合并共享公共元素的列表

Question

我的输入是一个列表列表。其中一些具有共同的元素，例如。

L = [['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g']]

我需要合并共享公共元素的所有列表，并重复此过程，只要没有更多列表具有相同的项目。我考虑过使用布尔运算和 while 循环，但无法想出一个好的解决方案。

最终结果应该是：

L = [['a','b','c','d','e','f','g','o','p'],['k']] 

Answer 1

您可以将列表视为图表的符号，即 ['a','b','c'] 是一个包含 3 个节点相互连接的图表。您试图解决的问题是找到此图中的连接组件。

您可以使用 NetworkX 来实现此目的，它的优点是几乎可以保证它是正确的：

l = [['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g']]

import networkx 
from networkx.algorithms.components.connected import connected_components


def to_graph(l):
    G = networkx.Graph()
    for part in l:
        # each sublist is a bunch of nodes
        G.add_nodes_from(part)
        # it also imlies a number of edges:
        G.add_edges_from(to_edges(part))
    return G

def to_edges(l):
    """ 
        treat `l` as a Graph and returns it's edges 
        to_edges(['a','b','c','d']) -> [(a,b), (b,c),(c,d)]
    """
    it = iter(l)
    last = next(it)

    for current in it:
        yield last, current
        last = current    

G = to_graph(l)
print connected_components(G)
# prints [['a', 'c', 'b', 'e', 'd', 'g', 'f', 'o', 'p'], ['k']]

要有效地解决这个问题无论如何，你自己都必须将列表转换成图形化的东西，所以你最好从一开始就使用networkX。

Answer 2

算法：

1. 从列表中取出第一个集合 A，
2. 并为列表中的每个其他集合 B 执行如果 B 与 A 具有共同元素，则将 B 加入到 A 中；从列表中删除 B
3. 重复 2. 直到不再与 A 重叠
4. 将 A 放入输出
5. 重复 1. 与列表的其余部分

因此您可能想使用集合而不是列表。下面的程序应该可以做到这一点。

l = [['a', 'b', 'c'], ['b', 'd', 'e'], ['k'], ['o', 'p'], ['e', 'f'], ['p', 'a'], ['d', 'g']]

out = []
while len(l)>0:
    first, *rest = l
    first = set(first)

    lf = -1
    while len(first)>lf:
        lf = len(first)

        rest2 = []
        for r in rest:
            if len(first.intersection(set(r)))>0:
                first |= set(r)
            else:
                rest2.append(r)     
        rest = rest2

    out.append(first)
    l = rest

print(out)

Answer 3

我需要对相当大的列表执行 OP 描述的聚类技术数百万次，因此想要确定上面建议的哪种方法最准确且性能最高。

我对上述每种方法的大小从 2^1 到 2^10 的输入列表进行了 10 次试验，每种方法使用相同的输入列表，并测量了上面提出的每种算法的平均运行时间（以毫秒为单位）。以下是结果：

这些结果帮助我发现，在始终返回正确结果的方法中，@jochen 的方法是最快的。在那些不能始终返回正确结果的方法中，mak 的解决方案通常不包含所有输入元素（即缺少列表成员列表），并且 braaksma、cmangla 和 asterisk 的解决方案不能保证最大程度地合并。

有趣的是，迄今为止，两种最快、正确的算法获得了最多的赞成票，并且按正确的排名顺序排列。

这是用于运行测试的代码：

from networkx.algorithms.components.connected import connected_components
from itertools import chain
from random import randint, random
from collections import defaultdict, deque
from copy import deepcopy
from multiprocessing import Pool
import networkx
import datetime
import os

##
# @mimomu
##

def mimomu(l):
  l = deepcopy(l)
  s = set(chain.from_iterable(l))
  for i in s:
    components = [x for x in l if i in x]
    for j in components:
      l.remove(j)
    l += [list(set(chain.from_iterable(components)))]
  return l

