Question

我接受过多层网络培训，但我仍然坚持如何对其他时间步骤进行预测。

我尝试通过创建此方法来跟踪字符迭代示例 -

public float[] sampleFromNetwork(INDArray testingData, int numTimeSteps, DataSetIterator iter){
    int inputCount = this.getNumOfInputs();
    int outputCount = this.getOutputCount();

    float[] samples = new float[numTimeSteps];

    //Sample from network (and feed samples back into input) one value at a time (for all samples)
    //Sampling is done in parallel here
    this.network.rnnClearPreviousState();
    INDArray output = this.network.rnnTimeStep(testingData);
    output = output.tensorAlongDimension(output.size(2)-1,1,0); //Gets the last time step output

    for( int i=0; i<numTimeSteps; ++i ){
        //Set up next input (single time step) by sampling from previous output
        INDArray nextInput = Nd4j.zeros(1,inputCount);

        //Output is a probability distribution. Sample from this for each example we want to generate, and add it to the new input
        double[] outputProbDistribution = new double[outputCount];
        for( int j=0; j<outputProbDistribution.length; j++ ) {
            outputProbDistribution[j] = output.getDouble(j);
        }
        int nextValue = sampleFromDistribution(outputProbDistribution, new Random());

        nextInput.putScalar(new int[]{0,nextValue}, 1.0f);      //Prepare next time step input
        samples[i] = (nextValue);   //Add sampled character to StringBuilder (human readable output)
        output = this.network.rnnTimeStep(nextInput);   //Do one time step of forward pass
    }

    return samples;
}

但是sampleFromDistribution（）没有意义，因为我没有使用离散类。

有什么想法？

Answer 1

我通过调整网络来使用IDENTITY激活并直接使用结果值来解决了这个问题。还有很多调整要做，但它的功能。

public float[] sampleFromNetwork(INDArray priori, int numTimeSteps){
    int inputCount = this.getNumOfInputs();
    float[] samples = new float[numTimeSteps];

    if(priori.size(1) != inputCount) {
        String format = String.format("the priori should have the same number of inputs [%s] as the trained network [%s]", priori.size(1), inputCount);
        throw new RuntimeException(format);
    }
    if(priori.size(2) < inputCount) {
        String format = String.format("the priori should have enough timesteps [%s] to prime the new inputs [%s]", priori.size(2), inputCount);
        throw new RuntimeException(format);
    }

    this.network.rnnClearPreviousState();
    INDArray output = this.network.rnnTimeStep(priori);

    output = output.ravel();
    // Store the output for use in the inputs
    LinkedList<Float> prevOutput = new LinkedList<>();
    for (int i = 0; i < output.length(); i++) {
        prevOutput.add(output.getFloat(0, i));
    }

    for( int i=0; i<numTimeSteps; ++i ){
        samples[i] = (prevOutput.peekLast());
        //Set up next input (single time step) by sampling from previous output
        INDArray nextInput = Nd4j.zeros(1,inputCount);

        float[] newInputs = new float[inputCount];
        newInputs[inputCount-1] = prevOutput.peekLast();
        for( int j=0; j<newInputs.length-1; j++ ) {
            newInputs[j] = prevOutput.get(prevOutput.size()-inputCount-j);
        }

        nextInput.assign(Nd4j.create(newInputs)); //Prepare next time step input
        output = this.network.rnnTimeStep(nextInput); //Do one time step of forward pass
        // Add the output to the end of the previous output queue
        prevOutput.addLast(output.ravel().getFloat(0, output.length()-1));
    }
    return samples;
}

DL4J中的回归 - 预测下一个时间步

1 个答案: