说我有一个列表如下：
l = [[1,2,3],[6,5,4,3,7,2],[4,3,2,9],[6,7],[5,1,0],[6,3,2,7]]
我如何编写python代码来检查是否有总是一起出现的元素？例如，在上面的例子中，2,3和6,7总是出现在相同的列表中。 （可能还有其他人，不确定）。
最容易理解的方法是什么？
我唯一的想法是将inner-list1转换为设置并检查与inner-list2的交叉点，但当我检查与inner-list3的交叉点时，这些元素可能根本不会出现在inner-list3中
我可以这样做：
for i in range(0,len(lists)):
a=set(lists[i]).intersection(lists[i+1])
if (len(a))==0:
continue
else:
a.intersection(lists[i+1])
这当然不起作用，但我怎么能正式编码呢？或者有更好的方法吗？
答案 0 :(得分：5)
itertools.combinations：我最初想过使用itertools.combination，的内容，但因为这允许来自 elements的{{1}}彼此不相邻，所以它不会起作用我想到的解决方案。
事实证明，在查看非数字输入list时，在两种情况下都需要lists 。我很困惑因为我认为itertools.combinations必须是groups。
我认为最适合这种方式的方法是生成可以工作的可能adjacent，然后针对{elements检查每一个<{1}} {1}}的{1}} - 而不是在function上进行某种组合工作并沿着这条路走下去。
因此要检查可能list的{{1}}是否“有效”，即如果所有sub-lists仅一起出现，我使用带有生成器的简单list list和elements内置elements来完成这部分工作。
现在这个工作正常，需要有一种方法来产生可能发生的潜在if。我刚刚使用两个嵌套all() - 一个any()覆盖functions elements，一个for-loops覆盖iterating width是。
然后，从这里开始，我们只检查window的{{1}}是否为iterating，如果是start则将其添加到另一个window。
elements将valid作为：
list这些import itertools
def valid(p):
for s in l:
if any(e in s for e in p) and not all(e in s for e in p):
return False
return True
l = [[1,2,3],[6,5,4,3,7,2],[4,3,2,9],[6,7],[5,1,0],[6,3,2,7]]
els = list(set(b for a in l for b in a))
sol = []
for w in range(2,len(els)+1):
for c in itertools.combinations(els, w):
if valid(c):
sol.append(c)
sol实际上可以拼凑成一个不错的[(2, 3), (6, 7)]]
（不确定其他人是否认为它是Pythonic）：
2工作原理相同但只是更短。
由于受欢迎的需求（nested for-loops），我已更新答案，以便它现在应该适用于除one-liner之外的其他sol = [c for w in range(2, len(els)+1) for c in itertools.combinations(els, w) if valid(c)]
。这是通过引入一个独特的@Arman elements（0-9）完成的。
来自elements的一些测试与上面的代码相同：
list将els作为：
@thanasisp再次与：
l = [[1, 3, 5, 7],[1, 3, 5, 7]]
给出：
sol我认为这是正确的，因为[(1, 3), (1, 5), (1, 7), (3, 5), (3, 7), (5, 7), (1, 3, 5), (1, 3, 7), (1, 5, 7), (3, 5, 7), (1, 3, 5, 7)]
不应该在任何群组中，因为所有其他 l = [[1, 2, 3, 5, 7], [1, 3, 5, 7]]
都在不同的 [(1, 3), (1, 5), (1, 7), (3, 5), (3, 7), (5, 7), (1, 3, 5), (1, 3, 7), (1, 5, 7), (3, 5, 7), (1, 3, 5, 7)]
中，所以它可以永远不要与另一个2组成一个小组。
答案 1 :(得分：3)
使用默认dicts的另一个线性解决方案（元组可用于制作可散列密钥）：
from collections import defaultdict
isin,contains = defaultdict(list),defaultdict(list)
for i,s in enumerate(l):
for k in s :
isin[k].append(i)
# isin is {1: [0, 4], 2: [0, 1, 2, 5], 3: [0, 1, 2, 5], 6: [1, 3, 5],
# 5: [1, 4], 4: [1, 2], 7: [1, 3, 5], 9: [2], 0: [4]}
# element 1 is in sets numbered 0 and 4, and so on.
for k,ss in isin.items():
contains[tuple(ss)].append(k)
# contains is {(0, 4): [1], (0, 1, 2, 5): [2, 3], (1, 3, 5): [6, 7],
# (1, 4): [5], (1, 2): [4], (2,): [9], (4,): [0]})
# sets 0 and 4 contains 1, and no other contain 1.
现在，如果您查找按n（n=2此处）组显示的元素，请输入：
print ([p for p in contains.values() if len(p)==n])
# [[2, 3], [6, 7]]
答案 2 :(得分：2)
这是我现在想到的强力选项，dct是每个数字的计数字典，然后我们检查dct中的相同列表，这意味着两个数字出现在相同的列表索引中：< / p>
l = [[1,2,3],[6,5,4,3,7,2,1],[4,3,2,9,1],[6,7],[5,1,2,3,0],[6,3,2,7,1]]
dct = defaultdict(list)
for i, v in enumerate(l):
for x in v:
dct[x].append(i)
dct # defaultdict(<class 'list'>, {0: [4], 1: [0, 1, 2, 4, 5], 2: [0, 1, 2, 4, 5], 3: [0, 1, 2, 4, 5], 4: [1, 2], 5: [1, 4], 6: [1, 3, 5], 7: [1, 3, 5], 9: [2]})
new_d = defaultdict(list)
for k, v in dct.items():
for k2, v2 in dct.items():
if(v == v2) and k != k2):
new_d[k].append(k2)
new_d # defaultdict(<class 'list'>, {1: [2, 3], 2: [1, 3], 3: [1, 2], 6: [7], 7: [6]})
这也是一项非常昂贵的操作，它是O(N*N*M)：N = list elements和M = longest sublist
答案 3 :(得分：2)
您可以使用设置交叉点执行此操作，并且每个组还可以很好地处理3个或更多元素：请注意，我向8组添加了6,7。
lists = [[1,2,3], [6,5,4,3,7,2,8], [4,3,2,9], [8,6,7], [5,1,0], [6,3,8,2,7]]
首先，我们将每个元素映射到它出现的所有其他元素的集合：
groups = {}
for lst in lists:
for x in lst:
if x not in groups:
groups[x] = set(lst)
else:
groups[x].intersection_update(lst)
# {0: {0, 1, 5}, 1: {1}, 2: {2, 3}, 3: {2, 3}, 4: {2, 3, 4}, 5: {5},
# 6: {8, 6, 7}, 7: {8, 6, 7}, 8: {8, 6, 7}, 9: {9, 2, 3, 4}}
接下来，我们只保留关系是双向的那些元素：
groups2 = {k: {v for v in groups[k] if k in groups[v]} for k in groups}
# {0: {0}, 1: {1}, 2: {2, 3}, 3: {2, 3}, 4: {4}, 5: {5},
# 6: {8, 6, 7}, 7: {8, 6, 7}, 8: {8, 6, 7}, 9: {9}}
最后，我们得到具有多个元素的唯一组：
groups3 = {frozenset(v) for v in groups2.values() if len(v) > 1}
# {frozenset({8, 6, 7}), frozenset({2, 3})}
答案 4 :(得分：1)
以下解决方案具有线性 O(n)复杂度，其中n是所有列表中的数字总数（展平后）。代码为Python2.x
我使用了所有可能模式的位图表示（使用python的无限数字）。例如，如果list0和list2但list1中存在数字，则相应的模式将为...000101。例如，在给定的输入中，值2将具有以下位图模式：100111，因此值3
l = [[1,2,3],[6,5,4,3,7,2],[4,3,2,9],[6,7],[5,1,0],[6,3,2,7]]
num_to_pattern = {}
for i, sublist in enumerate(l):
for num in sublist:
# turning ON the respective bit for each value
if not num in num_to_pattern:
num_to_pattern[num] = 1 << i
else:
num_to_pattern[num] |= (1 << i)
pattern_to_num_list = {}
# mapping patterns to all their respective numbers
for num, pattern in num_to_pattern.iteritems():
if not pattern in pattern_to_num_list:
pattern_to_num_list[pattern] = [num]
else:
pattern_to_num_list[pattern].append(num)
print pattern_to_num_list
此代码将打印：
{4: [9], 6: [4], 39: [2, 3], 42: [6, 7], 16: [0], 17: [1], 18: [5]}
您可以映射和过滤您喜欢的任何子列表（在您的情况下 - 列表等于或大于2）：
print filter(lambda x: len(x) >= 2, pattern_to_num_list.values())
答案 5 :(得分：1)
实现这一目标最容易理解的是什么？
我试图让我的解决方案尽可能短。我也尽可能地优化它。它可以根据您的喜好使用任何整数。
以下是包含解释的大量评论的代码，然后是更多基本解释：
注意：在以下代码中，我使用[[1, 2, 3], [2, 1, 4]]作为原始列表的示例，而不是您问题中的示例，以便于解释。
import itertools
# The original list of lists
org_list = [[1, 2, 3], [2, 1, 4]]
# Sort the lists of org_list to ensure that the resulting tuples of
# itertools.combinations below are sorted also, because later, we
# don't want (1, 2) to be not equal to (2, 1)
org_list = [sorted(l) for l in org_list]
# This list will contain the combinations of the original list
list_of_combinations = []
# --Building list_of_combinations--
# Looping through every list in the original list of lists (org_list)
for i, l in enumerate(org_list):
# Create a new set to hold the combinations for the i-th list of org_list
list_of_combinations.append(set())
# Starting with 2 because we want the combination to contain two
# items at least, and ending at len(org_list[i])+1 because we want
# the maximum length of the combination to be equal to the length
# of its original list
for comb_length in range(2, len(l) + 1):
# Update the set with its combinations of length comb_length
list_of_combinations[i].update(
tuple(itertools.combinations(org_list[i], comb_length))
)
# Now list_of_combinations = [
# {(1, 2), (1, 3), (2, 3), (1, 2, 3)},
# {(1, 2), (1, 2, 4), (2, 4), (1, 4)}
# ]
# This will hold the result. In our case: [2, 3], and [6, 7]
# It is a set because we don't want the result to contain duplicate items
combs = set()
# Looping through the sets in list_of_combinations
for s in list_of_combinations:
# s = {(1, 2), (1, 3), (2, 3), (1, 2, 3)} for example
# Looping through the combinations in the set s
for comb in s:
# comb = (1, 2) for example
# Set a flag (f) initially to 1
f = 1
# Loop through the sets in list_of_combinations
for ind, se in enumerate(list_of_combinations):
# See if comb exists in the set se
if comb not in se:
# If not, see if any number in comb exists in the ind-th list of
# the original list
for n in comb:
if n in org_list[ind]:
# If so, set f to 0
f = 0
break
# if f is still 1, then the current comb satisfy our conditions
# so we add it to the result
if f == 1:
combs.add(comb)
print(combs)
输出：
{(1, 2)}
正如所料。
对于您问题中的列表，此代码的输出为{(2, 3), (6, 7)}，这也是预期的。
itertools.combinations？ itertools.combinations(iterable, r)：从输入r返回元素的iterable长度元组。例如：
list(itertools.combinations([1, 2, 3], 2))
给出
[(1, 2), (1, 3), (2, 3)]
在上面的代码中，您可以注意到这些集用于保存原始列表中每个列表的组合。这是因为检查集合中某个值的成员资格是非常快，我们在代码中进行了很多这样的检查。
假设我们的原始列表是[[1, 2, 3], [2, 1, 4]]。
为原始列表中的每个列表获取所需的一组组合：
对于[1, 2, 3]：组合集合为(1, 2), (1, 3), (2, 3), (1, 2, 3)
