python - 查找字符串列表中最独特的字符

Question

最近加入了 python 代码俱乐部，所以不是学校作业。

任务:

Write a function coolest_word(words) that returns the coolest word if it exists.

The coolness is defined by how many unique characters a word has.

For example the word "decomposition" has a 10 unique letters. Your job is to find from the given list of words the coolest word and return it.

If the list is empty, return None. If there is a tie, return one of the words.

到目前为止我的解决方案:

words = ['expectation', 'discomfort', 'half', 'decomposition']
def coolest_word(words):
    length = 0
    unique = ['']
    for i in words:
        if (len(words) != 0 and len(set(i)) > length):
            length = len(set(i))
            unique[0] = i
        else:
            unique[0] = ('None')
    return(unique)  
print(coolest_word(words))

我选择将 list 更改为 set，然后将 set 的唯一字符相互比较。由于 set 没有考虑重复变量，这似乎是一个好主意。

每当 for 循环找到更长的集合(具有更多变量)时，它就会被覆盖为唯一的集合以供内存并稍后调用。

现在的问题是，如果给定的列表为空，唯一的返回['']，我不明白为什么。

如第二部分所示，如果列表为空，我希望唯一的被覆盖，但它不会这样做。

有什么想法吗？

Answer 1

max的简单应用具有key 和default。

def coolest_word(words):
    return max(words, key=lambda w: len(set(w)), default=None)

演示:

>>> coolest_word(['expectation', 'discomfort', 'half', 'decomposition'])
'decomposition'

>>> [coolest_word([])]   # Wrap in a list because None by itself doesn't get printed
[None]