##
# @Howard
##

def howard(l):
  out = []
  while len(l)>0:
      first, *rest = l
      first = set(first)

      lf = -1
      while len(first)>lf:
          lf = len(first)

          rest2 = []
          for r in rest:
              if len(first.intersection(set(r)))>0:
                  first |= set(r)
              else:
                  rest2.append(r)
          rest = rest2

      out.append(first)
      l = rest
  return out

##
# Nx @Jochen Ritzel
##

def jochen(l):
  l = deepcopy(l)

  def to_graph(l):
      G = networkx.Graph()
      for part in l:
          # each sublist is a bunch of nodes
          G.add_nodes_from(part)
          # it also imlies a number of edges:
          G.add_edges_from(to_edges(part))
      return G

  def to_edges(l):
      """
          treat `l` as a Graph and returns it's edges
          to_edges(['a','b','c','d']) -> [(a,b), (b,c),(c,d)]
      """
      it = iter(l)
      last = next(it)

      for current in it:
          yield last, current
          last = current

  G = to_graph(l)
  return list(connected_components(G))

##
# Merge all @MAK
##

def mak(l):
  l = deepcopy(l)
  taken=[False]*len(l)
  l=map(set,l)

  def dfs(node,index):
      taken[index]=True
      ret=node
      for i,item in enumerate(l):
          if not taken[i] and not ret.isdisjoint(item):
              ret.update(dfs(item,i))
      return ret

  def merge_all():
      ret=[]
      for i,node in enumerate(l):
          if not taken[i]:
              ret.append(list(dfs(node,i)))
      return ret

  result = list(merge_all())
  return result

##
# @cmangla
##

def cmangla(l):
  l = deepcopy(l)
  len_l = len(l)
  i = 0
  while i < (len_l - 1):
    for j in range(i + 1, len_l):
      # i,j iterate over all pairs of l's elements including new
      # elements from merged pairs. We use len_l because len(l)
      # may change as we iterate
      i_set = set(l[i])
      j_set = set(l[j])

      if len(i_set.intersection(j_set)) > 0:
        # Remove these two from list
        l.pop(j)
        l.pop(i)

        # Merge them and append to the orig. list
        ij_union = list(i_set.union(j_set))
        l.append(ij_union)

        # len(l) has changed
        len_l -= 1

        # adjust 'i' because elements shifted
        i -= 1

        # abort inner loop, continue with next l[i]
        break

      i += 1
  return l

##
# @pillmuncher
##

def pillmuncher(l):
  l = deepcopy(l)

  def connected_components(lists):
    neighbors = defaultdict(set)
    seen = set()
    for each in lists:
        for item in each:
            neighbors[item].update(each)
    def component(node, neighbors=neighbors, seen=seen, see=seen.add):
        nodes = set([node])
        next_node = nodes.pop
        while nodes:
            node = next_node()
            see(node)
            nodes |= neighbors[node] - seen
            yield node
    for node in neighbors:
        if node not in seen:
            yield sorted(component(node))

  return list(connected_components(l))

##
# @NicholasBraaksma
##

def braaksma(l):
  l = deepcopy(l)
  lists = sorted([sorted(x) for x in l]) #Sorts lists in place so you dont miss things. Trust me, needs to be done.

  resultslist = [] #Create the empty result list.

  if len(lists) >= 1: # If your list is empty then you dont need to do anything.
      resultlist = [lists[0]] #Add the first item to your resultset
      if len(lists) > 1: #If there is only one list in your list then you dont need to do anything.
          for l in lists[1:]: #Loop through lists starting at list 1
              listset = set(l) #Turn you list into a set
              merged = False #Trigger
              for index in range(len(resultlist)): #Use indexes of the list for speed.
                  rset = set(resultlist[index]) #Get list from you resultset as a set
                  if len(listset & rset) != 0: #If listset and rset have a common value then the len will be greater than 1
                      resultlist[index] = list(listset | rset) #Update the resultlist with the updated union of listset and rset
                      merged = True #Turn trigger to True
                      break #Because you found a match there is no need to continue the for loop.
              if not merged: #If there was no match then add the list to the resultset, so it doesnt get left out.
                  resultlist.append(l)
  return resultlist

