合并共享共同元素的列表
我的输入是一个列表列表。其中一些共享共同的元素，例如
L = [['a','b','c'],['b','d','e'],['k'],['o','p'],['e','f'],['p','a'],['d','g']]
我需要合并共享共同元素的所有列表，并且只要没有更多具有相同项目的列表，就重复此过程。我考虑过使用布尔运算和while循环，但无法找到一个好的解决方案。
最终结果应为：
L = [['a','b','c','d','e','f','g','o','p'],['k']]
15 个答案:
答案 0 :(得分：37)
您可以将列表视为图表的表示法，即['a','b','c']是一个图表，其中3个节点相互连接。您要解决的问题是找到connected components in this graph。
你可以使用NetworkX，这样做的好处是几乎可以肯定它是正确的：
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']]
要自己有效地解决这个问题，你必须将列表转换为图形化的东西，所以你不妨从一开始就使用networkX。
答案 1 :(得分：28)
算法：
- 从列表中取出第一组A
- 对于列表中的每个其他集合B，如果B具有将A连接B加入A的公共元素;从列表中删除B
- 重复2.直到与A 不再重叠
- 将A放入outpup
- 重复1.以列表的其余部分
所以你可能想要使用集而不是列表。以下程序应该这样做。
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)
答案 2 :(得分：7)
我遇到了尝试将列表与常用值合并的相同问题。这个例子可能就是你要找的东西。 它只循环遍历列表一次，并在结束时更新结果集。
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']]
答案 3 :(得分：6)
我认为这可以通过将问题建模为graph来解决。每个子列表都是一个节点，只有当两个子列表中有一些共同的元素时才与另一个节点共享边缘。因此，合并的子列表基本上是图中的connected component。合并所有这些只是找到所有连接组件并列出它们的问题。
这可以通过图表上的简单遍历来完成。可以使用BFS和DFS，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())
答案 4 :(得分：3)
作为Jochen Ritzel pointed out，您正在查找图表中的已连接组件。以下是如何在不使用图库的情况下实现它：
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))
答案 5 :(得分：2)
我的尝试。具有实用性。
#!/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])]
答案 6 :(得分：2)
我发现itertools是一个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，从最常见的元素到最少的频率对LL进行排序可以加快速度
答案 7 :(得分：2)
对于相当大的列表，我需要对OP描述的聚类技术进行数百万次，因此需要确定上面建议的哪种方法最准确，性能最高。
我针对上述每种方法的输入列表进行了10次试验，大小从2 ^ 1到2 ^ 10，使用每种方法的相同输入列表，并以毫秒为单位测量上面提出的每种算法的平均运行时间。结果如下：
这些结果帮助我看到了始终如一地返回正确结果的方法，@ jochen是最快的。在那些不能始终如一地返回正确结果的方法中，mak的解决方案通常不包括所有输入元素（即列表成员列表缺失），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，两个最快，最正确的算法在排名顺序上排名前两位。
以下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')
用于绘图：
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')
答案 8 :(得分：1)
这是一个没有依赖性的相当快速的解决方案。其工作原理如下：
-
为您的每个用户分配一个唯一的参考编号（在这种情况下，为子列表的初始索引）
-
为每个子列表以及每个子列表中的每个项目创建参考元素的字典。
-
重复以下过程，直到不引起任何变化：
3a。浏览每个子列表中的每个项目。如果该项目的当前参考编号与其子列表的参考编号不同，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号设置为新子列表的引用号。
此过程没有引起任何变化时，是因为所有元素都恰好属于一个列表。由于工作集每次迭代的大小都会减小，因此该算法必然会终止。
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，返回值是一组列表。
答案 9 :(得分：0)
在不知道你想要什么的情况下，我决定猜测你的意思：我想只找一次所有元素。
#!/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)
输出如下：
['a', 'c', 'b', 'e', 'd', 'g', 'f', 'k', 'o', 'p']
答案 10 :(得分：0)
这可能是一个更简单/更快的算法，似乎效果很好 -
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']]
答案 11 :(得分：0)
我想念一个非quirurgic版本。我张贴在2018年（7年后）
一种简单且不稳定的方法：
1）合并两个if元素，使笛卡尔积（交叉联接）
2）删除公仔
#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']]
答案 12 :(得分：0)
您可以使用networkx库，因为存在graph theory和connected components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']]
答案 13 :(得分：0)
简单来说，您可以使用快速查找。
关键是要使用两个临时列表。 第一个称为 elements ，它存储所有组中存在的所有元素。 第二个名称为标签。我从sklearn的kmeans算法中得到了启发。 “标签”存储元素的标签或质心。在这里，我只是让群集中的第一个元素为质心。最初，值从0到length-1递增。
对于每个小组，我都会在“元素”中得到他们的“指数”。 然后，我根据索引获得了组标签。 然后，我计算标签的最小值，这将是它们的新标签。 我将所有元素替换为新标签中的组标签中的标签。
或者说，对于每次迭代， 我尝试合并两个或多个现有组。 如果组的标签为0和2 我发现了新的标签0，这是两个的最小值。 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
我已打印出您问题的详细结果：
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-------------------------
小组是：
['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
[0，0，0，3，4，4，5，6，7，8，9]
--------------------组1 -------------------------
小组是：
['b'，'d'，'e']
元素中组的索引
[1、3、4]
组合前的标签
[0，0，0，3，4，4，5，6，7，8，9]
结合...
组合后的标签
[0，0，0，0，0，5，6，7，8，9]
--------------------组2 -------------------------
小组是：
['k']
--------------------组3 -------------------------
小组是：
['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
[0，0，0，0，0，5，6，7，8，8，8]
--------------------组4 -------------------------
小组是：
['e'，'f']
元素中组的索引
[4，5]
组合前的标签
[0，0，0，0，0，5，6，7，8，8，8]
结合...
组合后的标签
[0，0，0，0，0，0，6，7，8，8]
--------------------组5 -------------------------
小组是：
['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-------------------------
小组是：
['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']]）
有关详细信息，请参阅my github jupyter notebook
答案 14 :(得分：0)
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']]
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