给出两个相同长度的python列表.如何返回类似值的最佳匹配？

Question

给出了两个包含字符串的python列表(人名):

list_1 = ['J. Payne', 'George Bush', 'Billy Idol', 'M Stuart', 'Luc van den Bergen']
list_2 = ['John Payne', 'George W. Bush', 'Billy Idol', 'M. Stuart', 'Luc Bergen']

我想要一个最相似的名称映射.

'J. Payne'           -> 'John Payne'
'George Bush'        -> 'George W. Bush'
'Billy Idol'         -> 'Billy Idol'
'M Stuart'           -> 'M. Stuart'
'Luc van den Bergen' -> 'Luc Bergen'

在python中有一个简洁的方法吗？这些列表平均包含5个或6个名称.有时更多,但这很少.有时它只是每个列表中的一个名称,可能拼写略有不同.

Answer 1

使用此处定义的函数:http://hetland.org/coding/python/levenshtein.py

>>> for i in list_1:
...     print i, '==>', min(list_2, key=lambda j:levenshtein(i,j))
... 

J. Payne ==> John Payne
George Bush ==> George W. Bush
Billy Idol ==> Billy Idol
M Stuart ==> M. Stuart
Luc van den Bergen ==> Luc Bergen

您可以使用functools.partial而不是lambda

>>> from functools import partial
>>> for i in list_1:
...     print i, '==>', min(list_2, key=partial(levenshtein,i))
...

J. Payne ==> John Payne
George Bush ==> George W. Bush
Billy Idol ==> Billy Idol
M Stuart ==> M. Stuart
Luc van den Bergen ==> Luc Bergen

Answer 2

你可以试试difflib:

import difflib

list_1 = ['J. Payne', 'George Bush', 'Billy Idol', 'M Stuart', 'Luc van den Bergen']
list_2 = ['John Payne', 'George W. Bush', 'Billy Idol', 'M. Stuart', 'Luc Bergen']

mymap = {}
for elem in list_1:
    closest = difflib.get_close_matches(elem, list_2)
    if closest:
        mymap[elem] = closest[0]

print mymap

输出:

{'George Bush': 'George W. Bush', 
 'Luc van den Bergen': 'Luc Bergen', 
 'Billy Idol': 'Billy Idol', 
 'J. Payne': 'John Payne', 
 'M Stuart': 'M. Stuart'}