##
# @Rumple Stiltskin
##

def stiltskin(l):
  l = deepcopy(l)
  hashdict = defaultdict(int)

  def hashit(x, y):
      for i in y: x[i] += 1
      return x

  def merge(x, y):
      sums = sum([hashdict[i] for i in y])
      if sums > len(y):
          x[0] = x[0].union(y)
      else:
          x[1] = x[1].union(y)
      return x

  hashdict = reduce(hashit, l, hashdict)
  sets = reduce(merge, l, [set(),set()])
  return list(sets)

##
# @Asterisk
##

def asterisk(l):
  l = deepcopy(l)
  results = {}
  for sm in ['min', 'max']:
    sort_method = min if sm == 'min' else max
    l = sorted(l, key=lambda x:sort_method(x))
    queue = deque(l)

    grouped = []
    while len(queue) >= 2:
      l1 = queue.popleft()
      l2 = queue.popleft()
      s1 = set(l1)
      s2 = set(l2)

      if s1 & s2:
        queue.appendleft(s1 | s2)
      else:
        grouped.append(s1)
        queue.appendleft(s2)
    if queue:
      grouped.append(queue.pop())
    results[sm] = grouped
  if len(results['min']) < len(results['max']):
    return results['min']
  return results['max']

##
# Validate no more clusters can be merged
##

def validate(output, L):
  # validate all sublists are maximally merged
  d = defaultdict(list)
  for idx, i in enumerate(output):
    for j in i:
      d[j].append(i)
  if any([len(i) > 1 for i in d.values()]):
    return 'not maximally merged'
  # validate all items in L are accounted for
  all_items = set(chain.from_iterable(L))
  accounted_items = set(chain.from_iterable(output))
  if all_items != accounted_items:
    return 'missing items'
  # validate results are good
  return 'true'

##
# Timers
##

def time(func, L):
  start = datetime.datetime.now()
  result = func(L)
  delta = datetime.datetime.now() - start
  return result, delta

##
# Function runner
##

def run_func(args):
  func, L, input_size = args
  results, elapsed = time(func, L)
  validation_result = validate(results, L)
  return func.__name__, input_size, elapsed, validation_result

##
# Main
##

all_results = defaultdict(lambda: defaultdict(list))
funcs = [mimomu, howard, jochen, mak, cmangla, braaksma, asterisk]
args = []

for trial in range(10):
  for s in range(10):
    input_size = 2**s

    # get some random inputs to use for all trials at this size
    L = []
    for i in range(input_size):
      sublist = []
      for j in range(randint(5, 10)):
        sublist.append(randint(0, 2**24))
      L.append(sublist)
    for i in funcs:
      args.append([i, L, input_size])

pool = Pool()
for result in pool.imap(run_func, args):
  func_name, input_size, elapsed, validation_result = result
  all_results[func_name][input_size].append({
    'time': elapsed,
    'validation': validation_result,
  })
  # show the running time for the function at this input size
  print(input_size, func_name, elapsed, validation_result)
pool.close()
pool.join()

# write the average of time trials at each size for each function
with open('times.tsv', 'w') as out:
  for func in all_results:
    validations = [i['validation'] for j in all_results[func] for i in all_results[func][j]]
    linetype = 'incorrect results' if any([i != 'true' for i in validations]) else 'correct results'

    for input_size in all_results[func]:
      all_times = [i['time'].microseconds for i in all_results[func][input_size]]
      avg_time = sum(all_times) / len(all_times)

      out.write(func + '\t' + str(input_size) + '\t' + \
        str(avg_time) + '\t' + linetype + '\n')

以及用于绘图的代码：

library(ggplot2)
df <- read.table('times.tsv', sep='\t')

p <- ggplot(df, aes(x=V2, y=V3, color=as.factor(V1))) +
  geom_line() +
  xlab('number of input lists') +
  ylab('runtime (ms)') +
  labs(color='') +
  scale_x_continuous(trans='log10') +
  facet_wrap(~V4, ncol=1)

ggsave('runtimes.png')

