Comment 5 for bug 1545167

Auggy, IMO the nested list(or dicts) comparison is possible only when we have one extra list in the result as well as expected keys, but in case we have more than 1 in both, it is computationally not good to check the closest dictionary in the expected dictionaries and result dictionaries.

Assuming we get 3 not matching lists in result
and 2 not matching lists in expected result

we would have to match 2 lists for each of the 3 lists in the result dict, find the closest companion and print.

However it would be good if we have 1 not matching dict in result as well as expected result and then find the differing keys.I would try to work on this case.