在字符串列表中找到最常见的子字符串？

Question

我有一个字符串名称的Python列表，我想从所有名称中删除一个公共子字符串。你知道吗

在阅读了这个类似的answer之后，我几乎可以使用SequenceMatcher获得想要的结果。你知道吗

但仅当所有项都有一个公共子字符串时：

From List:
string 1 = myKey_apples
string 2 = myKey_appleses
string 3 = myKey_oranges

common substring = "myKey_"

To List:
string 1 = apples
string 2 = appleses
string 3 = oranges

不过，我有一个稍微有点嘈杂的列表，其中包含一些不符合相同命名约定的零散项。你知道吗

我想从多数中删除“最常见”的子字符串：

From List:
string 1 = myKey_apples
string 2 = myKey_appleses
string 3 = myKey_oranges
string 4 = foo
string 5 = myKey_Banannas

common substring = ""

To List:
string 1 = apples
string 2 = appleses
string 3 = oranges
string 4 = foo
string 5 = Banannas

我需要一种方法来匹配“myKey\”子字符串，这样我就可以从所有名称中删除它。你知道吗

但是当我使用SequenceMatcher时，项“foo”导致“longest match”等于blank。你知道吗

我认为解决这个问题的唯一方法是找到“最常见的子串”。但这怎么可能实现呢？你知道吗

---

基本示例代码：

from difflib import SequenceMatcher

names = ["myKey_apples",
"myKey_appleses",
"myKey_oranges",
#"foo",
"myKey_Banannas"]

string2 = names[0]
for i in range(1, len(names)):
    string1 = string2
    string2 = names[i]
    match = SequenceMatcher(None, string1, string2).find_longest_match(0, len(string1), 0, len(string2))

print(string1[match.a: match.a + match.size]) # -> myKey_

Answer 1

我会首先找到出现次数最多的起始字母。然后我会把每一个有起始字母的单词，当所有这些单词都有匹配的字母时。最后，我会删除每个起始词的前缀：

from collections import Counter
from itertools import takewhile

strings = ["myKey_apples", "myKey_appleses", "myKey_oranges", "berries"]

def remove_mc_prefix(words):
    cnt = Counter()
    for word in words:
        cnt[word[0]] += 1
    first_letter = list(cnt)[0]

    filter_list = [word for word in words if word[0] == first_letter]
    filter_list.sort(key = lambda s: len(s)) # To avoid iob

    prefix = ""
    length = len(filter_list[0])
    for i in range(length):
        test = filter_list[0][i]
        if all([word[i] == test for word in filter_list]):
            prefix += test
        else: break
    return [word[len(prefix):] if word.startswith(prefix) else word for word in words]

print(remove_mc_prefix(strings))

Out: ['apples', 'appleses', 'oranges', 'berries']

Answer 2

给定names = ["myKey_apples", "myKey_appleses", "myKey_oranges", "foo", "myKey_Banannas"]

我能想到的一个O(n^2)解决方案是找到所有可能的子字符串，并将它们存储在字典中，并记下它们出现的次数：

substring_counts={}

for i in range(0, len(names)):
    for j in range(i+1,len(names)):
        string1 = names[i]
        string2 = names[j]
        match = SequenceMatcher(None, string1, string2).find_longest_match(0, len(string1), 0, len(string2))
        matching_substring=string1[match.a:match.a+match.size]
        if(matching_substring not in substring_counts):
            substring_counts[matching_substring]=1
        else:
            substring_counts[matching_substring]+=1

print(substring_counts) #{'myKey_': 5, 'myKey_apples': 1, 'o': 1, '': 3}

然后选取最大发生子串

import operator
max_occurring_substring=max(substring_counts.iteritems(), key=operator.itemgetter(1))[0]
print(max_occurring_substring) #myKey_

Answer 3

下面是一个过于冗长的解决方案：

def find_matching_key(list_in, max_key_only = True):
  """
  returns the longest matching key in the list * with the highest frequency
  """
  keys = {}
  curr_key = ''

  # If n does not exceed max_n, don't bother adding
  max_n = 0

  for word in list(set(list_in)): #get unique values to speed up
    for i in range(len(word)):
      # Look up the whole word, then one less letter, sequentially
      curr_key = word[0:len(word)-i]
      # if not in, count occurance
      if curr_key not in keys.keys() and curr_key!='':
        n = 0
        for word2 in list_in:
          if curr_key in word2:
            n+=1
        # if large n, Add to dictionary
        if n > max_n:
          max_n = n
          keys[curr_key] = n
    # Finish the word
  # Finish for loop  
  if max_key_only:
    return max(keys, key=keys.get)
  else:
    return keys    

# Create your "from list"
From_List = [
             "myKey_apples",
             "myKey_appleses",
             "myKey_oranges",
             "foo",
             "myKey_Banannas"
]

# Use the function
key = find_matching_key(From_List, True)

# Iterate over your list, replacing values
new_From_List = [x.replace(key,'') for x in From_List]

print(new_From_List)
['apples', 'appleses', 'oranges', 'foo', 'Banannas']

不用说，使用递归，这个解决方案看起来会更整洁。我想我会为你画出一个粗略的动态规划解决方案。