I agree with Gael on this one and am happy to help with the PR if you need
any assistance.
Best,
Maciek
----
Pozdrawiam, | Best regards,
Maciek Wójcikowski
mac...@wojcikowski.pl
2017-12-29 18:14 GMT+01:00 Gael Varoquaux <gael.varoqu...@normalesup.org>:
> I think that a transform method would be good. We would have to add a
> parameter to the constructor to specify which layer is used for the
> transform. It should default to "-1", in my opinion.
>
> Cheers,
>
> Gaël
>
> Sent from my phone. Please forgive typos and briefness.
> On Dec 29, 2017, at 17:48, "Javier López" <jlo...@ende.cc> wrote:
>
>> Hi Thomas,
>>
>> it is possible to obtain the activation values of any hidden layer, but
>> the
>> procedure is not completely straight forward. If you look at the code of
>> the `_predict` method of MLPs you can see the following:
>>
>> ```python
>> def _predict(self, X):
>> """Predict using the trained model
>>
>> Parameters
>> ----------
>> X : {array-like, sparse matrix}, shape (n_samples, n_features)
>> The input data.
>>
>> Returns
>> -------
>> y_pred : array-like, shape (n_samples,) or (n_samples, n_outputs)
>> The decision function of the samples for each class in the
>> model.
>> """
>> X = check_array(X, accept_sparse=['csr', 'csc', 'coo'])
>>
>> # Make sure self.hidden_layer_sizes is a list
>> hidden_layer_sizes = self.hidden_layer_sizes
>> if not hasattr(hidden_layer_sizes, "__iter__"):
>> hidden_layer_sizes = [hidden_layer_sizes]
>> hidden_layer_sizes = list(hidden_layer_sizes)
>>
>> layer_units = [X.shape[1]] + hidden_layer_sizes + \
>> [self.n_outputs_]
>>
>> # Initialize layers
>> activations = [X]
>>
>> for i in range(self.n_layers_ - 1):
>> activations.append(np.empty((X.shape[0],
>> layer_units[i + 1])))
>> # forward propagate
>> self._forward_pass(activations)
>> y_pred = activations[-1]
>>
>> return y_pred
>> ```
>>
>> the line `y_pred = activations[-1]` is responsible for extracting the
>> values for the last layer,
>> but the `activations` variable contains the values for all the neurons.
>>
>> You can make this function into your own external method (changing the
>> `self` attribute by
>> a proper parameter) and add an extra argument which specifies the
>> layer(s) that you want.
>> I have done this myself in order to make an AutoEncoderNetwork out of the
>> MLP
>> implementation.
>>
>> This makes me wonder, would it be worth adding this to sklearn?
>> A very simple way would be to refactor the `_predict` method, with the
>> additional layer
>> argument, to a new method `_predict_layer`, then we can have the
>> `_predict` method
>> simply call `_predict_layer(..., layer=-1)` and add a new method (perhaps
>> a `transform`?)
>> that allows to get (raveled) values for an arbitrary subset of the layers.
>>
>> I'd be happy to submit a PR if you guys think it would be interesting for
>> the project.
>>
>> Javier
>>
>>
>>
>> On Thu, Dec 7, 2017 at 12:51 AM Thomas Evangelidis <teva...@gmail.com>
>> wrote:
>>
>>> Greetings,
>>>
>>> I want to train a MLPClassifier with one hidden layer and use it as a
>>> feature selector for an MLPRegressor.
>>> Is it possible to get the values of the neurons from the last hidden
>>> layer of the MLPClassifier to pass them as input to the MLPRegressor?
>>>
>>> If it is not possible with scikit-learn, is anyone aware of any
>>> scikit-compatible NN library that offers this functionality? For example
>>> this one:
>>>
>>> http://scikit-neuralnetwork.readthedocs.io/en/latest/index.html
>>>
>>> I wouldn't like to do this in Tensorflow because the MLP there is much
>>> slower than scikit-learn's implementation.
>>>
>> ------------------------------
>>
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>>
>>
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