I've been using MATLAB for my time series dataset (for an electricity dataset) as a part of my course. It consists of 40,000+ samples. After the formation of neural network, I wanted to test its accuracy. I've been curious more on MAPE(mean absolute percentage error) and RMS(Root Mean Square) errors. To calculate them, I've used following lines of code.
mape_res = zeros(N_TRAIN);
mse_res = zeros(N_TRAIN);
for i_train = 1:N_TRAIN
Inp = inputs_consumption(i_train );
Actual_Output = targets_consumption( i_train + 1 );
Observed_Output = sim( ann, Inp );
mape_res(i_train) = abs(Observed_Output - Actual_Output)/Actual_Output;
mse_res(i_train) = Observed_Output - Actual_Output;
end
mape = sum(mape_res)/N_TRAIN;
mse = sum(power(mse_res,2))/N_TRAIN;
sprintf( 'The MSE on training is %g', mse )
sprintf( 'The MAPE on training is %g', mape )
The problem with above coding is that, for a large dataset(40K samples), it takes almost 15 minutes to iterate through all those loops and it's quite a long waiting for getting result for the error rate; Isn't there any other efficient way to calculate them?
You could always do a rolling average that gets updated each iteration, as follows:
mape_res = abs(Observed_Output - Actual_Output) / Actual_Output;
mse_res = Observed_Output - Actual_Output;
alpha = 1 / i_train;
mape = mape * (1 - alpha) + mape_res * alpha;
mse = mes * (1 - alpha) + power(mse_res,2) * alpha;
Then you could either display the resulting values each iteration, use them for stopping criteria if the desired error rate is reached, or both. This also has the added benefit of not requiring the initialization and population of the mape_res and mse_res vectors unless they happen to be needed elsewhere...
Edit: Do make sure to initialize the mape and mse values to zero prior to entering the for loop :)
Related
I was following Siraj Raval's videos on logistic regression using gradient descent :
1) Link to longer video :
https://www.youtube.com/watch?v=XdM6ER7zTLk&t=2686s
2) Link to shorter video :
https://www.youtube.com/watch?v=xRJCOz3AfYY&list=PL2-dafEMk2A7mu0bSksCGMJEmeddU_H4D
In the videos he talks about using gradient descent to reduce the error for a set number of iterations so that the function converges(slope becomes zero).
He also illustrates the process via code. The following are the two main functions from the code :
def step_gradient(b_current, m_current, points, learningRate):
b_gradient = 0
m_gradient = 0
N = float(len(points))
for i in range(0, len(points)):
x = points[i, 0]
y = points[i, 1]
b_gradient += -(2/N) * (y - ((m_current * x) + b_current))
m_gradient += -(2/N) * x * (y - ((m_current * x) + b_current))
new_b = b_current - (learningRate * b_gradient)
new_m = m_current - (learningRate * m_gradient)
return [new_b, new_m]
def gradient_descent_runner(points, starting_b, starting_m, learning_rate, num_iterations):
b = starting_b
m = starting_m
for i in range(num_iterations):
b, m = step_gradient(b, m, array(points), learning_rate)
return [b, m]
#The above functions are called below:
learning_rate = 0.0001
initial_b = 0 # initial y-intercept guess
initial_m = 0 # initial slope guess
num_iterations = 1000
[b, m] = gradient_descent_runner(points, initial_b, initial_m, learning_rate, num_iterations)
# code taken from Siraj Raval's github page
Why does the value of b & m continue to update for all the iterations? After a certain number of iterations, the function will converge, when we find the values of b & m that give slope = 0.
So why do we continue iteration after that point and continue updating b & m ?
This way, aren't we losing the 'correct' b & m values? How is learning rate helping the convergence process if we continue to update values after converging? Thus, why is there no check for convergence, and so how is this actually working?
In practice, most likely you will not reach to slope 0 exactly. Thinking of your loss function as a bowl. If your learning rate is too high, it is possible to overshoot over the lowest point of the bowl. On the contrary, if the learning rate is too low, your learning will become too slow and won't reach the lowest point of the bowl before all iterations are done.
That's why in machine learning, the learning rate is an important hyperparameter to tune.
