, j, k in new_grammar:
    g[i][j] = k

但这不起作用，因为列表列表中只有两个元素。

我也尝试过：

grammar = {}
for rule in new_grammar:
    grammar[rule[0]] = rule[1:]

但是，这只给每个键一个值。

有什么办法吗？

Answer 1

您处在正确的轨道上，但是defaultdict是从字符串到列表的映射，而不是字典。试试这个：

g = defaultdict(list)
for i in new_grammar:
    g[i[0]].extend(i[1:])

Answer 2

这是使用collections.defaultdict的一种方式。

请注意，结果是一个字典，其中包含分配给每个键的值的列表。这不是您定义所需输出的方式，这不是有效的Python。

L = [['S', 'NP', 'VP'], ['NP', 'Det', 'N'], ['NP', 'NP', 'PP'], ['VP', 'V', 'NP'], ['VP', 'VP', 'PP'], ['PP', 'P', 'NP'], ['Det', "'the'"], ['N', "'pirate'"], ['N', "'sailor'"], ['N', "'telescope'"], ['V', "'sees'"], ['P', "'with'"]]

from collections import defaultdict

d = defaultdict(list)

for k, *v in L:
    d[k].extend(v)

print(d)

defaultdict(list,
            {'Det': ["'the'"],
             'N': ["'pirate'", "'sailor'", "'telescope'"],
             'NP': ['Det', 'N', 'NP', 'PP'],
             'P': ["'with'"],
             'PP': ['P', 'NP'],
             'S': ['NP', 'VP'],
             'V': ["'sees'"],
             'VP': ['V', 'NP', 'VP', 'PP']})

Answer 3

为什么不仅如此？还是您想要的？？

l = [['S', 'NP', 'VP'], ['NP', 'Det', 'N'], ['NP', 'NP', 'PP'], ['VP', 'V', 'NP'], ['VP', 'VP', 'PP'], ['PP', 'P', 'NP'], ['Det', "'the'"], ['N', "'pirate'"], ['N', "'sailor'"], ['N', "'telescope'"], ['V', "'sees'"], ['P', "'with'"]]
print(dict(zip([i[0] for i in l],[i[1:] for i in l])))

输出：

{'S': ['NP', 'VP'], 'NP': ['NP', 'PP'], 'VP': ['VP', 'PP'], 'PP': ['P', 'NP'], 'Det': ["'the'"], 'N': ["'telescope'"], 'V': ["'sees'"], 'P': ["'with'"]}

Answer 4

如何？如果element不存在，则添加到字典中。如果元素存在，则追加到字典。

l = [['S', 'NP', 'VP'], ['NP', 'Det', 'N'], ['NP', 'NP', 'PP'], ['VP', 'V', 'NP'], ['VP', 'VP', 'PP'], ['PP', 'P', 'NP'], ['Det', "'the'"], ['N', "'pirate'"], ['N', "'sailor'"], ['N', "'telescope'"], ['V', "'sees'"], ['P', "'with'"]]


grammar = {}
for rule in l:
    if(grammar.get(rule[0],None) == None):
        grammar[rule[0]] = rule[1:]
    else:
        for elem in rule[1:]:
            grammar[rule[0]].append(elem)

print grammar

Answer 5

也可以使用itertools.groupby()：

from itertools import groupby
from itertools import chain
from operator import itemgetter
from pprint import pprint

L = [['S', 'NP', 'VP'], ['NP', 'Det', 'N'], ['NP', 'NP', 'PP'], ['VP', 'V', 'NP'], ['VP', 'VP', 'PP'], ['PP', 'P', 'NP'], ['Det', "'the'"], ['N', "'pirate'"], ['N', "'sailor'"], ['N', "'telescope'"], ['V', "'sees'"], ['P', "'with'"]]

result = {k: list(chain(*(x[1:] for x in g))) for 
          k, g in groupby(sorted(L, key=itemgetter(0)), key=itemgetter(0))}

pprint(result)

哪些输出：

{'Det': ["'the'"],
 'N': ["'pirate'", "'sailor'", "'telescope'"],
 'NP': ['Det', 'N', 'NP', 'PP'],
 'P': ["'with'"],
 'PP': ['P', 'NP'],
 'S': ['NP', 'VP'],
 'V': ["'sees'"],
 'VP': ['V', 'NP', 'VP', 'PP']}

如何将具有相同第一个元素的单独列表的列表转换为字典？

5 个答案: