Enable sane naming of registered objects with defaults (#429)

# What does this PR do? 

This is a follow-up to #425. That PR allows for specifying models in the
registry, but each entry needs to look like:

```yaml
- identifier: ...
  provider_id: ...
  provider_resource_identifier: ...
```

This is headache-inducing.

The current PR makes this situation better by adopting the shape of our
APIs. Namely, we need the user to only specify `model-id`. The rest
should be optional and figured out by the Stack. You can always override
it.

Here's what example `ollama` "full stack" registry looks like (we still
need to kill or simplify shield_type crap):
```yaml
models:
- model_id: Llama3.2-3B-Instruct
- model_id: Llama-Guard-3-1B
shields:
- shield_id: llama_guard
  shield_type: llama_guard
```

## Test Plan

See test plan for #425. Re-ran it.
This commit is contained in:
Ashwin Bharambe 2024-11-12 11:18:05 -08:00 committed by GitHub
parent d9d271a684
commit 09269e2a44
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17 changed files with 295 additions and 207 deletions

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@ -7,6 +7,8 @@
import pytest
import pytest_asyncio
from llama_stack.apis.models import ModelInput
from llama_stack.distribution.datatypes import Api, Provider
from llama_stack.providers.tests.resolver import resolve_impls_for_test_v2
@ -76,20 +78,14 @@ async def scoring_stack(request, inference_model):
[Api.scoring, Api.datasetio, Api.inference],
providers,
provider_data,
)
provider_id = providers["inference"][0].provider_id
await impls[Api.models].register_model(
model_id=inference_model,
provider_id=provider_id,
)
await impls[Api.models].register_model(
model_id="Llama3.1-405B-Instruct",
provider_id=provider_id,
)
await impls[Api.models].register_model(
model_id="Llama3.1-8B-Instruct",
provider_id=provider_id,
models=[
ModelInput(model_id=model)
for model in [
inference_model,
"Llama3.1-405B-Instruct",
"Llama3.1-8B-Instruct",
]
],
)
return impls