Answer 4

我遇到了尝试合并具有共同值的列表的相同问题。这个例子可能就是您正在寻找的。
它只循环列表一次并更新结果集。

lists = [['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g']]
lists = sorted([sorted(x) for x in lists]) #Sorts lists in place so you dont miss things. Trust me, needs to be done.

resultslist = [] #Create the empty result list.

if len(lists) >= 1: # If your list is empty then you dont need to do anything.
    resultlist = [lists[0]] #Add the first item to your resultset
    if len(lists) > 1: #If there is only one list in your list then you dont need to do anything.
        for l in lists[1:]: #Loop through lists starting at list 1
            listset = set(l) #Turn you list into a set
            merged = False #Trigger
            for index in range(len(resultlist)): #Use indexes of the list for speed.
                rset = set(resultlist[index]) #Get list from you resultset as a set
                if len(listset & rset) != 0: #If listset and rset have a common value then the len will be greater than 1
                    resultlist[index] = list(listset | rset) #Update the resultlist with the updated union of listset and rset
                    merged = True #Turn trigger to True
                    break #Because you found a match there is no need to continue the for loop.
            if not merged: #If there was no match then add the list to the resultset, so it doesnt get left out.
                resultlist.append(l)
print resultlist

#

resultset = [['a', 'b', 'c', 'd', 'e', 'g', 'f', 'o', 'p'], ['k']]

Answer 5

我发现 itertools 是合并列表的快速选项，它为我解决了这个问题：

import itertools

LL = set(itertools.chain.from_iterable(L)) 
# LL is {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'k', 'o', 'p'}

for each in LL:
  components = [x for x in L if each in x]
  for i in components:
    L.remove(i)
  L += [list(set(itertools.chain.from_iterable(components)))]

# then L = [['k'], ['a', 'c', 'b', 'e', 'd', 'g', 'f', 'o', 'p']]

对于大型集合，按频率从最常见的元素到最少的元素对 LL 进行排序可以加快速度

Answer 6

我认为这可以通过将问题建模为图表来解决。每个子列表都是一个节点，并且仅当两个子列表具有某些共同元素时才与另一个节点共享边。因此，合并的子列表基本上是图中的连接组件。合并所有组件只需找到所有连接的组件并列出它们即可。

这可以通过对图的简单遍历来完成。 BFS 和 DFS 可以使用，但我在这里使用 DFS，因为它对我来说有点短。

l = [['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g']]
taken=[False]*len(l)
l=[set(elem) for elem in l]

def dfs(node,index):
    taken[index]=True
    ret=node
    for i,item in enumerate(l):
        if not taken[i] and not ret.isdisjoint(item):
            ret.update(dfs(item,i))
    return ret

def merge_all():
    ret=[]
    for i,node in enumerate(l):
        if not taken[i]:
            ret.append(list(dfs(node,i)))
    return ret

print(merge_all())

Answer 7

正如 Jochen Ritzel 指出的，您正在寻找图中的连通分量。以下是在不使用图形库的情况下实现它的方法：

from collections import defaultdict

def connected_components(lists):
    neighbors = defaultdict(set)
    seen = set()
    for each in lists:
        for item in each:
            neighbors[item].update(each)
    def component(node, neighbors=neighbors, seen=seen, see=seen.add):
        nodes = set([node])
        next_node = nodes.pop
        while nodes:
            node = next_node()
            see(node)
            nodes |= neighbors[node] - seen
            yield node
    for node in neighbors:
        if node not in seen:
            yield sorted(component(node))

L = [['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g']]
print list(connected_components(L))

Answer 8

您可以使用networkx库，因为它是图论和连接组件问题：

import networkx as nx

L = [['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g']]

G = nx.Graph()

#Add nodes to Graph    
G.add_nodes_from(sum(L, []))

