Python Comparing List Values to Keys in List of Dicts

Comparing list of dictionaries when the keys are out of order

I'm pretty sure you made a mistake somehow. I and others get True:

>>> a = [{"name": "hello", "city": "xyz"},{"city": "rty", "name": "ert"}]
>>> b = [{"city": "xyz","name": "hello"},{"name": "ert", "city": "rty"}]
>>> a == b
True

And that's what it should do.

The documentation about OrderedDict says (emphasis mine):

Equality tests between OrderedDict objects and other Mapping objects are order-insensitive like regular dictionaries.

And the documentation about Value comparisons says this, which is the case for those dicts:

Mappings (instances of dict) compare equal if and only if they have equal (key, value) pairs.

How do i compare items in a python LIST to DICTIONARY keys and then construct a dictionary with all elements of the LIST

You can't have a dictionary with multiple values for the same key. A better way to represent this might be to construct a list of tuples:

[(score, b[score]) for score in a]

Sort and Compare List of Dicts Python

You can sort both lists before comparing them and compare the sorted results:

>>> list_dict_a = [
        {'expiration_date': None, 'identifier_country': None, 'identifier_number': 'Male', 'identifier_type': 'Gender', 'issue_date': None},
        {'expiration_date': None, 'identifier_country': 'VE', 'identifier_number': '1234567', 'identifier_type': 'Foo No.', 'issue_date': None}]

>>> list_dict_b = [
        {'identifier_country': 'VE', 'expiration_date': None, 'identifier_type': 'Foo No.', 'issue_date': None, 'identifier_number': '1234567'},
        {'identifier_country': None, 'expiration_date': None, 'identifier_type': 'Gender', 'issue_date': None, 'identifier_number': 'Male'}]

>>> list_dict_a == list_dict_b
False
>>> def key_func(d):
        items = ((k, v if v is not None else '') for k, v in d.items())
        return sorted(items)
>>> sorted(list_dict_a, key=key_func) == sorted(list_dict_b, key=key_func)
True

The order of the dicts within each list will then not matter.

Passing the key function is needed, because dicts are not orderable, thus we need to tell the sorting function what key to use for each pair of dict objects when comparing them. A key for each dictionary is simply a sorted list of its (key, value) pairs.

The key function calculates a key for each dict as follows:

>>> dict_a0 = list_dict_a[0]
>>> key_func(dict_a0)
[('expiration_date', ''), ('identifier_country', ''), ('identifier_number', 'Male'), ('identifier_type', 'Gender'), ('issue_date', '')]

Footnotes

In order for this list of (key, value) pairs to be comparable with other dicts' lists, None values had to be converted to an empty string. This allows None values to be comparable with other non-None values.

The underlying assumption in the solution above is that all dictionary values in your case are either strings or None, and that "empty" values are consistently represented as None (and not e.g. by an empty string). If this is not the case, key_func() would have to be adjusted accordingly to assure that the resulting lists are always comparable to each other for any dict value expected in the data.

Also, for large dicts this key function might not be ideal, because comparisons of key pairs would be too slow. It would thus be better to instead calculate a unique hash value for each dict (but the same hash for dicts that compare equal).

Comparing a dictionary value with a list and returning keys in list's order in Python

You should invert the dictionary:

inverse = {index: key for key, index in dictionary.items()}

Now you can look up the keys in the correct order:

[inverse[index] for index in top_indices]

Another way would be

list(map(inverse.__getitem__, top_indices))

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