我需要将具有相同元素的子列表组合在一起，并且子列表中的所有元素在list2中都没有元素，并且在list3中有一个元素
例如：
list1 = [[1,4],[4,5],[5,7],[6,7],[9,7],[10,9],[8,10],[8,11],[8,13],[13,15]]
list2 = [7,8]
list3 = [1,6,11,13]
我会将[4,5]和[1,4]联系在一起，因为它们都包含1个相同的数字，而这两个数字会合并为[1,4,5]，并且1中包含list3并且7
list2
所以在链接之后，新列表应该是：
new_list1 = [[1,4,5],[6],[11],[13,15]]
IE：子列表中不应该有相同的数字，顺序并不重要。
更长的例子：
list1 = [[1,4],[4,5],[5,7],[6,7],[9,7],[10,9],[8,10],[8,11],[8,13],[6,8],[20,2],[20,3],[11,14],[14,16],[13,15]]
list2 = [7,8,20]
list3 = [1,2,3,16,15]
将是
new_list = [[1,4,5],[2,3],[11,14,16],[13,15]]
如何以一般方式完成？
修改 最终算法应包括以下三个基本步骤：
list1 list2所有子列表的所有元素
list1子列表list1 list3子列表
答案 0 :(得分：2)
如果托马斯·库恩能够正确地读懂你的想法，那就是我的看法：
def subgroup_join(data, exclude, include):
exclude = set(exclude) # turn into set for faster lookup/compare
include = set(include) # turn into set for faster lookup/compare
data = [set(element) - exclude for element in data] # remove excluded elements
results = [set()] # init with an empty set
for element in data: # loop through our remaining elements
groups = [] # store elements / current results filtered by exclude list
ignore_element = False # flag if we should add the element as a standalone
for result in results: # go through each subgroup in the results
if element & result: # if the current element has common items with the result
result |= element # ... concatenate both into a subgroup
ignore_element = True
groups.append(result) # add the current result subgroup
if not ignore_element: # add only if the element wasn't concatenated
groups.append(element) # add the current element
results = groups # our element store becomes our new results set
return sorted([sorted(res) for res in results if result & include]) # sort & return
至于测试：
list1 = [[1, 4], [4, 5], [5, 7], [6, 7], [7, 8], [9, 7], [10, 9], [8, 10], [8, 11], [8, 13], [13, 15]]
list2 = [7, 8]
list3 = [1, 6, 11, 13]
print(subgroup_join(list1, list2, list3))
# prints: [[1, 4, 5], [6], [11], [13, 15]]
list1 = [[1, 4], [4, 5], [5, 7], [6, 7], [9, 7], [10, 9], [8, 10], [8, 11], [8, 13], [6, 8], [20, 2], [20, 3], [11, 14], [14, 16], [13, 15]]
list2 = [7, 8, 20]
list3 = [1, 2, 3, 16, 15]
print(subgroup_join(list1, list2, list3))
# prints: [[1, 4, 5], [2], [3], [11, 14, 16], [13, 15]]
这可能是所提出的最快的方法，但同样 - 它与您的示例不完全匹配 - 请检查上一个结果集以及[2]和[3]结果。
<强>更新：
在性能方面，使用第二个列表组：
zwer_join - 100,000 loops: 2.849 s; per loop: 28.399 µs
kuhn_join - 100,000 loops: 3.071 s; per loop: 30.706 µs
nuag_join - 1,000 loops: 15.82 s; per loop: 15.819 ms (had to reduce the number of loops)
答案 1 :(得分：1)
此代码应该完成这项工作：
list1 = [[1,4],[4,5],[5,7],[6,7],[9,7],[10,9],[8,10],[8,11],[8,13],[6,8],[20,2],[20,3],[11,14],[14,16],[13,15]]
list2 = [7,8,20]
list3 = [1,2,3,16,15]
list1a = [set(l) for l in list1]
#removing elements from list1 that contain numbers of list2:
for x in list2:
for l in list(list1a):
if x in l:
l.remove(x)
#joining sub-lists in list1:
list1b = [set(l) for l in list1a]
list1c = []
while list1b:
s1 = list1b.pop(0)
for s2 in list(list1b):
if s1 & s2:
s1 |= s2
list1b.remove(s2)
list1c.append(s1)
#generating final list with only sub-lists that contain elements of list2
list1_new = sorted([sorted(list(s)) for s in list1c if s & set(list3)])
对于第一个例子，这给出了：
[[1, 4, 5], [6], [11], [13, 15]]
和第二个例子
[[1, 4, 5], [2], [3], [11, 14, 16], [13, 15]]
希望这有帮助。
答案 2 :(得分：0)
首先，您首先要编写join函数。我添加了第二个列表来删除不需要的元素。
然后，您遍历加入列表，查看列表中是否有任何元素。如果是，则查找它们所在的位置，然后添加元素（我使用set以避免重复）。
最后给出了输出。
def join(list1, list2):
l = []
for ee in list1:
# We consider here that list1 only have pairs
if ee[0] not in list2 and ee[1] not in list2:
flat_l = [x for e in l for x in e]
if ee[0] in flat_l or ee[1] in flat_l:
for i, e in enumerate(l):
if ee[0] in e:
l[i].append(ee[1])
if ee[1] in e:
l[i].append(ee[0])
else:
l.append(ee)
return l
def f(list1,list2,list3):
l = [[e] for e in list3]
list1 = join(list1, list2)
for ee in list1:
flat_l = [x for e in l for x in e]
for e in ee:
if e in flat_l:
for i in range(len(l)):
if e in l[i]:
l[i] = list(set(l[i]+ee))
print(l)
list1 = [[1,4],[4,5],[5,7],[6,7],[9,7],[10,9],[8,10],[8,11],[8,13],[13,15]]
list2 = [7,8]
list3 = [1,6,11,13]
f(list1,list2,list3)
# [[1, 4, 5], [6], [11], [13, 15]]
list1 = [[1,4],[4,5],[5,7],[6,7],[9,7],[10,9],[8,10],[8,11],[8,13],[6,8],[20,2],[20,3],[11,14],[14,16],[13,15]]
list2 = [7,8,20]
list3 = [1,2,3,16,15]
f(list1,list2,list3)
# [[1, 4, 5], [2], [3], [16, 11, 14], [13, 15]]