#Create edges from list of nodes
q = [[(s[i],s[i+1]) for i in range(len(s)-1)] for s in L]

for i in q:

    #Add edges to Graph
    G.add_edges_from(i)

#Find all connnected components in graph and list nodes for each component
[list(i) for i in nx.connected_components(G)]

输出：

[['p', 'c', 'f', 'g', 'o', 'a', 'd', 'b', 'e'], ['k']]

Answer 9

我怀念一个非古怪的版本。我在 2018 年（7 年后）发布了它

简单且不稳定方法：

1）使笛卡尔积（交叉连接）合并两个共同的 if 元素
2）删除重复项

#your list
l=[['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g']]

#import itertools
from itertools import product, groupby

#inner lists to sets (to list of sets)
l=[set(x) for x in l]

#cartesian product merging elements if some element in common
for a,b in product(l,l):
    if a.intersection( b ):
       a.update(b)
       b.update(a)

#back to list of lists
l = sorted( [sorted(list(x)) for x in l])

#remove dups
list(l for l,_ in groupby(l))

#result
[['a', 'b', 'c', 'd', 'e', 'f', 'g', 'o', 'p'], ['k']]

Answer 10

我的尝试。具有实用的外观。

#!/usr/bin/python
from collections import defaultdict
l = [['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g']]
hashdict = defaultdict(int)

def hashit(x, y):
    for i in y: x[i] += 1
    return x

def merge(x, y):
    sums = sum([hashdict[i] for i in y])
    if sums > len(y):
        x[0] = x[0].union(y)
    else:
        x[1] = x[1].union(y)
    return x


hashdict = reduce(hashit, l, hashdict)
sets = reduce(merge, l, [set(),set()])
print [list(sets[0]), list(sets[1])]

Answer 11

这是一个相当快的解决方案，没有依赖性。它的工作原理如下：

1. 为每个 subsists 分配一个唯一的参考号（在本例中为子列表的初始索引）

2. 为每个子列表以及每个子列表中的每个项目创建引用元素的字典。

3. 重复以下过程，直到没有任何变化：

   3a。浏览每个子列表中的每个项目。如果该项目的当前引用编号与其子列表的引用编号不同，则该元素必须是两个列表的一部分。合并两个列表（从引用中删除当前子列表）并将当前子列表中所有项目的引用编号设置为新子列表的引用编号。

当此过程没有引起任何变化时，这是因为所有元素都属于一个列表。由于工作集的大小在每次迭代中都会减小，因此算法必然终止。

   def merge_overlapping_sublists(lst):
    output, refs = {}, {}
    for index, sublist in enumerate(lst):
        output[index] = set(sublist)
        for elem in sublist:
            refs[elem] = index
    changes = True
    while changes:
        changes = False
        for ref_num, sublist in list(output.items()):
            for elem in sublist:
                current_ref_num = refs[elem]
                if current_ref_num != ref_num:
                    changes = True
                    output[current_ref_num] |= sublist
                    for elem2 in sublist:
                        refs[elem2] = current_ref_num
                    output.pop(ref_num)
                    break
    return list(output.values())

以下是对此代码的一组测试：

def compare(a, b):
    a = list(b)
    try:
        for elem in a:
            b.remove(elem)
    except ValueError:
        return False
    return not b

import random
lst = [["a", "b"], ["b", "c"], ["c", "d"], ["d", "e"]]
random.shuffle(lst)
assert compare(merge_overlapping_sublists(lst), [{"a", "b", "c", "d", "e"}])
lst = [["a", "b"], ["b", "c"], ["f", "d"], ["d", "e"]]
random.shuffle(lst)
assert compare(merge_overlapping_sublists(lst), [{"a", "b", "c",}, {"d", "e", "f"}])
lst = [["a", "b"], ["k", "c"], ["f", "g"], ["d", "e"]]
random.shuffle(lst)
assert compare(merge_overlapping_sublists(lst), [{"a", "b"}, {"k", "c"}, {"f", "g"}, {"d", "e"}])
lst = [["a", "b", "c"], ["b", "d", "e"], ["k"], ["o", "p"], ["e", "f"], ["p", "a"], ["d", "g"]]
random.shuffle(lst)
assert compare(merge_overlapping_sublists(lst), [{"k"}, {"a", "c", "b", "e", "d", "g", "f", "o", "p"}])    
lst = [["a", "b"], ["b", "c"], ["a"], ["a"], ["b"]]
random.shuffle(lst)
assert compare(merge_overlapping_sublists(lst), [{"a", "b", "c"}])

