查找最匹配字母的单词

Question

我有两份清单：

l1 = ['UK', 'GER', 'POL']

及

l2 = ['Germany', 'Poland', 'United Kingdom']

这些列表是指国家，但顺序不匹配

有没有办法根据l2中的字母数将l1中的值与l2中的值进行匹配

因此，输出为：

dic = {'UK': 'United Kingdom',
       'GER': 'Germany',
       'POL': 'Poland'}

Answer 1

像这样的东西可能适合你：

#!/usr/bin/env python3

l1 = ['UK', 'GER', 'POL']
l2 = ['Germany', 'Poland', 'United Kingdom', 'Ukraine']

d = {}
for short in l1:
    lower = short.lower()

    # Match prefix or initials.
    matches = [x for x in l2 if
               x.lower().startswith(lower) or
               ''.join(w[0] for w in x.split()).lower() == lower]

    if len(matches) == 0:
        print('no match', short)
    elif len(matches) > 1:
        print('ambiguous', short, matches)
    else:
        d[short] = matches[0]

print(d)

$ ./test.py 
ambiguous UK ['United Kingdom', 'Ukraine']
{'GER': 'Germany', 'POL': 'Poland'}

我添加了“乌克兰”来测试处理模糊匹配