Check if elements occur together in all lists?

Question

Say I have a list of lists like so:

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

How would I write python code to check if there are elements that always occur together? For example, in the above example 2,3 and 6,7 always occur in the same lists. (there may be others, not sure).

What's the easiest to understand way of achieving this?

My only idea is to convert inner-list1 to set and check intersection with inner-list2 but when I check intersection with inner-list3, those elements may not occur at all in inner-list3.

Could I do something like:

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])

This of course doesn't work but how could I formally code this or is there a better approach to this?

Answer 1

Using itertools.combinations:

I initially thought of using something with itertools.combination, but as this allows elements from a list which are not next to each other, it wasn't going to work for the solution I had in mind.

Turns out that when looking at non-numerical input lists, itertools.combinations is necessary in both cases. I had been confused because I assumed the groups had to be adjacent.

The way I thought would work best for this would be to generate the possible elements that could work and then check each one of these with a function against the list of sub-lists - as opposed to doing some kind of combinatoric work on the list and going down that path.

So to check if a list of possible elements is 'valid' i.e. if all the elements only occur together, I used a simple if with a generator with the all() and any() built-in functions to do this part of the job.

Now this was working, there needed to be a way of generating the potential elements that could occur. I just did this with two nested for-loops - one iterating over the width of the window, and one iterating over where the start of the window is.

Then from here, we just check if that set of elements is valid and add it to another list if it is!

---

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)

which gives sol as:

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

---

These 2 nested for-loops can actually be thrown together into a nice one-liner (not sure if others think it is Pythonic):

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

which works just the same but is simply shorter.

---

Due to popular demand (@Arman), I have updated the answer so that it should now work for other elements apart from 0-9. This was done with the introduction of a unique elements list (els).

---

And some tests from @thanasisp with the same code from above:

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

gives sol as:

[(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)]

and again with:

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

gives:

 [(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)]

which I believe is correct as the 2 shouldn't be in any groups as all other elements are in a different sub-list, so it can never make a group with another element.

Answer 2

An other linear solution with default dicts (tuple is there to make hashable keys) :

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. 

Now if you look for elements which appear by group of n (n=2 here), type :

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

Answer 3

You can do this using set intersections, and it also works nicely for 3 or more elements per group: Note that I added an 8 to the 6,7 group.

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]]

First, we map each element to sets of all the other elements it appears together with:

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}}

Next, we retain only those elements where the relationship is bidirectional:

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}}

Finally, we get the unique groups with more than one element:

groups3 = {frozenset(v) for v in groups2.values() if len(v) > 1}
# {frozenset({8, 6, 7}), frozenset({2, 3})}

Answer 4

First, the data

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

Generating combinations is expensive, so I wanted to avoid that as much as possible.

My "Eureka!" moment came when I realized I don't have to generate all the pairs. Instead, I can map each number to all the lists that contain it.

appears_in = defaultdict(set)
for g in groups:
    for number in g:
        appears_in[number].add(tuple(g))

The resulting dictionary is

{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)}}

Look at the entries for 2 and 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)},

The set of lists containing 2 is identical to the set of lists containing 3. So I conclude that 2 and 3 always appear together.

Contrast this with 3 and 4

 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)},

Notice the gaps where (6, 3, 2, 7) and (1, 2, 3) should be. I conclude that 3 and 4 do NOT always appear together.

Here is the complete code

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
    ]

Running this gives

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

Answer 5

This is brute force option that comes in my mind now, dct is counter dictionary for each digit , then we check for same lists in dct that means both digit occurs in same list indexes:

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]})

also it is a very expensive operation , it's O(N*N*M) : N = list elements and M = longest sublist

Answer 6

The following solution has a linear O(n) complexity where n is the total number of numbers in all lists (after flattening). The code is Python2.x

I'm using a bitmap representation (made easy with python's endless numbers) of all the possible patterns. For example if a number is present in list0 and list2 but not list1, the respective pattern will be ...000101. For example in the given input, the value 2 will have the following bitmap pattern: 100111 and so will the value 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

This code will print:

{4: [9], 6: [4], 39: [2, 3], 42: [6, 7], 16: [0], 17: [1], 18: [5]}

And you can map and filter whichever sublists you'd like (in your case - lists equal or larger than 2):

print filter(lambda x: len(x) >= 2, pattern_to_num_list.values())

Answer 7

What's the easiest to understand way of achieving this?

