How to map list items which are in groups in the dictionary values and return keys corrosponding it?

Question

dict1 = {'chr1:1000-2000': ['A', 'B', 'C'], 'chr2:999-2000': ['A', 'X', 'Y'], 'chr3:1200-3000': ['B', 'A', 'C']}

list1 = [('A', 'B'), ('A', 'C'), ('A', 'X'), ('A', 'Y'), ('B', 'C'), ('B', 'X'), ('B', 'Y'), ('C', 'X'), ('C', 'Y'), ('X', 'Y')]

I want to map the items in list1 eg('A','B') both in dict1 values and return keys in wich this group occur

Answer 1

I am assuming you only want keys where both tuple items are present. If so, this should work:

no_match = []
for item in list1:
    item1, item2 = item
    wanted_keys = [
        key for key in dict1.keys() if
        item1 in dict1[key] and
        item2 in dict1[key]
    ]
    if len(wanted_keys) > 0:
        print("%s:  %s" % (item, wanted_keys))
    else:
        no_match.append(item)
print("No matches found for items:", no_match)

The output looks like this:

('A', 'B'):  ['chr1:1000-2000', 'chr3:1200-3000']
('A', 'C'):  ['chr1:1000-2000', 'chr3:1200-3000']
('A', 'X'):  ['chr2:999-2000']
('A', 'Y'):  ['chr2:999-2000']
('B', 'C'):  ['chr1:1000-2000', 'chr3:1200-3000']
('X', 'Y'):  ['chr2:999-2000']
No matches found for items: [('B', 'X'), ('B', 'Y'), ('C', 'X'), ('C', 'Y')]

There is probably a better, more elegant way of doing this, I am not a python person, but this is the best I could come up with.