我正在使用NLTK - 一个用于操作语料库文本的特定工具包,我已经定义了一个函数来将用户输入与莎士比亚的单词相交叉。
def shakespeareOutput(userInput):
user = userInput.split()
user = random.sample(set(user), 3)
#here is NLTK's method
play = gutenberg.sents('shakespeare-hamlet.txt')
#all lowercase
hamlet = map(lambda sublist: map(str.lower, sublist), play)
print hamlet
返回:
[ ['[', 'the', 'tragedie', 'of', 'hamlet', 'by', 'william', 'shakespeare', '1599', ']'],
['actus', 'primus', '.'],
['scoena', 'prima', '.'],
['enter', 'barnardo', 'and', 'francisco', 'two', 'centinels', '.'],
['barnardo', '.'],
['who', "'", 's', 'there', '?']...['finis', '.'],
['the', 'tragedie', 'of', 'hamlet', ',', 'prince', 'of', 'denmarke', '.']]
我想找到包含最多用户词的句子并返回句子。我在尝试:
bestCount = 0
for sent in hamlet:
currentCount = len(set(user).intersection(sent))
if currentCount > bestCount:
bestCount = currentCount
answer = ' '.join(sent)
return ''.join(answer).lower(), bestCount
调用函数:
shakespeareOutput("The Actus Primus")
返回:
['The', 'Actus', 'Primus']
None
我做错了什么?
提前感谢。
答案 0 :(得分:2)
您评估currentCount
的方式是错误的。设置交集返回匹配的不同元素的数量,而不是匹配元素的数量。
>>> s = [1,1,2,3,3,4]
>>> u = set([1,4])
>>> u.intersection(s)
set([1, 4]) # the len is 2, however the total number matched elements are 3
使用以下代码。
bestCount = 0
for sent in hamlet:
currentCount = sum([sent.count(i) for i in set(user)])
if currentCount > bestCount:
bestCount = currentCount
answer = ' '.join(sent)
return answer.lower(), bestCount