对于[2, 1, 4]：组合的集合为(1, 2), (1, 2, 4), (2, 4), (1, 4)
对于每个组合，并且为了在我们的代码输出中（意味着它满足我们的条件），我们要确保每个组合，
让我们从第一个组合中获取(1, 3)。我们遍历组合集：
对于第一个集，我们可以看到(1, 3)存在于其中，因此我们继续前进。
对于第二个集，我们可以看到它不存在，所以我们想知道相应列表中是否存在任何项目（即原始列表的第二个列表：[2, 1, 4]）：
从1开始，我们可以看到它存在于相应的列表中 - ＆gt; (1, 3)不能出现在输出中，因为它不满足所需的条件。
答案 6 :(得分：1)
首先，数据
data = [[1,2,3],[6,5,4,3,7,2],[4,3,2,9],[6,7],[5,1,0],[6,3,2,7]]
生成组合很昂贵，所以我想尽可能地避免这种情况。
我和＃34;尤里卡！＆＃34;当我意识到我不必产生所有配对时，那一刻就来了。相反，我可以将每个数字映射到包含它的所有列表。
appears_in = defaultdict(set)
for g in groups:
for number in g:
appears_in[number].add(tuple(g))
结果字典是
{0: {(5, 1, 0)},
1: {(5, 1, 0), (1, 2, 3)},
2: {(4, 3, 2, 9), (6, 3, 2, 7), (6, 5, 4, 3, 7, 2), (1, 2, 3)},
3: {(4, 3, 2, 9), (6, 3, 2, 7), (6, 5, 4, 3, 7, 2), (1, 2, 3)},
4: {(4, 3, 2, 9), (6, 5, 4, 3, 7, 2)},
5: {(5, 1, 0), (6, 5, 4, 3, 7, 2)},
6: {(6, 3, 2, 7), (6, 7), (6, 5, 4, 3, 7, 2)},
7: {(6, 3, 2, 7), (6, 7), (6, 5, 4, 3, 7, 2)},
9: {(4, 3, 2, 9)}}
查看2和3的条目
2: {(4, 3, 2, 9), (6, 3, 2, 7), (6, 5, 4, 3, 7, 2), (1, 2, 3)},
3: {(4, 3, 2, 9), (6, 3, 2, 7), (6, 5, 4, 3, 7, 2), (1, 2, 3)},
包含2的列表集与包含3的列表集相同。所以我得出结论，2和3总是一起出现。
将此与3和4对比
3: {(4, 3, 2, 9), (6, 3, 2, 7), (6, 5, 4, 3, 7, 2), (1, 2, 3)},
4: {(4, 3, 2, 9), (6, 5, 4, 3, 7, 2)},
请注意(6, 3, 2, 7)和(1, 2, 3)应该有的差距。我得出结论，3和4并不总是一起出现。
这是完整的代码
from collections import defaultdict
from itertools import combinations
from pprint import pprint
def always_appear_together(groups):
appears_in = defaultdict(set)
for g in groups:
for number in g:
appears_in[number].add(tuple(g))
#pprint(appears_in) # for debugging
return [
(i,j)
for (i,val_i),(j,val_j) in combinations(appears_in.items(),2)
if val_i == val_j
]
运行此功能
print(always_appear_together(data))
[(2, 3), (6, 7)]
答案 7 :(得分：0)
这更像是一个强力解决方案，但是，它会生成一个包含所有元素的大型列表，这些元素通过生成l中每个子列表的排列并过滤以查找其元素全部的所有排列出现在l的子列表中。如果任何排列通过该条件，则排列将添加到final_pairs：
l = [[1,2,3],[6,5,4,3,7,2],[4,3,2,9],[6,7],[5,1,0],[6,3,2,7]]
import itertools
final_pairs = []
for i in l:
combos = [list(itertools.permutations(i, b)) for b in range(2, len(i))]
for combo in combos:
for b in combo:
if any(all(c in a for c in b) for a in l):
final_pairs.append(combo)
final_data = list(set(itertools.chain.from_iterable(final_pairs)))
输出：
[(2, 5, 6, 7, 3), (7, 3), (2, 6, 3, 7), (5, 3, 2, 6, 7), (5, 6, 4, 7), (7, 2, 5, 4, 6), (6, 7, 3, 4), (5, 2, 3, 7, 6), (7, 4, 3, 2), (6, 4, 7, 2), (4, 7, 6), (7, 3, 4, 6, 2), (5, 3, 7, 2, 6), (5, 7, 6, 4), (7, 4, 6, 2, 5), (7, 5, 4, 6, 3), (4, 2, 7, 3, 5), (4, 7, 3, 2), (2, 5, 4, 7, 3), (6, 5, 7, 2, 4), (4, 6, 7, 2), (2, 7, 5, 6, 3), (2, 6, 7), (5, 4, 2, 3, 7), (2, 3, 4, 6, 5), (5, 7, 2, 3), (3, 2, 4, 7, 6), (2, 6, 3, 5, 7), (3, 6, 5, 4, 7), (6, 