Does anyone know how we can place the `is_strict` flag?

[Neel Gupta]

I have been going mad about not figuring out where I should place the is_strict flag. Error dictates its somewhere near override but for the life of me, I can't really find where it's supposed to be.

Does anyone have any idea? This is the error message that brought me here:-

Traceback (most recent call last):
  File "/kaggle/working/tpu/models/official/detection/main.py", line 192, in <module>
    tf.app.run(main)
  File "/opt/conda/lib/python3.7/site-packages/tensorflow_core/python/platform/app.py", line 40, in run
    _run(main=main, argv=argv, flags_parser=_parse_flags_tolerate_undef)
  File "/opt/conda/lib/python3.7/site-packages/absl/app.py", line 303, in run
    _run_main(main, args)
  File "/opt/conda/lib/python3.7/site-packages/absl/app.py", line 251, in _run_main
    sys.exit(main(argv))
  File "/kaggle/working/tpu/models/official/detection/main.py", line 85, in main
    params, FLAGS.params_override, is_strict=True)
  File "/kaggle/working/tpu/models/hyperparameters/params_dict.py", line 416, in override_params_dict
    params.override(params_dict, is_strict)
  File "/kaggle/working/tpu/models/hyperparameters/params_dict.py", line 157, in override
    self._override(override_params, is_strict)  # pylint: disable=protected-access
  File "/kaggle/working/tpu/models/hyperparameters/params_dict.py", line 169, in _override
    '`override` with `is_strict` = False.'.format(k))
KeyError: 'The key `is_strict=False` does not exist. To extend the existing keys, use `override` with `is_strict` = False.'