请注意，返回值是一个集合列表。

Answer 12

在不太了解你想要什么的情况下，我决定猜测你的意思：我想只找到每个元素一次。

#!/usr/bin/python


def clink(l, acc):
  for sub in l:
    if sub.__class__ == list:
      clink(sub, acc)
    else:
      acc[sub]=1

def clunk(l):
  acc = {}
  clink(l, acc)
  print acc.keys()

l = [['a', 'b', 'c'], ['b', 'd', 'e'], ['k'], ['o', 'p'], ['e', 'f'], ['p', 'a'], ['d', 'g']]

clunk(l)

输出看起来像：

['a', 'c', 'b', 'e', 'd', 'g', 'f', 'k', 'o', 'p']

Answer 13

这可能是一个更简单/更快的算法，并且似乎运行良好 -

l = [['a', 'b', 'c'], ['b', 'd', 'e'], ['k'], ['o', 'p'], ['e', 'f'], ['p', 'a'], ['d', 'g']]

len_l = len(l)
i = 0
while i < (len_l - 1):
    for j in range(i + 1, len_l):

        # i,j iterate over all pairs of l's elements including new 
        # elements from merged pairs. We use len_l because len(l)
        # may change as we iterate
        i_set = set(l[i])
        j_set = set(l[j])

        if len(i_set.intersection(j_set)) > 0:
            # Remove these two from list
            l.pop(j)
            l.pop(i)

            # Merge them and append to the orig. list
            ij_union = list(i_set.union(j_set))
            l.append(ij_union)

            # len(l) has changed
            len_l -= 1

            # adjust 'i' because elements shifted
            i -= 1

            # abort inner loop, continue with next l[i]
            break

    i += 1

print l
# prints [['k'], ['a', 'c', 'b', 'e', 'd', 'g', 'f', 'o', 'p']]

Answer 14

简单来说，你可以使用快速查找。

关键是要使用两个临时列表。
第一个称为元素，它存储所有组中存在的所有元素。
第二个名为标签。我从sklearn的kmeans算法中得到了灵感。 “labels”存储元素的标签或质心。这里我简单地让簇中的第一个元素作为质心。最初，这些值从 0 到 length-1，按升序排列。

对于每个组，我在“元素”中获取他们的“索引”。
然后，我根据索引获取组标签。
我计算标签的最小值，这将是它们的新标签。
我将组标签中带有标签的所有元素替换为新标签。

或者说，对于每次迭代，
我尝试合并两个或多个现有小组。
如果该组的标签为 0 和 2
我找到了新标签 0，这是两者中的最小值。
我将它们替换为 0。

def cluser_combine(groups):
    n_groups=len(groups)

    #first, we put all elements appeared in 'gruops' into 'elements'.
    elements=list(set.union(*[set(g) for g in groups]))
    #and sort elements.
    elements.sort()
    n_elements=len(elements)

    #I create a list called clusters, this is the key of this algorithm.
    #I was inspired by sklearn kmeans implementation.
    #they have an attribute called labels_
    #the url is here:
    #https://scikit-learn.org/stable/modules/generated/sklearn.cluster.KMeans.html
    #i called this algorithm cluster combine, because of this inspiration.
    labels=list(range(n_elements))


    #for each group, I get their 'indices' in 'elements'
    #I then get the labels for indices.
    #and i calculate the min of the labels, that will be the new label for them.
    #I replace all elements with labels in labels_for_group with the new label.

