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_dict

def 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 set

resultdict = findOverlappingLists(similiarStrings)

Answer 2

这是一个可能更简单/更快的实现

from collections import Counter

from itertools import groupby

import pprint

# Strategy:

# 1.  Find common words in strings in list

# 2.  Group strings which have the same common words together

def 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]