python difflib.get_close_matches()函数如何工作？

Question

以下是两个数组:

import difflib
import scipy
import numpy

a1=numpy.array(['198.129.254.73','134.55.221.58','134.55.219.121','134.55.41.41','198.124.252.101'], dtype='|S15')
b1=numpy.array(['198.124.252.102','134.55.41.41','134.55.219.121','134.55.219.137','134.55.220.45', '198.124.252.130'],dtype='|S15')

difflib.get_close_matches(a1[-1],b1,2)

输出:

['198.124.252.130', '198.124.252.102']

不应该'198.124.252.102'是最接近的匹配'198.124.252.101'？

我查看了他们已经指定了一些浮动类型权重的文档,但没有关于算法使用的信息.

我需要找出最后两个八位位组之间的绝对差值是否为1(假设前三个八位位组相同).

所以我首先找到最接近的字符串然后检查上述条件的最接近的字符串.

有没有其他功能或方法来实现这一目标？另外get_close_matches()表现如何？

ipaddr 似乎对ips没有这样的操纵.

Answer 1

好吧,文档中有一部分解释了您的问题:

这不会产生最小的编辑序列,但确实会产生对人们"看起来正确"的匹配.

为了获得您期望的结果,您可以使用Levenshtein_distance.

但是为了比较IP,我建议使用整数比较:

>>> parts = [int(s) for s in '198.124.252.130'.split('.')]
>>> parts2 = [int(s) for s in '198.124.252.101'.split('.')]
>>> from operator import sub
>>> diff = sum(d * 10**(3-pos) for pos,d in enumerate(map(sub, parts, parts2)))
>>> diff
29

您可以使用此样式创建比较函数:

from functools import partial
from operator import sub

def compare_ips(base, ip1, ip2):
    base = [int(s) for s in base.split('.')]
    parts1 = (int(s) for s in ip1.split('.'))
    parts2 = (int(s) for s in ip2.split('.'))
    test1 = sum(abs(d * 10**(3-pos)) for pos,d in enumerate(map(sub, base, parts1)))
    test2 = sum(abs(d * 10**(3-pos)) for pos,d in enumerate(map(sub, base, parts2)))
    return cmp(test1, test2)

base = '198.124.252.101'
test_list = ['198.124.252.102','134.55.41.41','134.55.219.121',
             '134.55.219.137','134.55.220.45', '198.124.252.130']
sorted(test_list, cmp=partial(compare_ips, base))
# yields:
# ['198.124.252.102', '198.124.252.130', '134.55.219.121', '134.55.219.137', 
#  '134.55.220.45', '134.55.41.41']