如何按子字符串模式对列表进行排序，使其成为 dict 的 dict

Question

我正在尝试对基于类似子字符串的值列表进行排序。我想将其分组到列表的字典中，其中键是相似的子字符串，值是这些分组值的列表。

例如（实际列表有 24k 个条目）：

test_list = [ 'Doghouse Amsterdam', 'Doghouse Antwerp', 'Doghouse Vienna', 

        'House by KatSkill', 'Garden by KatSkill', 'Meadow by KatSkill']

至：

resultdict = { 

'Doghouse' : ['Doghouse Amsterdam', 'Doghouse Antwerp', 'Doghouse Vienna'],

'by KatSkill' : [ 'House by KatSkill', 'Garden by KatSkill', 'Meadow by KatSkill' ]

}

我尝试了以下方法，但这根本不起作用。

from itertools import groupby 

test_list = [ 'Doghouse Amsterdam', 'Doghouse Antwerp', 'Doghouse Vienna', 

            'House by KatSkill', 'Garden by KatSkill', 'Meadow by KatSkill']

res = [list(i) for j, i in groupby(test_list, 

                          lambda a: a.partition('_')[0])]

Answer 1

最初，查找出现在输入列表的另一个字符串中的所有以“”分隔的子字符串。在此过程中，构建一个字典，其中包含所有相应的子字符串作为键，输入字符串作为值。这将返回一个只有单个子字符串作为键的字典。使用该示例返回：{'by': ['Garden by KatSkill', 'Meadow by KatSkill', 'House by KatSkill'], 'KatSkill': ['Garden by KatSkill', 'Meadow by KatSkill', 'House by KatSkill'], 'Doghouse': ['Doghouse Antwerp', 'Doghouse Vienna', 'Doghouse Amsterdam']}为了获得预期的结果，需要进行压实。对于压缩，利用每个字典键也是字典字符串列表的一部分这一事实是有益的。因此迭代字典值并将字符串再次拆分为子字符串。然后按照子串列表的顺序遍历子串，确定包含字典键的子串列表范围。将相应的范围添加到新的字典中。对于 24k 条目，这可能需要一段时间。请参阅下面的源代码：mylist = [ 'Doghouse Amsterdam', 'Doghouse Antwerp', 'Doghouse Vienna',         'House by KatSkill', 'Garden by KatSkill', 'Meadow by KatSkill']def findSimilarSubstrings(list):    res_dict = {}    for string in list:        substrings = string.split(" ")        for otherstring in list:            # Prevent check with the same string            if otherstring == string:                continue            for substring in substrings:                if substring in otherstring:                   if not(substring in res_dict):                       res_dict[substring] = []                   # Prevent duplicates                   if not(otherstring in res_dict[substring]):                       res_dict[substring].append(otherstring)    return res_dictdef findOverlappingLists(dict):    res_dict = {}    for list in dict.values():        for string in list:            substrings = string.split(" ")            lastIndex = 0            lastKeyInDict = False            substring = ""            numsubstrings = len(substrings)            for i in range(len(substrings)):               substring = substrings[i]               if substring in dict:                    if not(lastKeyInDict):                        lastIndex = i                        lastKeyInDict = True               elif lastKeyInDict:                   commonstring = " ".join(substrings[lastIndex:i])                   # Add key string to res_dict                   if not(commonstring in res_dict):                      res_dict[commonstring] = []                   # Prevent duplicates                   if not(string in res_dict[commonstring]):                      res_dict[commonstring].append(string)                   lastKeyInDict = False            # Handle last substring            if lastKeyInDict:                commonstring = " ".join(substrings[lastIndex:numsubstrings])                if not(commonstring in res_dict):                    res_dict[commonstring] = []                if not(string in res_dict[commonstring]):                    res_dict[commonstring].append(string)    return res_dict# Initially find all the substrings (seperated by " ") returning:# {'by': ['Garden by KatSkill', 'Meadow by KatSkill', 'House by KatSkill'],#  'KatSkill': ['Garden by KatSkill', 'Meadow by KatSkill', 'House by KatSkill'],#  'Doghouse': ['Doghouse Antwerp', 'Doghouse Vienna', 'Doghouse Amsterdam']}similiarStrings = findSimilarSubstrings(mylist)# Perform a compaction on similiarStrings.values() by lookup in the dictionary's key setresultdict = findOverlappingLists(similiarStrings)

Answer 2

这是一个可能更简单/更快的实现from collections import Counterfrom itertools import groupbyimport pprint# Strategy:# 1.  Find common words in strings in list# 2.  Group strings which have the same common words togetherdef find_common_words(lst):  " finds strings with common words "  cnt = Counter()  for s in lst:    cnt.update(s.split(" "))  # return words which appear in more than one string  words = set([k for k, v in cnt.items() if v > 1])  return words  def grouping_key(s, words):  " Key function for grouping strings with common words in the same sequence"  k = []  for i in s.split():    if i in words:      k.append(i)  return ' '.join(k)def calc_groupings(lst):  " Generate the string groups based upon common words "  common_words = find_common_words(lst)  # Group strings with common words  g = groupby(lst, lambda x: grouping_key(x, common_words))  # Result  return {k: list(v) for k, v in g}t = ['Doghouse Amsterdam', 'Doghouse Antwerp', 'Doghouse Vienna',         'House by KatSkill', 'Garden by KatSkill', 'Meadow by KatSkill']pp = pprint.PrettyPrinter(indent=4)pp.pprint(calc_groupings(t))输出{   'Doghouse': ['Doghouse Amsterdam', 'Doghouse Antwerp', 'Doghouse Vienna'],'by KatSkill': [   'House by KatSkill',                   'Garden by KatSkill',                   'Meadow by KatSkill']}

Answer 3

mylist = [ 'Doghouse Amsterdam', 'Doghouse Antwerp', 'Doghouse Vienna',             'House by KatSkill', 'Garden by KatSkill', 'Meadow by KatSkill']test = ['Doghouse', 'by KatSkill']使用 dict 和列表理解：res = { i: [j for j in mylist if i in j] for i in test}或设置您的 dict 并使用带有列表理解的循环resultdict = {}for i in test:     resultdict[i] = [j for j in mylist if i in j]