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Simplify and generalize llama api build yay

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This commit is contained in:
Ashwin Bharambe 2024-08-30 14:51:40 -07:00
parent 297d51b183
commit f8517e4688
9 changed files with 103 additions and 151 deletions
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40
llama_toolchain/cli/api/configure.py
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@ -31,36 +31,16 @@ class ApiConfigure(Subcommand):
self.parser.set_defaults(func=self._run_api_configure_cmd)
def _add_arguments(self):
from llama_toolchain.core.distribution import stack_apis
from llama_toolchain.core.package import BuildType
allowed_args = [a.name for a in stack_apis()]
self.parser.add_argument(
"api",
choices=allowed_args,
help="Stack API (one of: {})".format(", ".join(allowed_args)),
)
self.parser.add_argument(
"--build-name",
type=str,
help="(Fully qualified) name of the API build to configure. Alternatively, specify the --provider and --name options.",
required=False,
)
self.parser.add_argument(
"--provider",
type=str,
help="The provider chosen for the API",
required=False,
help="Name of the build",
required=True,
)
self.parser.add_argument(
"--name",
type=str,
help="Name of the build target (image, conda env)",
required=False,
)
self.parser.add_argument(
"--type",
"--build-type",
type=str,
default="conda_env",
choices=[v.value for v in BuildType],
@ -69,15 +49,11 @@ class ApiConfigure(Subcommand):
def _run_api_configure_cmd(self, args: argparse.Namespace) -> None:
from llama_toolchain.core.package import BuildType
if args.build_name:
name = args.build_name
if name.endswith(".yaml"):
name = name.replace(".yaml", "")
else:
build_type = BuildType(args.type)
name = f"{build_type.descriptor()}-{args.provider}-{args.name}"
config_file = BUILDS_BASE_DIR / args.api / f"{name}.yaml"
build_type = BuildType(args.build_type)
name = args.build_name
config_file = (
BUILDS_BASE_DIR / "adhoc" / build_type.descriptor() / f"{name}.yaml"
)
if not config_file.exists():
self.parser.error(
f"Could not find {config_file}. Please run `llama api build` first"