Change the number of outputs by adding a parser#

By default, sklearn-onnx assumes that a classifier has two outputs (label and probabilities), a regressor has one output (prediction), a transform has one output (the transformed data). What if it is not the case? The following example creates a custom converter and a custom parser which defines the number of outputs expected by the converted model.

Example A new converter with options shows a converter which selects two ways to compute the same outputs. In this one, the converter produces both. That would not be a very efficient converter but that’s just for the sake of using a parser. By default, a transformer only returns one output but both are needed.

A new transformer#

import numpy
from onnxruntime import InferenceSession
from sklearn.base import TransformerMixin, BaseEstimator
from sklearn.datasets import load_iris
from skl2onnx import update_registered_converter
from skl2onnx.common.data_types import guess_numpy_type
from skl2onnx.algebra.onnx_ops import OnnxSub, OnnxMatMul, OnnxGemm
from skl2onnx import to_onnx, get_model_alias

class DecorrelateTransformer(TransformerMixin, BaseEstimator):
    Decorrelates correlated gaussian features.

    :param alpha: avoids non inversible matrices
        by adding *alpha* identity matrix


    * `self.mean_`: average
    * `self.coef_`: square root of the coveriance matrix

    def __init__(self, alpha=0.0):
        self.alpha = alpha

    def fit(self, X, y=None, sample_weights=None):
        if sample_weights is not None:
            raise NotImplementedError("sample_weights != None is not implemented.")
        self.mean_ = numpy.mean(X, axis=0, keepdims=True)
        X = X - self.mean_
        V = X.T @ X / X.shape[0]
        if self.alpha != 0:
            V += numpy.identity(V.shape[0]) * self.alpha
        L, P = numpy.linalg.eig(V)
        Linv = L ** (-0.5)
        diag = numpy.diag(Linv)
        root = P @ diag @ P.transpose()
        self.coef_ = root
        return self

    def transform(self, X):
        return (X - self.mean_) @ self.coef_

data = load_iris()
X =

dec = DecorrelateTransformer()
pred = dec.transform(X[:5])
[[ 0.0167562   0.52111756 -1.24946737 -0.56194325]
 [-0.0727878  -0.80853732 -1.43841018 -0.37441392]
 [-0.69971891 -0.09950908 -1.2138161  -0.3499275 ]
 [-1.13063404 -0.13540568 -0.79087008 -0.73938966]
 [-0.35790036  0.91900236 -1.04034399 -0.6509266 ]]

Conversion into ONNX with two outputs#

Let’s try to convert it and see what happens.

def decorrelate_transformer_shape_calculator(operator):
    op = operator.raw_operator
    input_type = operator.inputs[0].type.__class__
    input_dim = operator.inputs[0].type.shape[0]
    output_type = input_type([input_dim, op.coef_.shape[1]])
    operator.outputs[0].type = output_type

def decorrelate_transformer_converter(scope, operator, container):
    op = operator.raw_operator
    opv = container.target_opset
    out = operator.outputs

    X = operator.inputs[0]

    dtype = guess_numpy_type(X.type)

    Y1 = OnnxMatMul(
        OnnxSub(X, op.mean_.astype(dtype), op_version=opv),

    Y2 = OnnxGemm(
        (-op.mean_ @ op.coef_).astype(dtype),

    Y1.add_to(scope, container)
    Y2.add_to(scope, container)

def decorrelate_transformer_parser(scope, model, inputs, custom_parsers=None):
    alias = get_model_alias(type(model))
    this_operator = scope.declare_local_operator(alias, model)

    # inputs

    # outputs
    cls_type = inputs[0].type.__class__
    val_y1 = scope.declare_local_variable("nogemm", cls_type())
    val_y2 = scope.declare_local_variable("gemm", cls_type())

    # ends
    return this_operator.outputs

The registration needs to declare the parser as well.


And conversion.

onx = to_onnx(dec, X.astype(numpy.float32), target_opset=14)

sess = InferenceSession(onx.SerializeToString(), providers=["CPUExecutionProvider"])

exp = dec.transform(X.astype(numpy.float32))
results =, {"X": X.astype(numpy.float32)})
y1 = results[0]
y2 = results[1]

def diff(p1, p2):
    p1 = p1.ravel()
    p2 = p2.ravel()
    d = numpy.abs(p2 - p1)
    return d.max(), (d / numpy.abs(p1)).max()

print(diff(exp, y1))
print(diff(exp, y2))
(6.04657619085458e-07, 0.0002951417065406967)
(2.01757041717876e-06, 0.0005483764980468156)

Total running time of the script: (0 minutes 0.021 seconds)

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