Actually, once we reach a slope 0; b_gradient and m_gradient will become 0;
thus, for :
new_b = b_current - (learningRate * b_gradient)
new_m = m_current - (learningRate * m_gradient)
new_b and new_m will remain the old correct values; as nothing will be subtracted from them.
The total error for the network did not change on over 100,000 iterations.
The input is 22 values and the output is a single value. the input array is [195][22] and the output array is [195][1].
BasicNetwork network = new BasicNetwork();
network.addLayer(new BasicLayer(null,true,22));
network.addLayer(new BasicLayer(new ActivationSigmoid(),true,10));
network.addLayer(new BasicLayer(new ActivationSigmoid(),false,1));
network.getStructure().finalizeStructure();
network.reset();
MLDataSet training_data = new BasicMLDataSet(input, target_output);
final Backpropagation train = new Backpropagation(network, training_data);
int epoch = 1;
do {
train.iteration();
System.out.println("Epoch #" + epoch + " Error:" + train.getError());
epoch++;
}
while(train.getError() > 0.01);
{
train.finishTraining();
}
What is wrong with this code?
Depending on what the data you are trying to classify your network may be too small to transform the search space into a linearly separable problem. So try adding more neurons or layers - this will probably take longer to train. Unless it is already linearly separable and then a NN may be an inefficient way to solve this.
Also you don't have a training strategy, if the NN falls into local minima on the error surface it will be stuck there. See the encog user guide https://s3.amazonaws.com/heatonresearch-books/free/Encog3Java-User.pdf pg 166 has a list of training strategy's.
final int strategyCycles = 50;
final double strategyError = 0.25;
train.addStrategy(new ResetStrategy(strategyError,strategyCycles));
There are several experiments that rely on gradient ascent rather than gradient descent. I have looked into some approaches to using "cost" and the minimize function to simulate the "maximize" function, but I am still not certain I know how to properly implement a maximize() function. Also, in most of these cases, I would say they are closer to an unsupervised learning. So given this code concept for a cost function:
cost = (Yexpected - Ycalculated)^2
train_step = tf.train.AdamOptimizer(0.5).minimize(cost)
I would like to write something were I am following the positive gradient and there may not be a Yexpected value:
maxMe = Function(Ycalculated)
train_step = tf.train.AdamOptimizer(0.5).maximize(maxMe)
A good example of this need is "http://cs229.stanford.edu/proj2009/LvDuZhai.pdf" with Recurrent Reinforcement Learning.
I have read a few papers and references that state changing the sign will flip the direction of movement to increasing gradient, but given TensorFlow's internal calculation of the gradient, I am not sure if this will work to Maximize as I don't know of a way to validate the results:
maxMe = Function(Ycalculated)
train_step = tf.train.AdamOptimizer(0.5).minimize( -1 * maxMe )
The intuition is simple, the minimize() function keeps squashing the given value, for example, if you start with 5, then for every iteration (for example and depending on the learning rate), the value will become say, 4, then 3, then 2, 1, 0 and so on if possible to bring it down more. Now if you pass -5 at the beginning (which is in fact a +5 but you changed the sign explicitly), the gradient will try to change the parameters to bring the number down more, as for example, -5, -6, -7, -8, ...etc. But in fact, the function is increasing because we changed the sign, and the actual sign is (+). In other words, the gradient, in the latter case, is changing the parameters of the neural network in a way that maximizes the function, not minimizing it.
Toy example with arbitrary numbers:
The input x = 1.5, The weight parameter at time (t) w_t = 0.1,
The observed response y = 3.0, The learning rate lr = 0.1.
x * w = 0.15 (this is y predicted for the current w)
loss function = (3.0 - 0.15)^2 = 8.1
Applying gradient descent:
w_(t+1) = w_t - lr * (derivative of loss function with respect to w)
w_(t+1) = 0.1 - (0.1 * [1.5 * 2(0.15 - 3.0)]) = 0.1 - (-0.855) = 0.955
If we use the new w_(t+1) we will have:
1.5 * 0.955 = 1.49 (which is closer to the correct answer 3.0)
and the new loss is: (3.0 - 1.49)^2 = 2.27 (smaller error).
If we keep iterating, we will adjust w to a value that gives us the minimum cost possible.
