python - 计算字符串中最小长度的唯一单词

Question

我必须编写一个函数，它需要两个变量，即一个句子和一个数字。该函数应返回字符串中等于或大于该数字的唯一词的数量。示例结果应该是：

>>> unique_func("The sky is blue and the ocean is also blue.",3)
    6

我能想到的解决方案是

def unique_func(sentence,number):
    sentence_split = sentence.lower().split()
    for w in sentence_split:
        if len(w) >= number:

现在我不知道如何继续我的解决方案。谁能帮我？

Answer 1

尝试这个：

from string import punctuation

def unique_func(sentence, number):
    cnt = 0
    sentence = sentence.translate(None, punctuation).lower()
    for w in set(sentence.split()):
        if len(w) >= number:
            cnt += 1
    return cnt 

或者：

def unique_func(sentence, number):
    sentence = sentence.translate(None, punctuation).lower()
    return len([w for w in set(sentence.split()) if len(w) >= number])

Answer 2

这里有一个提示：

>>> set('The sky is blue and the ocean is also blue'.lower().split())
{'is', 'also', 'blue', 'and', 'the', 'sky', 'ocean'}
>>> len(set('The sky is blue and the ocean is also blue'.lower().split()))
7

Answer 3

>>> from string import punctuation
>>> def unique_func(text, n):
        words = (w.strip(punctuation) for w in text.lower().split())
        return len(set(w for w in words if len(w) >= n))


>>> unique_func("The sky is blue and the ocean is also blue.",3)
6