remove duplicates from dicts with list values

[wardancer84]

I'm not sure if this is right since you never clarified what "duplicate" means in this particular case, but here's your band-aid.

dict1 = {"SAG01112_SSAP_HA_LPM": [["OS_TYPE", "AIX"], ["IS_COBOL", "1"]], "SAP": [], "C11_RG": [], "W11_RG": []}
dict2 = {
    "SAG01112_SSAP_HA_LPM": [
        ["OS_TYPE", "AIX"],
        ["IP", "172.17.10.112"],
        ["IP", "10.111.160.119"],
        ["IP", "10.111.160.68"],
        ["IP", "10.111.160.66"],
        ["IP", "10.95.0.112"],
        ["IP", "10.111.162.119"],
    ],
    "SAP": [],
    "C11_RG": [],
    "W11_RG": [],
}


def remove_common_items(dict_a, dict_b):
    """Remove values that are common to a and b."""
    # For common keys
    for key in set(dict_a) & set(dict_b):
        a = dict_a[key]
        b = dict_b[key]
        # Remove items common to both value lists.
        for item in [x for x in a if x in b]:  # Make list of common before iterating
            a.remove(item)
            b.remove(item)


remove_common_items(dict1, dict2)
print(dict1)
print(dict2)

I still think you dictionaries are messed up and should look like this:

hm...looks also usable..but not sure if sets will work with embedded lists as lists are not hashable afaik. will try...
and no need to nitpick on the term "duplicate"...if a value is in both dicts i call this duplicate, and i want to remove it in the second dict, easy as that.