我如何从一个大列表中最多提取14个名字，这些名字加在一起，所有的字母都在一起？

Question

我是个新程序员。正如标题所述，这是我试图完成的：

def alphabet_set(countrylist):
    alphabetical_list = []
    alphabet = 'abcdefghijklmnopqrstuvwxyz'
   
    for country in countries:
        for i in country:
            if i in alphabet:
                alphabetical_list.append(country)
            if i == 26:
                break    
    return alphabetical_list

print(alphabet_set(countries))

再一次，我从这张清单中画出的清单很大。我需要14个国家的名字，这会给我所有的字母。Alphabetical_list将是我列出的14个国家的名单，其中包含所有26个字母。

Answer 1

你可以用set()

def alphabet_set(countrylist:list[str]):
    alphabetical_list = []
    alphabet = 'abcdefghijklmnopqrstuvwxyz'
    countriescontains = []
    letters = set()
    for country in countrylist:
        if len(letters)==26 or len(alphabetical_list) == 14:break
        elif set(country.lower()).intersection(alphabet):
            alphabetical_list.append(country)
            print({x.lower() for x in country},letters.union({x.lower() for x in country}))
            letters=letters.union({x.lower() for x in country if x != ' '})
    print(sorted(list(letters)))
    
    return alphabetical_list
print(alphabet_set(countries))

这将降低alphabet_set从O(n^2)到O(n)的复杂性，这将大大提高性能。

Answer 2

好的，这就是我想出的答案：

def alphabet_set(countries):   
    alphabetical_list = []
    alphabet = 'abcdefghijklmnopqrstuvwxyz'
    for country in countries:
        for i in country:
            if i.lower() in alphabet and country not in alphabetical_list:
                alphabet = alphabet.replace(i.lower(),'')
                alphabetical_list.append(country)
            if alphabet == '':
                break    
    return alphabetical_list
print(alphabet_set(countries))