I have a list of lists. I want to remove lists if the lenght of intersection of values in the list with previous lists is more than one.

For example

A=[(1,2,3,4), (2,3,5,6),(1,7,8,9),(2,4,6,8)]

We need to keep first list because there is no previous list. We remove (2,3,5,6) as its intersection of previous list is (2,3) and its length is 2. We keep (1,7,8,9) and remove (2,4,6,8) likewise.

At the end our list should become B=[(1,2,3,4), (1,7,8,9)]

CodePudding user response：

A potentially more efficient solution, in case you have (and keep) many more but still short tuples. This checks pairs of numbers in the lists. Keep the current tuple a if none of its pairs occur in an already kept tuple.

from itertools import combinations

A = [(1,2,3,4), (2,3,5,6), (1,7,8,9), (2,4,6,8)]

B = []
B_pairs = set()
for a in A:
    pairs = set(map(frozenset, combinations(a, 2)))
    if not B_pairs & pairs:
        B.append(a)
        B_pairs |= pairs
print(B)

CodePudding user response：

Here is a direct implementation:

A = [(1,2,3,4), (2,3,5,6), (1,7,8,9), (2,4,6,8)]

B = []
intersect = 0
for a in A:
    for b in B:
        intersect = 0
        for num in b:
            if num in a:
                intersect  = 1
        if intersect > 1:
            break
    if intersect < 2:
        B.append(a)

which gives:

B = [(1, 2, 3, 4), (1, 7, 8, 9)]

CodePudding user response：

Here is a possible solution:

A=[(1,2,3,4),(2,3,5,6),(1,7,8,9),(2,4,6,8)]
B=[]
P=[]

# iterate over sublists a of the list A
for (index, a) in enumerate(A):
    # define and reset the intersections counter
    intersections=0
    # iterate of sublists p of P
    for p in P:
        # reset the intersections counter
        intersections=0
        # iterate over the numbers num of sublist a
        for num in a:
            # perform checks
            if num in p:
                intersections =1
            if intersections>1:
                break
        # break the iteration over sublists p if check fulfilled
        if intersections>1:
            break
    # Append a to the subset P only if intersection were not > 1 
    if intersections>1:
        continue
    P.append(a)
    # repeat

print(P)
#P now contains the result... but for completion
B=P
Print(B)

CodePudding user response：

Normally, this piece of code should do what you are looking for

A=[(1,2,3,4), (2,3,5,6),(1,7,8,9),(2,4,6,8)]

B = []

precedent = []
for list_ in A:
    intersection = 0

    for num in list_:
        if num in precedent:
            intersection  = 1

    if intersection < 1:
        B.append(list_)

    precedent = list_

CodePudding user response：

I think this is the one way you could try.

def intersection_length(lst1, lst2):
    intersection_list = [value for value in lst1 if value in lst2]
    return len(intersection_list)

def filter_lists(lists):
    filtered_lists = []
    filtered_lists.append(lists[0])
    del lists[0]

    for list_item in lists:
        is_ok = True
        for prev_list in filtered_lists:
            if intersection_length(list_item, prev_list) > 1:
                is_ok = False
                continue
        if is_ok:
            filtered_lists.append(list_item)
        return filtered_lists

def main():
    filtered = filter_lists(
        [(1, 2, 3, 4), (2, 3, 5, 6), (1, 7, 8, 9), (2, 4, 6, 8)]
    )
    print(filtered)

if __name__ == "__main__":
    main()