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pythonjsondictionarycomparison
Keyan Kazemian
Keyan Kazemian

Reputation: 80

How to compare two json files disregarding a certain key

I have two json files, each in the form of a dictionary. I would like to compare them but ignore the 'ver' key when doing so. I've looked at the following question and implemented the answer as my function: Compare dictionaries ignoring specific keys

However, the function is still returning false when comparing two files that only have a difference in the 'ver' key. 

def compare_json(file_1, file_2, ignore_key):
    ignored = set(ignore_key)
    for k1, v1 in file_1.iteritems():
        if k1 not in ignored and (k1 not in file_2 or file_2[k1] != v1):
            return False

    for k2, v2 in file_2.iteritems():
        if k2 not in ignored and k2 not in file_1:
            return False

    return True



if not compare_json(data, latest_file, ('ver')):
        print 'not equal'
        data['ver'] += 1
        ver_number = data['ver']
        with open(('json/{0}.v{1}.json').format(name, ver_number)) as new_json:
            json.dump(data, new_json)
    else:
        print 'equal'

Here is what printing the json dicts look like:

{'ver': 1, 'data': 0}

{'ver': 2, 'data': 0}

Comparing the above should return true; however, it returns false. When I change the version numbers to the same number, it returns true.

Upvotes: 1

Views: 1957

Answers (1)

DYZ
DYZ

Reputation: 57033

Change ('ver') to ('ver',).

('ver') is not a tuple, it is simply 'ver' in the parentheses. Respectively, set(('ver')) is {'e','r','v'}, which are the keys that your function ignores - but they are not the keys that you want to ignore.

On the contrary, ('ver',) is a one-element tuple, and set(('ver',)) is {'ver'}.

Upvotes: 2

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