python - 如何递归地生成多词术语？

Question

假设我有一串单词:'a b c d e f'。我想从这个字符串中生成一个多词术语列表。

词序很重要。术语 'fe d' 不应从上面的示例中生成。

编辑:此外，不应跳过单词。不应生成 'a c' 或 'b d f'。

我现在拥有的:

doc = 'a b c d e f'
terms= []
one_before = None
two_before = None
for word in doc.split(None):
    terms.append(word)
    if one_before:
        terms.append(' '.join([one_before, word]))
    if two_before:
        terms.append(' '.join([two_before, one_before, word]))
    two_before = one_before
    one_before = word

for term in terms:
    print term

打印:

a
b
a b
c
b c
a b c
d
c d
b c d
e
d e
c d e
f
e f
d e f

我如何使它成为一个递归函数，以便我可以向它传递每个术语的可变最大单词数？

应用:

我将使用它从 HTML 文档中的可读文本生成多词术语。总体目标是对大型语料库(约 200 万个文档)进行潜在语义分析。这就是为什么保持词序很重要(自然语言处理等)。

Answer 1

这不是递归的，但我认为它可以满足您的需求。

doc = 'a b c d e f'
words = doc.split(None)
max = 3          


for index in xrange(len(words)):    
    for n in xrange(max):
        if index + n < len(words):           
            print ' '.join(words[index:index+n+1])   

这是一个递归的解决方案:

def find_terms(words, max_words_per_term):       
    if len(words) == 0: return []
    return [" ".join(words[:i+1]) for i in xrange(min(len(words), max_words_per_term))] + find_terms(words[1:], max_words_per_term)


doc = 'a b c d e f'
words = doc.split(None) 
for term in find_terms(words, 3):
    print term

这里又是递归函数，有一些解释变量和注释。

def find_terms(words, max_words_per_term):   

    # If there are no words, you've reached the end. Stop.    
    if len(words) == 0:
        return []      

    # What's the max term length you could generate from the remaining 
    # words? It's the lesser of max_words_per_term and how many words 
    # you have left.                                                         
    max_term_len = min(len(words), max_words_per_term)       

    # Find all the terms that start with the first word.
    initial_terms = [" ".join(words[:i+1]) for i in xrange(max_term_len)]

    # Here's the recursion. Find all of the terms in the list 
    # of all but the first word.
    other_terms = find_terms(words[1:], max_words_per_term)

    # Now put the two lists of terms together to get the answer.
    return initial_terms + other_terms