Question

我要说的是,我有两条str。

AAABBBCCCCC

以及

AAAABBBBCCCC

鉴于我只能去除我应当删除的特性,因此,尽可能使这种说法相似。

delete the last C from the first string
delete the last A 以及 the last B from the second string,

以便

AAABBBCCCC

找到从每一条指示中删除哪些特性的高效算法是什么?

目前,我大米.忙地思考一下,有人把扼子放在一边的孤独,看着他们,一把其他扼子。

Answer 1

Levenshtein gap,可计算你需要多少改动才能将一个扼杀变成另一个。对来源的微小改动,而且你可能不仅有距离,而且需要转换。

Answer 2

How about using difflib?

import difflib

s1 =  AAABBBCCCCC 
s2 =  AAAABBBBCCCC 

for difference in difflib.ndiff(s1, s2):
    print difference,
    if difference[0] ==  + :
        print  remove this char from s2 
    elif difference[0] ==  - :
        print  remove this char from s1 
    else:
        print  no change here

这将勾销你可用来消除分歧的两条插图之间的差别。产出如下:

  A no change here
  A no change here
  A no change here
+ A remove this char from s2
+ B remove this char from s2
  B no change here
  B no change here
  B no change here
  C no change here
  C no change here
  C no change here
  C no change here
- C remove this char from s1

Answer 3

Don t know if it s the fastest, but as code goes, it is at least short:

import difflib
  .join([c[-1] for c in difflib.Differ().compare( AAABBBCCCCC , AAAABBBBCCCC ) if c[0] ==    ])

Answer 4

I think regular expression can do this.It s a string distance problem. I mean. Let s have two string:

str1 =  abc 
str2 =  aabbcc

首先,我选择简短,并作如下常规表述:

regex =  (w*) + (w*) .join(list(str1))+ (w*)

Then, we can search:

matches = re.search(regex,str2)

I use round brackets to group the section I am interested. these groups of matches.group() is the distance of two strings.Next, I can figure out what characters should be removed. It s my idea, I hope it can help you.

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