dictionary with lists custom equality check

Question

Say, I have two lists:

a = [1,2,3] and b = [2,3,1]

If I do a a == b it returns False,

If I check sorted(a) == sorted(b), it returns True.

Now, I have two objects:

obj1 = {'a': 1, 'b': 2, 'c': [1, 2]} and obj2 = {'b': 2, 'a': 1, 'c': [1, 2]}

obj1 == obj2 is True, irrespective of the order of keys.

But if obj2 = {'b': 2, 'a': 1, 'c': [2, 1]}

how do I test the equality? Obviously, obj1 == obj2 returns False. sorted(obj1) will have ['a', 'b', 'c'], so sorted(obj1) == sorted(obj2) is kind of waste check.

I should have probably overridden the equality method for the object, or use some library. Is, is there any way to write idiomatic python code for deep equality?

I need `obj1 == obj2` return True, if `obj1 = {'a': 1, 'b': 2, 'c': [1, 2]}` and `obj2 = {'b': 2, 'a': 1, 'c': [2, 1]}` — inquisitive, Aug 14 '18 at 11:59
`sorted(['a', 'b', 'c'])` isn't as wasteful as you might think since Python's timsort uses a clever hybrid sorting algorithm that performs well on already sorted (or almost sorted) data — Chris_Rands, Aug 14 '18 at 12:08

Sunitha · Accepted Answer · 2018-08-14T12:22:31.027

6

sort each element in the dict if it is of type list and then compare

>>> def sorted_element_dict(d):
...     return {k:sorted(v) if isinstance(v, list) else v for k,v in d.items()}
...
>>> sorted_element_dict(obj1) == sorted_element_dict(obj2)
True

edited Aug 14 '18 at 12:22

answered Aug 14 '18 at 11:59

Sunitha

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1

Also use `isinstance` instead of `type`. – timgeb Aug 14 '18 at 12:00
3

@timgeb. I would prefer `sorted` as the list may contain unhashable elements – Sunitha Aug 14 '18 at 12:02
4

@timgeb And the list may contain duplicates (which sets cannot) – Tim Klein Aug 14 '18 at 12:03
1

@timgeb, convert to set is wrong because sets cannot have duplicates while list can do – Massimo Costa Aug 14 '18 at 12:03
@timgeb. Changed `type` check to `isinstance`. Thanks for the suggestion – Sunitha Aug 14 '18 at 12:05
1

There is no reason to use `lambda` here, it should be `def` – Chris_Rands Aug 14 '18 at 12:05
@Chris_Rands. I believe `lambda` vs `def` is a personal preference – Sunitha Aug 14 '18 at 12:08
1

@Sunitha If you're assigning to `lambda` you're doing it wrong, `lambda` functions should be anonymous, see https://stackoverflow.com/questions/134626/which-is-more-preferable-to-use-in-python-lambda-functions-or-nested-functions – Chris_Rands Aug 14 '18 at 12:08
@Chris_Rands.. Ok you win :) – Sunitha Aug 14 '18 at 12:22

Mad Physicist · Answer 2 · 2018-08-14T12:27:52.497

If you are only interested in an equality check, consider using all. You could just apply a custom comparator to all the dictionary values:

def check_equal(d1, d2):
    if d1.keys() != d2.keys():
        return False
    return all(sorted(v) == sorted(d2[k]) if isinstance(v, list) else v == d2[k] for k, v in d1.items())

A more elegant way might be to factor out the comparator into a separate function and check against something more general, like collections.abc.Sequence:

def check_equal(d1, d2):
    if d1.keys() != d2.keys():
        return False
    def cmp(v1, v2):
        if isinstance(v1, Sequence) and isinstance(v2, sequence):
            return sorted(v1) == sorted(v2)
        return v1 == v2
    return all(cmp(v, d2[k]) for k, v in d1.items())

This has the advantage of not storing all the intermediate sorted products. If, on the other hand, you need to do the comparison frequently, it may be better to transform your dictionaries before using regular ==:

def normalize(d):
    for k, v in d.items():
        if isinstance(v, Sequence):
            d[k] = sorted(v)

Notice that I used a loop instead of a comprehension here. That way, your dictionary is transformed in-place, rather than allocating a whole new hash table.

dictionary with lists custom equality check

2 Answers2