Now lets repeat the same experiment but with the sign flipped to negative:
loss function = - (3.0 - 0.15)^2 = -8.1
Applying gradient descent:
w_(t+1) = w_t - lr * (derivative of loss function with respect to w)
w_(t+1) = 0.1 - (0.1 * [1.5 * -2(0.15 - 3.0)]) = 0.1 - 0.855 = −0.755
If we apply the new w_(t+1) we will have:
1.5 * −0.755 = −1.1325 and the new loss is: (3.0 - (-1.1325))^2 = 17.07
(the loss function is maximizing!).
That is also applicable to any differentiable function, but this is just a simple naive example to demonstrate the idea.
So, you can do, as you suggested already:
optimizer.minimize( -1 * value)
Or if you like, create a wrapper function (which in fact is needless, but just to mention it):
def maximize(optimizer, value, **kwargs):
return optimizer.minimize(-value, **kwargs)
I'm trying to learn theano and decided to implement linear regression (using their Logistic Regression from the tutorial as a template). I'm getting a wierd thing where T.grad doesn't work if my cost function uses .sum(), but does work if my cost function uses .mean(). Code snippet:
(THIS DOESN'T WORK, RESULTS IN A W VECTOR FULL OF NANs):
x = T.matrix('x')
y = T.vector('y')
w = theano.shared(rng.randn(feats), name='w')
b = theano.shared(0., name="b")
# now we do the actual expressions
h = T.dot(x,w) + b # prediction is dot product plus bias
single_error = .5 * ((h - y)**2)
cost = single_error.sum()
gw, gb = T.grad(cost, [w,b])
train = theano.function(inputs=[x,y], outputs=[h, single_error], updates = ((w, w - .1*gw), (b, b - .1*gb)))
predict = theano.function(inputs=[x], outputs=h)
for i in range(training_steps):
pred, err = train(D[0], D[1])
(THIS DOES WORK, PERFECTLY):
x = T.matrix('x')
y = T.vector('y')
w = theano.shared(rng.randn(feats), name='w')
b = theano.shared(0., name="b")
# now we do the actual expressions
h = T.dot(x,w) + b # prediction is dot product plus bias
single_error = .5 * ((h - y)**2)
cost = single_error.mean()
gw, gb = T.grad(cost, [w,b])
train = theano.function(inputs=[x,y], outputs=[h, single_error], updates = ((w, w - .1*gw), (b, b - .1*gb)))
predict = theano.function(inputs=[x], outputs=h)
for i in range(training_steps):
pred, err = train(D[0], D[1])
The only difference is in the cost = single_error.sum() vs single_error.mean(). What I don't understand is that the gradient should be the exact same in both cases (one is just a scaled version of the other). So what gives?
The learning rate (0.1) is way to big. Using mean make it divided by the batch size, so this help. But I'm pretty sure you should make it much smaller. Not just dividing by the batch size (which is equivalent to using mean).
Try a learning rate of 0.001.
Try dividing your gradient descent step size by the number of training examples.
I have this Backpropagation implementation in MATLAB, and have an issue with training it. Early on in the training phase, all of the outputs go to 1. I have normalized the input data(except the desired class, which is used to generate a binary target vector) to the interval [0, 1]. I have been referring to the implementation in Artificial Intelligence: A Modern Approach, Norvig et al.
Having checked the pseudocode against my code(and studying the algorithm for some time), I cannot spot the error. I have not been using MATLAB for that long, so have been trying to use the documentation where needed.
I have also tried different amounts of nodes in the hidden layer and different learning rates (ALPHA).
The target data encodings are as follows: when the target is to classify as, say 2, the target vector would be [0,1,0], say it were 1, [1, 0, 0] so on and so forth. I have also tried using different values for the target, such as (for class 1 for example) [0.5, 0, 0].
I noticed that some of my weights go above 1, resulting in large net values.
%Topological constants
NUM_HIDDEN = 8+1;%written as n+1 so is clear bias is used
NUM_OUT = 3;
%Training constants
ALPHA = 0.01;
TARG_ERR = 0.01;
MAX_EPOCH = 50000;
%Read and normalize data file.
