检查元素是否在所有列表中一起出现？

Question

假设我有一个这样的列表：

l = [[1,2,3],[6,5,4,3,7,2],[4,3,2,9],[6,7],[5,1,0],[6,3,2,7]]

我将如何编写 python 代码来检查是否有元素总是一起出现？例如，在上面的示例中，2,3 和 6,7 总是出现在同一个列表中。 （可能还有其他的，不确定）。

实现这一目标的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 的交集，但是当我检查与inner-list3 的交集时，inner-list3 中可能根本不会出现这些元素。

我可以这样做吗：

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])

这当然行不通，但我该如何正式编码呢？或者有更好的方法吗？

Answer 1

一、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，所以我想尽可能避免这种情况。

我的“尤里卡！”那一刻我意识到我不必生成所有的配对。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 进行对比

 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)},

注意(6, 3, 2, 7) 和(1, 2, 3) 应该存在的差距。我的结论是 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
    ]

运行它会给出

print(always_appear_together(data))
[(2, 3), (6, 7)]

Answer 2

使用itertools.combinations:

我最初考虑使用 itertools.combination 的东西，但是因为这允许 elements 来自 list 彼此不相邻，所以它不适用于我的解决方案介意。

事实证明，在查看非数字输入 lists 时，itertools.combinations 在这两种情况下都是必要的。我一直很困惑，因为我认为 groups 必须是 adjacent。

我认为对此最有效的方法是生成可能的elements，可以工作，然后用function 对照@ 的list 检查每一个987654333@ - 而不是在list 上进行某种组合工作并沿着这条路走下去。

所以要检查可能的elements 中的list 是否“有效”，即如果所有elements 只一起出现，我使用了一个简单的if 和带有all() 和any() 的生成器内置functions 来完成这部分工作。

现在这正在起作用，需要一种方法来生成可能发生的elements。我只是用两个嵌套的for-loops 做到了这一点——一个iterating 在window 的width 上，一个iterating 在window 的start 的位置上。

然后从这里开始，我们只检查elements 的集合是否为valid，如果是则将其添加到另一个list！

---

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)

将sol 设为：

[(2, 3), (6, 7)]]

---

这些2nested for-loops实际上可以组合成一个漂亮的one-liner（不确定其他人是否认为它是Pythonic）：

sol = [c for w in range(2, len(els)+1) for c in itertools.combinations(els, w) if valid(c)]

效果相同，只是更短。

---

由于大众需求 (@Arman)，我更新了答案，现在它应该适用于除 0-9 之外的其他 elements。这是通过引入一个独特的 elements list (els) 来完成的。

---

还有来自@thanasisp 的一些测试，使用与上面相同的代码：

l = [[1, 3, 5, 7],[1, 3, 5, 7]]

将sol 设为：

[(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)]

再次：

 l = [[1, 2, 3, 5, 7], [1, 3, 5, 7]]

给予：

 [(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)]

我认为这是正确的，因为2 不应该在任何组中，因为所有其他elements 都在不同的sub-list 中，因此它永远不能与另一个组element.

Answer 3

您可以使用集合交集来做到这一点，它也适用于每组 3 个或更多元素：请注意，我在 6,7 组中添加了一个 8。

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与它一起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})}

Answer 4

另一种带有默认字典的线性解决方案（元组用于制作可散列键）：

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 组（此处为n=2）出现的元素，请输入：

print ([p for p in contains.values() if len(p)==n])    
# [[2, 3], [6, 7]]

Answer 5

实现这一目标的最容易理解的方法是什么？

我试图使我的解决方案尽可能简短。我还尽可能地优化它。它适用于您喜欢的任何整数。

这里有很多cmets的代码解释，后面是更多基本解释：

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。

代码

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)}

正如预期的那样。

对于您问题中的列表，此代码的输出是 {(2, 3), (6, 7)}，这也符合预期。

itertools.combinations?

itertools.combinations(iterable, r)：从输入iterable 返回长度为r 的元素元组。例如：

list(itertools.combinations([1, 2, 3], 2))

给予

[(1, 2), (1, 3), (2, 3)]

为什么要使用集合？

在上面的代码中，您可以注意到集合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身份非常快，我们在代码中做了很多这样的检查。

解释主要思想

假设我们的原始列表是[[1, 2, 3], [2, 1, 4]]。

1. 为原始列表中的每个列表获取所需的组合集：

   对于[1, 2, 3]：组合集是(1, 2), (1, 3), (2, 3), (1, 2, 3)

   对于[2, 1, 4]：组合集是(1, 2), (1, 2, 4), (2, 4), (1, 4)

2. 对于每个的组合，为了在我们的代码输出中（意味着它满足我们的条件），我们要确保对于每个 一组组合，任一

   • 它存在在这个集合中（即这个组合的项目一起出现在这个集合中）
   • 或者它不存在于这个集合中 -> 没有它的项目应该出现在相应的列表中

   
   例如

   让我们从 first 组合中选取(1, 3)。我们遍历组合集：

   对于 first 集合，我们可以看到 (1, 3) 存在于其中，因此我们继续前进。

   对于second集合，我们可以看到它并不存在于其中，所以我们想看看它的any项是否存在于对应的列表中（即原列表的第二个列表：[2, 1, 4])：

   从1开始，我们可以看到它存在于对应的列表中->(1, 3)不能在输出中，因为它不满足要求的条件。

Answer 6

以下解决方案具有 线性 O(n) 复杂性，其中 n 是所有列表中的数字总数（展平后）。代码是Python2.x

我正在使用所有可能模式的位图表示（使用 python 的无穷无尽的数字很容易）。例如，如果list0 和list2 中存在一个数字，但list1 中没有，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，值 3 也是如此

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想要的任何子列表（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())

Answer 7

这是我现在想到的蛮力选项，dct 是每个数字的计数器字典，然后我们检查 dct 中的相同列表，这意味着两个数字都出现在相同的列表索引中：

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]})

这也是一个非常昂贵的操作，它是 O(N*N*M) : N = list elements 和 M = longest sublist

Answer 8

这更像是一种蛮力解决方案，但是，它将生成一个包含所有元素的大列表，方法是生成l 中每个子列表的排列并过滤以查找其元素全部出现在中的任何排列l 的子列表。如果任何排列满足该条件，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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