Compare dictionaries ignoring specific keys

Question

How can I test if two dictionaries are equal while taking some keys out of consideration. For example,

equal_dicts(
    {\'foo\':1, \'bar\':2, \'x\':55, \'y\         


        

Answer 1

in case your dictionary contained lists or other dictionaries:

def equal_dicts(d1, d2, ignore_keys, equal):
    # print('got d1', d1)
    # print('got d2', d2)
    if isinstance(d1, str):
        if not isinstance(d2, str):
            return False
        return d1 == d2
    for k in d1:
        if k in ignore_keys:
            continue
        if not isinstance(d1[k], dict) and not isinstance(d1[k], list) and d2.get(k) != d1[k]:
            print(k)
            equal = False
        elif isinstance(d1[k], list):
            if not isinstance(d2.get(k), list):
                equal = False
            if len(d1[k]) != len(d2[k]):
                return False
            if len(d1[k]) > 0 and isinstance(d1[k][0], dict):
                if not isinstance(d2[k][0], dict):
                    return False
                d1_sorted = sorted(d1[k], key=lambda item: item.get('created'))
                d2_sorted = sorted(d2[k], key=lambda item: item.get('created'))
                equal = all(equal_dicts(x, y, ignore_keys, equal) for x, y in zip(d1_sorted, d2_sorted)) and equal
            else:
                equal = all(equal_dicts(x, y, ignore_keys, equal) for x, y in zip(d1[k], d2[k])) and equal
        elif isinstance(d1[k], dict):
            if not isinstance(d2.get(k), dict):
                equal = False
            print(k)
            equal = equal_dicts(d1[k], d2[k], ignore_keys, equal) and equal
    return equal