Thank you for the report. At this moment, you would need to work around by loading the model directly using MLModel API. To prevent Xcode from generating the model access code, you can select the model file in the project navigator, open "Target Membership", and select "No generated class".

You are on the right track that you precompiled a source model (.mlpackage or .mlmodel) to .mlmodelc manually before adding it to Swift Package. Currently SPM doesn't support compiling the source model in .process(:) rule. Let's say, you have a source model named model.mlpackage. The first step is to compile it to .mlmodelc. xcrun coremlcompiler compile model.mlpackage /tmp/ Then, copy the compiled model to your Swift Package. cp -r /tmp/model.mlmodelc /path/to/ModelInSwiftPackage/Sources/ModelInSwiftPackage/ In my test, the resultant directory structure is something like this. . ├── Package.swift ├── Sources │   └── ModelInSwiftPackage │   ├── ModelInSwiftPackage.swift │   └── model.mlmodelc │   ├── analytics │   │   └── coremldata.bin │   ├── coremldata.bin │   └── model.mil └── Tests └── ModelInSwiftPackageTests └── ModelInSwiftPackageTests.swift Now, edit Package.swift as follows, using copy rule. import PackageDescription let package = Package( : targets: [ .target( name: "ModelInSwiftPackage", resources: [ .copy("model.mlmodelc")] ), ) Now, you should be able to build the package with the model. To access the model, you can use Bundle.module.url(). public func loadModel() throws -> MLModel? { guard let modelURL = Bundle.module.url(forResource: "model", withExtension: "mlmodelc") else { return nil } return try MLModel(contentsOf: modelURL) } Hope it works!