I tried to make my solution as short as possible. I also worked to optimize it as much as I could. It works for any integer as you prefer.

Here is the code with a lot of comments that explain it, followed by more essential explanations:

Note: In the following code, I used [[1, 2, 3], [2, 1, 4]] as an example of the original list instead of the one in your question, to make explanation easier.

Code

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)

Output:

{(1, 2)}

as expected.

For the list in your question, the output of this code is {(2, 3), (6, 7)} which is as expected also.

itertools.combinations?

itertools.combinations(iterable, r): Returns r length tuples of elements from the input iterable. For example:

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

gives

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

Why using sets?

In the code above, you can notice that sets are used to hold the combinations of each list from the original list. This is because checking for membership of a value in a set is very fast, and we do a lot of such checks in the code.

Explaining the main idea

Assume that our original list is [[1, 2, 3], [2, 1, 4]].

1. Get the required set of combinations for each list in the original:

   For [1, 2, 3]: the set of combinations is (1, 2), (1, 3), (2, 3), (1, 2, 3)

   For [2, 1, 4]: the set of combinations is (1, 2), (1, 2, 4), (2, 4), (1, 4)

2. For each of the combinations, and in order to be in the output of our code (meaning that it satisfies our condition), we want to make sure that for each set of combinations, either

   • it exists in this set (i.e. the items of this combination occur together in this set)
   • or it does not exist in this set -> none of its items should appear in the corresponding list

   
   For example

   Let's take (1, 3) from the first set of combinations. We iterate through the combination sets:

   For the first set, we can see that (1, 3) exists into it, so we move forward.

   For the second set, we can see that it does not exist into it, so we want to see if any of its items exists in the corresponding list (i.e the second list of the original list: [2, 1, 4]):

   Starting with 1, we can see that it exists in the corresponding list -> (1, 3) can't be in the output because it does not satisfy the required condition.

Answer 8

This is more of a brute-force solution, however, it will generate a large list of all elements that occur together by generating permutations of every sublist in l and filtering to find any permutations whose elements of which all appear in a sublist of l. If any permutations pass that condition, the permutation will be added to 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)))

Output:

[(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), (4, 7, 3, 2, 6), (2, 6, 7, 4, 5), (2, 3, 5, 6), (3, 2, 5, 4), (5, 7, 6, 4, 2), (2, 4, 5, 7, 3), (7, 5, 4, 2, 3), (7, 6, 3, 5), (6, 5, 4), (3, 6, 5, 7, 4), (2, 7, 3, 6, 5), (4, 5, 2, 7), (7, 3, 5, 6, 4), (5, 7, 4, 2, 6), (7, 4, 3, 5, 6), (3, 4, 6, 2, 7), (2, 5, 4, 7), (2, 7, 6, 3, 4), (5, 7, 3, 2, 4), (2, 6, 7, 3), (3, 4, 2, 5), (3, 7, 2, 4, 6), (7, 6, 4, 2, 3), (3, 2, 7), (7, 6, 5, 2, 3), (7, 6, 4, 3), (5, 6, 3, 2, 4), (6, 5, 3, 4, 7), (9, 2, 4), (6, 7, 3, 5), (2, 3, 4, 7, 5), (7, 6, 4, 5), (6, 2, 5, 4), (5, 6, 7, 2, 4), (4, 6, 5, 7, 3), (4, 2, 3, 5, 6), (4, 5, 7, 3, 2), (4, 2, 6, 7), (6, 3, 4, 5, 7), (4, 7, 6, 2, 5), (7, 6, 3), (2, 6, 7, 3, 4), (6, 7, 3, 4, 5), (4, 6, 7, 5), (7, 5, 3, 4), (5, 6, 7, 3, 2), (5, 2, 6, 7), (3, 4), (7, 5, 3, 4, 6), (5, 7, 3, 6, 2), (7, 3, 6, 2, 4), (4, 7), (4, 5, 7, 6), (5, 6, 4, 2, 7), (3, 6, 7, 4), (5, 6), (7, 2, 5, 6, 4), (4, 5, 6, 7, 3), (2, 4, 3, 6, 5), (2, 3, 7), (7, 6, 3, 2, 4), (6, 4, 3, 5, 7), (6, 2, 7), (6, 3, 2, 7), (3, 5, 4, 6), (7, 6, 5, 4, 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