JSON encoding for dict of dicts

You have several options:

• Serialize with a custom key prefix that you can pick out and unserialize again:

  tuple_key = '__tuple__({})'.format(','.join(key))

  would produce '__tuple__(node1,node2,nodeN)' as a key, which you could parse back into a tuple on the other side:

  if key.startswith('__tuple__('):     key = tuple(key[10:-1].split(','))

  Demo:

  >>> key = ('node1', 'node2', 'node3')>>> '__tuple__({})'.format(','.join(key))'__tuple__(node1,node2,node3)'>>> mapped_key = '__tuple__({})'.format(','.join(key))>>> tuple(mapped_key[10:-1].split(','))('node1', 'node2', 'node3')

• Don't use dictionaries, use a list of lists:

  {'Path': [[[node1, node2 .. nodeN], [float1, float2, float3]], [...]]}

  You can build such a list simply from the dict.items() result:

  >>> json.dumps({(1, 2, 3): ('foo', 'bar')}.items())'[[[1, 2, 3], ["foo", "bar"]]]'

  and when decoding, feed the whole thing back into dict() while mapping each key-value list to tuples:

  >>> dict(map(tuple, kv) for kv in json.loads('[[[1, 2, 3], ["foo", "bar"]]]')){(1, 2, 3): (u'foo', u'bar')}

The latter approach is more suitable for custom classes as well, as the JSONEncoder.default() method will still be handed these custom objects for you to serialize back to a suitable dictionary object, which gives means that a suitable object_hook passed to JSONDecoder() a chance to return fully deserialized custom objects again for those.