I am trying to fit the model from Frome(1983) in R using poisson Regression, but i dont know how add the exposure: ln(person-years), the model is:
ln E(Y|X)= ln(person-years)+ B0+ B1X1+ B2X2.
Do you know how fit that ?
You can perform poisson regression analysis using the below R code.
m1<- glm(y ~ offset(log(person-years))+X1+X2, family="poisson", data=dat)
summary(m1)
Because you use an offset of 'log(person-years)', the coefficient of 'log(person-years)' is forced to be 1.
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I am beginner in deep learning.
I am using this dataset and I want my network to detect keypoints of a hand.
How can I make my output layer's nodes to be in range [-1, 1] (range of normalized 2D points)?
Another problem is when I train for more than 1 epoch the loss gets negative values
criterion: torch.nn.MultiLabelSoftMarginLoss() and optimizer: torch.optim.SGD()
Here u can find my repo
net = nnModel.Net()
net = net.to(device)
criterion = nn.MultiLabelSoftMarginLoss()
optimizer = optim.SGD(net.parameters(), lr=learning_rate)
lr_scheduler = torch.optim.lr_scheduler.ExponentialLR(optimizer=optimizer, gamma=decay_rate)
You can use the Tanh activation function, since the image of the function lies in [-1, 1].
The problem of predicting key-points in an image is more of a regression problem than a classification problem (especially if you're making your model outputs + targets fall within a continuous interval). Therefore, I suggest you use the L2 Loss.
In fact, it could be a good exercise for you to determine which loss function that is appropriate for regression problems provides the lowest expected generalization error using cross-validation. There's several such functions available in PyTorch.
One way I can think of is to use torch.nn.Sigmoid which produces outputs in [0,1] range and scale outputs to [-1,1] using 2*x-1 transformation.
I am new to using Eviews. I want to fit my data to an exponential curve: y = a*e^(bx). I am running the regression using these two equations:
MICE_1= COEF(1)*(EXP(COEF(2)*OBSERVATION))
LOG(MICE_1) = COEF(3) + COEF(4)*OBSERVATION
Since, both equations are the same, coef(2)=coef(4) and coef(1)= exp(coef(3))
However, I am getting different coefficients using the two equations. Can someone please explain why is this happening?
I measured the spectrum of an X-ray tube with a Geiger counter. Such data aren't normally distributed, but instead they follow the Poisson distribution. I want to fit a curve to the measured data:
a = 1
b = 1
f(x) = a*(x/b - 1)/x**2
fit f(x) 'data.txt' via a,b
This will, however, perform the least-squares regression, which is theoretically wrong in this case. Instead I need to perform the Poisson regression. How do I achieve this?
The link function of the Poisson distribution is ln(μ). Is there a way to use this information, eg. with the gnuplot's link command?
I am trying to implement discriminant condition codes in Keras as proposed in
Xue, Shaofei, et al., "Fast adaptation of deep neural network based
on discriminant codes for speech recognition."
The main idea is you encode each condition as an input parameter and let the network learn dependency between the condition and the feature-label mapping. On a new dataset instead of adapting the entire network you just tune these weights using backprop. For example say my network looks like this
X ---->|----|
|DNN |----> Y
Z --- >|----|
X: features Y: labels Z:condition codes
Now given a pretrained DNN, and X',Y' on a new dataset I am trying to estimate the Z' using backprop that will minimize prediction error on Y'. The math seems straightforward except I am not sure how to implement this in keras without having access to the backprop itself.
For instance, can I add an Input() layer with trainable=True with all other layers set to trainable= False. Can backprop in keras update more than just layer weights? Or is there a way to hack keras layers to do this?
Any suggestions welcome.
thanks
I figured out how to do this (exactly) in Keras by looking at fchollet's post here
Using the keras backend I was able to compute the gradient of my loss w.r.t to Z directly and used it to drive the update.
Code below:
import keras.backend as K
import numpy as np
model.summary() #Pretrained model
loss = K.categorical_crossentropy(Y, Y_out)
grads = K.gradients(loss, Z)
grads /= (K.sqrt(K.mean(K.square(grads)))+ 1e-5)
iterate = K.function([X,Z],[loss,grads])
step = 0.1
Z_adapt = Z_in.copy()
for i in range(100):
loss_val, grads_val = iterate([X_in,Z_adapt])
Z_adapt -= grads_val[0] * step
print "iter:",i,np.mean(loss_value)
print "Before:"
print model.evaluate([X_in, Z_in],Y_out)
print "After:"
print model.evaluate([X_in, Z_adapt],Y_out)
X,Y,Z are nodes in the model graph. Z_in is an initial value for Z'. I set it to an average value from the train set. Z_adapt is after 100 iterations of gradient descent and should give you a better result.
Assume that the size of Z is m x n. Then you can first define an input layer of size m * n x 1. The input will be an m * n x 1 vector of ones. You can define a dense layer containing m * n neurons and set trainable = True for it. The response of this layer will give you a flattened version of Z. Reshape it appropriately and give it as input to the rest of the network that can be appended ahead of this.
Keep in mind that if the size of Z is too large, then network may not be able to learn a dense layer of that many neurons. In that case, maybe you need to put additional constraints or look into convolutional layers. However, convolutional layers will put some constraints on Z.
I would like to make a prediction by using Least Squares Support Vector Machine for Regression, which is proposed by Suykens et al. I am using LS-SVMlab, which you can find the MATLAB toolbox here. Let's consider I have an independent variable X and a dependent variable Y, that both are simulated. I am following the instructions in the tutorial.
>>X = linspace(-1,1,50)’;
>>Y = (15*(X.^2-1).^2.*X.^4).*exp(-X)+normrnd(0,0.1,length(X),1);
>>type = ’function estimation’;
>>[gam,sig2] = tunelssvm({X,Y,type,[], [],’RBF_kernel’},’simplex’,...’leaveoneoutlssvm’,’mse’});
>>[alpha,b] = trainlssvm({X,Y,type,gam,sig2,’RBF_kernel’});
>>plotlssvm({X,Y,type,gam,sig2,’RBF_kernel’},{alpha,b});
The code above finds the best parameters using simplex method and leave-one-out cross validation and trains the model and give me alphas (support vector values for all the data points in the training set) and b coefficients. However, it does not give me the predictions of the variable Y. It only draws the plot. In some articles, I saw plots like the one below,
As I said before, the LS-SVM toolbox does not give me the predicted values of Y, it only draws the plot but no values in the workspace. How can I get these values and draw a graph of predicted values together with actual values?
There is one solution that I think of. By using X values in the training set, I re-run the model and get the prediction of values Y by using simlssvm command but it does not seem reasonable to me. Any solution that you can offer? Thanks in advance.
I am afraid you have answered your own question. The only way to obtain the prediction for the training points in LS-SVMLab is by simulating the training points after training your model.
[yp,alpha,b,gam,sig2,model] = lssvm(x,y,'f')
when u use this function yp is the predicted value