Implement index-agnostic list comparison

[jonapich]

Sometimes, I want to ensure that a value is in a list but I don't care if the object is at a different index in the list.

I'd like to be able to pass a flag so that [1, 2, 3, 4] returns no difference against [3, 4, 2, 1].

However, [1,2,3,4] compared to [3,2,1,4,1] should be able to tell me that an extra '1' is present in the 2nd list.

[seperman]

Ok, do you think it should be a flag that you set and it makes all the iterables comparison index agnostic, or should it be a new key in the returned dictionary that is like "index_agnostic_iterable_item_added" ?

[jonapich]

I guess it depends on the scenario. For my specific case, I just want DeepDiff to completely stop reporting a difference if the index is the only difference.

[seperman]

I added the ignore_order flag. v0.5.7
Example:

List difference ignoring order or duplicates: (with the same dictionaries as above)
    >>> t1 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 2, 3]}}
    >>> t2 = {1:1, 2:2, 3:3, 4:{"a":"hello", "b":[1, 3, 2, 3]}}
    >>> ddiff = DeepDiff(t1, t2, ignore_order=True)
    >>> print (ddiff)
    {}

[jonapich]

That's useful, thank you :)