    #or to say, for each iteration,
    #i try to combine two or more existing groups.
    #if the group has labels of 0 and 2
    #i find out the new label 0, that is the min of the two.
    #i than replace them with 0.
    for i in range(n_groups):

        #if there is only zero/one element in the group, skip
        if len(groups[i])<=1:
            continue

        indices=list(map(elements.index, groups[i]))

        labels_for_group=list(set([labels[i] for i in indices]))
        #if their is only one label, all the elements in group are already have the same label, skip.
        if len(labels_for_group)==1:

            continue

        labels_for_group.sort()
        label=labels_for_group[0]

        #combine
        for k in range(n_elements):
            if labels[k] in labels_for_group[1:]:
                labels[k]=label


    new_groups=[]
    for label in set(labels):
        new_group = [elements[i] for i, v in enumerate(labels) if v == label]
        new_groups.append(new_group)

    return new_groups

我打印出了您问题的详细结果：

cluser_combine([['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g']])

元素：
['a'、'b'、'c'、'd'、'e'、'f'、'g'、'k'、'o'、'p']
标签：
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
--------------------第 0 组------------------------
该组是：
['a'、'b'、'c']
元素中组的索引
[0, 1, 2]
组合前的标签
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
结合...
组合后的标签
[0, 0, 0, 3, 4, 5, 6, 7, 8, 9]
--------------------第 1 组-------------------------
该组是：
['b'、'd'、'e']
元素中组的索引
[1,3,4]
组合前的标签
[0, 0, 0, 3, 4, 5, 6, 7, 8, 9]
结合...
组合后的标签
[0, 0, 0, 0, 0, 5, 6, 7, 8, 9]
--------------------第2组------------------------------------
该组是：
['k']
--------------------第3组------------------------------------
该组是：
['o', 'p']
元素中组的索引
[8, 9]
组合前的标签
[0, 0, 0, 0, 0, 5, 6, 7, 8, 9]
结合...
组合后的标签
[0, 0, 0, 0, 0, 5, 6, 7, 8, 8]
--------------------第 4 组------------------------
该组是：
['e', 'f']
元素中组的索引
[4, 5]
组合前的标签
[0, 0, 0, 0, 0, 5, 6, 7, 8, 8]
结合...
组合后的标签
[0, 0, 0, 0, 0, 0, 6, 7, 8, 8]
------------------第5组------------------------------------
该组是：
['p', 'a']
元素中组的索引
[9, 0]
组合前的标签
[0, 0, 0, 0, 0, 0, 6, 7, 8, 8]
结合...
组合后的标签
[0, 0, 0, 0, 0, 0, 6, 7, 0, 0]
--------------------第 6 组------------------------------------
该组是：
['d', 'g']
元素中组的索引
[3, 6]
组合前的标签
[0, 0, 0, 0, 0, 0, 6, 7, 0, 0]
结合...
组合后的标签
[0, 0, 0, 0, 0, 0, 0, 7, 0, 0]
([0, 0, 0, 0, 0, 0, 0, 7, 0, 0],
[['a', 'b', 'c', 'd', 'e', 'f', 'g', 'o', 'p'], ['k']])
< /p>

请有关详细信息，请参阅我的github jupyter笔记本

Answer 15

这是我的答案。

orig = [['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g'], ['k'],['k'],['k']]

def merge_lists(orig):
    def step(orig): 
        mid = []
        mid.append(orig[0])
        for i in range(len(mid)):            
            for j in range(1,len(orig)):                
                for k in orig[j]:
                    if k in mid[i]:                
                        mid[i].extend(orig[j])                
                        break
                    elif k == orig[j][-1] and orig[j] not in mid:
                        mid.append(orig[j])                        
        mid = [sorted(list(set(x))) for x in mid]
        return mid

    result = step(orig)
    while result != step(result):                    
        result = step(result)                  
    return result

merge_lists(orig)
[['a', 'b', 'c', 'd', 'e', 'f', 'g', 'o', 'p'], ['k']]