Model_url question in config.yml file

Rasa Open Source
1

Hello. This is Ryohei.

I’m new to RASA and I’m trying to make chatbot with this framework.

When I tried “rasa train”, I got following error.

“”" RasaException: Parameter “model_url” was not specified in the configuration of “ConveRTFeaturizer”. It is mandatory to pass a value for this parameter. You can either use a community hosted URL of the model or if you have a local copy of the model, pass the path to the directory containing the model files. “”"

And my config.yml is like following.

“”" language: en

pipeline:

  • name: HFTransformersNLP model_weights: “bert-base-uncased” model_name: “bert”
  • name: ConveRTTokenizer
  • name: ConveRTFeaturizer
  • name: LanguageModelTokenizer
  • name: LanguageModelFeaturizer
  • name: WhitespaceTokenizer
  • name: CountVectorsFeaturizer
  • name: CountVectorsFeaturizer analyzer: char_wb min_ngram: 1 max_ngram: 4
  • name: DIETClassifier epochs: 200

policies: “”"

I tried “pip install rasa[full]” and “pip install rasa[convert]” but it didn’t work.

Would anyone teach any solutions to this please?

best regards

Ryohei

1 Like
2

As per Components,

You need to add a hosted model URL.

You can add the following line below the declaration of ConveRTTokenizer:

model_url: "https://github.com/connorbrinton/polyai-models/releases/download/v1.0/model.tar.gz"

3

Thank you very much. Your comment helped so much!

1 Like
4

It goes under ConveRTFeaturizer for anyone needing this.

EX:

pipeline:
  - name: ConveRTTokenizer
  - name: ConveRTFeaturizer
    model_url: "https://github.com/connorbrinton/polyai-models/releases/download/v1.0/model.tar.gz"
  - name: DIETClassifier
    epochs: 200