5, 7), (2, 4, 6, 7, 5), (4, 3, 5, 2), (2, 3, 5, 7), (4, 5, 7, 3), (4, 6, 7, 2, 5), (3, 4, 5, 7, 2), (2, 4, 5, 6, 3), (3, 5, 2, 7, 6), (6, 3, 5, 7, 2), (5, 2, 7, 3, 6), (6, 3, 5, 4, 2), (2, 7, 4, 5), (2, 5, 3), (3, 2), (3, 2, 6, 7), (5, 3, 7, 6, 4), (4, 5), (2, 7, 3, 6, 4), (6, 4, 2, 5), (7, 5, 4, 2, 6), (2, 4, 3, 7, 6), (3, 2, 6), (4, 5, 3, 6), (7, 4, 3, 6, 5), (7, 3, 4), (5, 3, 4, 6, 7), (6, 5, 3, 2, 4), (6, 4, 2, 3), (5, 2, 7, 6, 3), (5, 4, 6, 3, 7), (3, 2, 6, 5, 7), (6, 5, 4, 3, 7), (3, 5, 2, 6, 4), (7, 3, 6, 2, 5), (2, 3, 7, 6, 4), (3, 4, 5, 2, 7), (7, 3, 5, 2), (2, 4, 5, 7), (2, 3, 6, 4), (7, 5, 6, 4), (7, 6, 2), (3, 9, 4), (4, 6, 5), (6, 4, 5, 3, 2), (6, 7, 3, 2, 5), (3, 5, 7, 6), (2, 5, 3, 4, 6), (5, 3, 6), (2, 3, 4, 6, 7), (6, 5, 2, 3, 7), (6, 3, 5, 2, 4), (5, 4, 2, 3), (5, 7, 6, 3, 2), (4, 6, 5, 2, 7), (7, 5, 2, 3), (4, 5, 2, 6, 3), (5, 7, 6, 3), (2, 7, 3, 4, 6), (2, 3, 6), (7, 4, 3, 5), (4, 3, 5, 6, 7), (7, 3, 6, 5, 2), (6, 2, 5, 3, 7), (5, 6, 4), (5, 2, 7, 6), (4, 6, 2, 3), (4, 3, 2, 6, 7), (3, 2, 7, 5), (6, 7, 2, 4, 5), (4, 3, 6, 2), (4, 3, 6, 7, 2), (6, 7, 4, 3, 2), (5, 1), (5, 7, 4, 3, 2), (6, 3, 7), (6, 7, 3, 4, 2), (7, 6, 3, 5, 2), (4, 9, 3), (4, 7, 5, 2), (5, 4, 2, 7, 6), (5, 3, 7, 2, 4), (3, 2, 5, 4, 7), (4, 2, 5, 7, 6), (3, 7, 6, 4), (7, 3, 2, 6, 4), (7, 2, 5, 3, 6), (2, 3, 5, 6, 4), (4, 5, 2, 3, 6), (5, 6, 7, 4, 3), (4, 2, 6, 5, 7), (6, 2, 3, 7), (7, 4, 5, 3), (5, 3, 4, 2, 7), (5, 7, 3), (5, 7, 3, 2, 6), (3, 5, 2, 7), (2, 7, 6, 5, 4), (4, 6, 5, 7), (3, 4, 7, 6, 5), (6, 2, 3, 5, 7), (6, 5, 3, 4, 2), (5, 4, 7, 2), (5, 7, 4, 6), (7, 6, 2, 5), (3, 4, 9), (6, 4, 5, 7, 2), (4, 7, 5, 3, 2), (3, 5, 6, 2), (4, 7, 2, 6, 3), (5, 4, 7), (5, 3, 7, 6, 2), (2, 4, 3, 5, 7), (1, 0), (3, 2, 6, 7, 5), (2, 3, 4, 7, 6), (6, 5, 2, 7), (7, 5, 2, 4, 3), (5, 3, 6, 2, 4), (2, 7), (2, 3, 6, 5, 7), (5, 3, 2, 6), (2, 6, 3, 4, 5), (6, 3, 7, 4, 5), (5, 6, 4, 2, 3), (2, 6, 5, 3, 4), (3, 4, 2, 7, 5), (5, 7, 3, 6, 4), (6, 3, 4, 5), (7, 4), (6, 7, 5), (7, 4, 6, 2), (6, 4, 3, 2, 7), (3, 5, 6), (3, 5, 6, 4, 2), (7, 2, 4), (2, 3, 6, 4, 5), (4, 2, 3), (2, 5, 3, 4, 7), (5, 2, 3, 6, 7), (4, 7, 6, 2), (3, 4, 6), (4, 3, 7, 6, 5), (7, 2, 4, 6, 5), (5, 3, 6, 7), (4, 6, 2, 5, 7), (6, 4, 3, 7, 2), (7, 4, 5, 2, 6), (3, 6, 7, 4, 5), (3, 6, 2, 5, 7), (3, 6, 2, 5), (5, 3, 4, 2, 6), (6, 5, 4, 3), (7, 4, 2, 3, 5), (2, 4, 5, 6, 7), (3, 7, 4, 5, 2), (2, 4, 7, 5), (5, 7, 3, 4), (7, 5, 4, 6), (4, 7, 6, 5, 3), (4, 3, 2, 6, 5), (7, 6, 2, 4, 5), (6, 3, 4), (3, 4, 6, 2, 5), (2, 5, 4, 6, 3), (2, 6, 3, 7, 5), (6, 7, 2, 5, 4), (6, 5, 7, 3, 2), 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