Question

我正在使用此代码来获取bigrams的频率：

text1='the cat jumped over the dog in the dog house'
text=text1.split()

counts = defaultdict(int)
for pair in nltk.bigrams(text):
    counts[pair] +=1

for c, pair in ((c, pair) for pair, c in counts.iteritems()):
    print pair, c

输出结果为：

('the', 'cat') 1
('dog', 'in') 1
('cat', 'jumped') 1
('jumped', 'over') 1
('in', 'the') 1
('over', 'the') 1
('dog', 'house') 1
('the', 'dog') 2

我需要的是列出的双胞胎，但是我需要将单词的等级打印出来，而不是每个单词。当我的意思是“等级”时，我的意思是频率最高的单词有等级1，第二等级有等级2等等......这里的等级是：1.the 2.dog和频率相等的等级按降序排列等级。 3.cat 4.jumped 5.over等..

例如

1 3 1

而不是

('the', 'cat') 1

我相信要做到这一点，我需要一本带有单词和等级的字典，但我被困住了，不知道如何继续。我所拥有的是：

fd=FreqDist()
ranks=[]
rank=0
for word in text:
    fd.inc(word)
for rank, word in enumerate(fd):
    ranks.append(rank+1)

word_rank = {}
for word in text:
    word_rank[word] = ranks

print ranks

Answer 1

假设已经创建了counts，以下内容应该得到您想要的结果：

freq = defaultdict(int)
for word in text:
    freq[word] += 1

ranks = sorted(freq.keys(), key=lambda k: (-freq[k], text.index(k)))
ranks = dict(zip(ranks, range(1, len(ranks)+1)))

for (a, b), count in counts.iteritems():
    print ranks[a], ranks[b], count

输出：

以下是一些可能有助于理解其工作原理的中间值：

>>> dict(freq)
{'house': 1, 'jumped': 1, 'over': 1, 'dog': 2, 'cat': 1, 'in': 1, 'the': 3}
>>> sorted(freq.keys(), key=lambda k: (-freq[k], text.index(k)))
['the', 'dog', 'cat', 'jumped', 'over', 'in', 'house']
>>> dict(zip(ranks, range(1, len(ranks)+1)))
{'house': 7, 'jumped': 4, 'over': 5, 'dog': 2, 'cat': 3, 'in': 6, 'the': 1}

Answer 2

text1='the cat jumped over the dog in the dog house'.split(' ')
word_to_rank={}
for i,word in enumerate(text1):
    if word not in word_to_rank:
        word_to_rank[word]=i+1

from collections import Counter
word_to_frequency=Counter(text1)

word_to_tuple={}
for word in word_to_rank:
    word_to_tuple[word]=(-word_to_frequency[word],word_to_rank[word])

tuple_to_word=dict(zip(word_to_tuple.values(),word_to_tuple.keys()))

sorted_by_conditions=sorted(tuple_to_word.keys())

word_to_true_rank={}
for i,_tuple in enumerate(sorted_by_conditions):
    word_to_true_rank[tuple_to_word[_tuple]]=i+1

def fix(pair,c):
    return word_to_true_rank[pair[0]],word_to_true_rank[pair[1]],c

pair=('the', 'cat')
c=1
print fix(pair,c)

pair=('the', 'dog')
c=2
print fix(pair,c)


>>>
(1, 3, 1)
(1, 2, 2)

Bigrams和词组

2 个答案: