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，并将它们与它们出现的次数一起存储在字典中：

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}

然后选择出现次数最多的子串

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

以下是您的问题的详细解决方案：

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，这个解决方案使用递归会看起来更简洁。不过，我还是为您勾画出一个粗略的动态编程解决方案。*

我会先找出出现次数最多的起始字母，然后取每个有该起始字母的单词，并在所有这些单词都有匹配字母的情况下取，最后去掉每个起始单词的前缀：

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))

出局：[“苹果”、“苹果”、“橙子”、“浆果”]

从python-string列表中查找most-common-substring
我已经在python-3.10.5上测试过了，我希望它能为你工作。我有相同的用例，但任务不同，我只需要从超过100s的文件列表中找到一个common-pattern-string。要作为regular-expression使用。
你的Basic示例代码在我的例子中不起作用。因为第一个检查第二个，第二个检查第三个，第三个检查第四个，等等。所以，我把它改为最常见的子字符串，并将检查每一个。
这段代码的缺点是，如果最常见的子字符串中有不常见的地方，那么最后一个最常见的子字符串将是空的。

from difflib import SequenceMatcher
for i in range(1, len(names)):
    if i==1:
        string1, string2 = names[0], names[i]
    else:
        string1, string2 = most_common_substring, names[i]
    match = SequenceMatcher(None, string1, string2).find_longest_match(0, len(string1), 0, len(string2))
    most_common_substring = string1[match.a: match.a + match.size]

print(f"most_common_substring : {most_common_substring}")

pythonpython-3python-difflib