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

  • 1
    $\begingroup$ Please edit your question and show us the expected output. Should both keys be there or is one enough? Can they be in any order? $\endgroup$
    – terdon
    Nov 16, 2021 at 17:51
  • $\begingroup$ edited thank you $\endgroup$
    – amardeep
    Sep 20, 2023 at 9:02

1 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))
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.


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