X = normdata(dlmread('iris.data'));
X = shuffle(X);
%X_test = normdata(dlmread('iris2.data'));
%epocherrors = fopen('epocherrors.txt', 'w');
%Weight matrices.
%Features constitute size(X, 2)-1, however size is (X, 2) to allow for
%appending bias.
w_IH = rand(size(X, 2), NUM_HIDDEN)-(0.5*rand(size(X, 2), NUM_HIDDEN));
w_HO = rand(NUM_HIDDEN+1, NUM_OUT)-(0.5*rand(NUM_HIDDEN+1, NUM_OUT));%+1 for bias
%Layer nets
net_H = zeros(NUM_HIDDEN, 1);
net_O = zeros(NUM_OUT, 1);
%Layer outputs
out_H = zeros(NUM_HIDDEN, 1);
out_O = zeros(NUM_OUT, 1);
%Layer deltas
d_H = zeros(NUM_HIDDEN, 1);
d_O = zeros(NUM_OUT, 1);
%Control variables
error = inf;
epoch = 0;
%Run the algorithm.
while error > TARG_ERR && epoch < MAX_EPOCH
for n=1:size(X, 1)
x = [X(n, 1:size(X, 2)-1) 1]';%Add bias for hiddens & transpose to column vector.
o = X(n, size(X, 2));
%Forward propagate.
net_H = w_IH'*x;%Transposed w.
out_H = [sigmoid(net_H); 1]; %Append 1 for bias to outputs
net_O = w_HO'*out_H;
out_O = sigmoid(net_O); %Again, transposed w.
%Calculate output deltas.
d_O = ((targetVec(o, NUM_OUT)-out_O) .* (out_O .* (1-out_O)));
%Calculate hidden deltas.
for i=1:size(w_HO, 1);
delta_weight = 0;
for j=1:size(w_HO, 2)
delta_weight = delta_weight + d_O(j)*w_HO(i, j);
end
d_H(i) = (out_H(i)*(1-out_H(i)))*delta_weight;
end
%Update hidden-output weights
for i=1:size(w_HO, 1)
for j=1:size(w_HO, 2)
w_HO(i, j) = w_HO(i, j) + (ALPHA*out_H(i)*d_O(j));
end
end
%Update input-hidden weights.
for i=1:size(w_IH, 1)
for j=1:size(w_IH, 2)
w_IH(i, j) = w_IH(i, j) + (ALPHA*x(i)*d_H(j));
end
end
out_O
o
%out_H
%w_IH
%w_HO
%d_O
%d_H
end
end
function outs = sigmoid(nets)
outs = zeros(size(nets, 1), 1);
for i=1:size(nets, 1)
if nets(i) < -45
outs(i) = 0;
elseif nets(i) > 45
outs(i) = 1;
else
outs(i) = 1/1+exp(-nets(i));
end
end
end
From what we've established in the comments the only thing that comes in my mind are all recipes written down together in this great NN archive:
ftp://ftp.sas.com/pub/neural/FAQ2.html#questions
First things you could try are:
1) How to avoid overflow in the logistic function? Probably that's the problem - many times I've implemented NNs the problem was with such an overflow.
2) How should categories be encoded?
And more general:
3) How does ill-conditioning affect NN training?
4) Help! My NN won't learn! What should I do?
After the discussion it turns out the problem lies within the sigmoid function:
function outs = sigmoid(nets)
%...
outs(i) = 1/1+exp(-nets(i)); % parenthesis missing!!!!!!
%...
end
It should be:
function outs = sigmoid(nets)
%...
outs(i) = 1/(1+exp(-nets(i)));
%...
end
The lack of parenthesis caused that the sigmoid output was larger than 1 sometimes. That made the gradient calculation incorrect (because it wasn't a gradient of this function). This caused the gradient to be negative. And this caused that the delta for the output layer was most of the time in the wrong direction. After the fix (the after correctly maintaining the error variable - this seems to be missing in your code) all seems to work fine.
Beside that, there are two other main problems with this code:
1) No bias. Without the bias each neuron can only represent a line which crosses the origin. If data is normalized (i.e. values are between 0 and 1), some configurations are inseparable.
2) Lack of guarding against high gradient values (point 1 in my previous answer).