snippet

rmdict.py

def rm_dict(dicts, keyword, recursive=True):
    """ remove a keyword(k-v pair) from (nest) dict.  default is recursived.
    :param: dicts   - an dict (nested allow).
    :param: keyword - an tuple with 2 values, means (key, value).
    :param: recursive - recursive search, default is True.
    ... admonition : doctest string below.
    pretty a as:
    a = \
    {
        'name':'adam',
        'child':{
            'name':'adam',
            'father':'adam',
            'child':{
                'name':'eve',
                'father':'adam',
                'child':{
                    'name':"adam"
                }
            }
        }
    }
    >>> a = {'child': {'child': {'child': {'name': 'adam'}, 'name': 'eve', 'father': \
    'adam'}, 'name': 'adam', 'father': 'adam'}, 'name': 'adam'}
    >>> rm_dict(a, ('name', 'eve'))
    1
    >>> rm_dict(a, ('father', 'eve'))
    0
    >>> rm_dict(a, ('father', 'adam'), recursive=False)
    0
    >>> rm_dict(a, ('father', 'adam'), recursive=True)
    2
    >>> a
    {'child': {'child': {'child': {'name': 'adam'}}, 'name': 'adam'}, 'name': 'adam'}
    >>> rm_dict(a, ('name', 'adam'), recursive=False)
    1
    >>> a
    {'child': {'child': {'child': {'name': 'adam'}}, 'name': 'adam'}}
    >>> rm_dict(a, ('name', 'adam'))
    2
    >>> a
    {'child': {'child': {'child': {}}}}
    >>>
    """
    k, v = keyword
    count = 0
    if k in dicts.keys():
        if dicts.get(k) == v:
            dicts.pop(k)
            count += 1
        # verbose output.
        # elif not isinstance(dicts.get(k), dict):
        #     print("found key in {}, but value {} equal {}.".format(k, v, dicts.get(k)))
    if recursive:
        for val in dicts.values():
            if isinstance(val, dict):
                count += rm_dict(val, keyword, recursive)
    return count

if __name__ == '__main__':
    import doctest
    doctest.testmod(verbose=True)