I have an input and target series. However, the target series lags 3 steps behind the input. Can I still use narx or some other network?
http://www.mathworks.co.uk/help/toolbox/nnet/ref/narxnet.html
Predict: y(t+1)
Input:
x(t) |?
x(t-1)|?
x(t-2)|?
x(t-3)|y(t-3)
x(t-4)|y(t-4)
x(t-5)|y(t-5)
...
During my training, I have y(t-2), y(t-1), y(t) in advance, but when I do the prediction in real life, those values are only available 3 steps later, because I calculate y from the next 3 inputs.
Here are some options
1) Also, you could have two inputs and one output as
x(t), y(t-3) -> y(t)
x(t-1),y(t-4) -> y(t-1)
x(t-2),y(t-5) -> y(t-2)
...
and predict the single output y(t)
2) You could also use ar or arx with na = 0, nb > 0, and nk = 3.
3) Also, you could have four inputs, where 2 of the inputs are estimated and one output as
x(t), y(t-3), ye(t-2), ye(t-1) -> y(t)
x(t-1),y(t-4), y(t-3), ye(t-2) -> y(t-1)
x(t-2),y(t-5), y(t-4), y(t-3) -> y(t-2)
...
and predict the single output y(t), using line 3 and higher as training data
4) You could setup the input/output as in steps one or three and use s4sid
I have a similar problem, but without any measurable inputs. And I'm trying to see how much error there is as the forecast distance and model complexity are increased. But I've only tried approach 2 and set nb = 5 to 15 by 5 and varied nk from 20 to 150 by 10 and plotted contours of maximum error. In my case, I'm not interested in predictions of less than 20 time steps.
Define a window of your choice( you need to try different sizes to see which is the best value). Now make this problem a regression problem. Use values of xt and yt from t=T-2... T-x where x-2 is the size of window. Now use regress() to train a regression model and use it for prediction.
Related
I've extracted certain data from an Excel file.
It involves two columns : one for certain periods and another for corresponing
daily prices. Followings are my codes.(t1 and t2 are user inputs.)
row_1 = find(period==t1)
row_2 = find(period==t2)
f_0 = period(row_1:row_2, 1)
f_1 = price(row_1:row_2 , 1)
y_1 = plot(handles.axes2, f_0, f_1)
f_0 : period (x-axis), f_1 : price(y-axis)
My goal is to express the trend of price fluctuations by using sounds.
So the way I came up with this is as follows.
Step1: Find the maximum and minimum value of the price corresponding to the given period. Step2: Divide the distances between these two points into eight sections. Step3: Allocate eight musical scales(C D E F G A B C) to each
eight sections and play it.
At my level, I achieved to find the min/max values of the given period.
But from the next stage, I can't come up with any ideas.
Please help me with any advice.
If I understand you correctly, you want to allocate eight musical scales to divided period, and such codes may help.
%% let's play some music~
clc; clear;
%% Set the Sampling frequency & time period
fs=44100;
t=0:1/fs:0.5;
%% eight musical scales
Cscale{1}=sin(2*pi*262*t); %c-do
Cscale{2}=sin(2*pi*294*t); %c-re
Cscale{3}=sin(2*pi*330*t); %c-mi
Cscale{4}=sin(2*pi*349*t); %c-fa
Cscale{5}=sin(2*pi*392*t); %c-so
Cscale{6}=sin(2*pi*440*t); %c-la
Cscale{7}=sin(2*pi*494*t); %c-ti
Cscale{8}=sin(2*pi*523*t); %c-do-high
%you could call "sound(Cscale{i},fs)" to paly each scales
%% Divide the distances between these two points
% the highest point must be special treated
Min_p=0;
Max_p=8;
Sample_p=[0 1 2 3 4 5 6 7 8];
for i=1:length(Sample_p)
S_p=Sample_p(i);
if (S_p == Max_p)
sound(Cscale{end},fs);
else
%Find the correct music scale and play it
sound(Cscale{1+floor(8*(Sample_p(i)-Min_p)/(Max_p-Min_p))},fs);
end
pause(0.5)
end
Here is what I looked at(you may need google translation because it is written in Chinese)
http://blog.csdn.net/weaponsun/article/details/46695255
I'm attempting to model fMRI data so I can check the efficacy of an experimental design. I have been following a couple of tutorials and have a question.
I first need to model the BOLD response by convolving a stimulus input time series with a canonical haemodynamic response function (HRF). The first tutorial I checked said that one can make an HRF that is of any amplitude as long as the 'shape' of the HRF is correct so they created the following HRF in matlab:
hrf = [ 0 0 1 5 8 9.2 9 7 4 2 0 -1 -1 -0.8 -0.7 -0.5 -0.3 -0.1 0 ]
And then convolved the HRF with the stimulus by just using 'conv' so:
hrf_convolved_with_stim_time_series = conv(input,hrf);
This is very straight forward but I want my model to eventually be as accurate as possible so I checked a more advanced tutorial and they did the following. First they created a vector of 20 timepoints then used the 'gampdf' function to create the HRF.
t = 1:1:20; % MEASUREMENTS
h = gampdf(t,6) + -.5*gampdf(t,10); % HRF MODEL
h = h/max(h); % SCALE HRF TO HAVE MAX AMPLITUDE OF 1
Is there a benefit to doing it this way over the simpler one? I suppose I have 3 specific questions.
The 'gampdf' help page is super short and only says the '6' and '10' in each function call represents 'A' which is a 'shape' parameter. What does this mean? It gives no other information. Why is it 6 in the first call and 10 in the second?
This question is directly related to the above one. This code is written for a situation where there is a TR = 1 and the stimulus is very short (like 1s). In my situation my TR = 2 and my stimulus is quite long (12s). I tried to adapt the above code to make a working HRF for my situation by doing the following:
t = 1:2:40; % 2s timestep with the 40 to try to equate total time to above
h = gampdf(t,6) + -.5*gampdf(t,10); % HRF MODEL
h = h/max(h); % SCALE HRF TO HAVE MAX AMPLITUDE OF 1
Because I have no idea what the 'gampdf' parameters mean (or what that line does, in all actuality) I'm not sure this gives me what I'm looking for. I essentially get out 20 values where 1-14 have SOME numeric value in them but 15-20 are all 0. I'm assuming there will be a response during the entire 12s stimulus period (first 6 TRs so values 1-6) with the appropriate rectification which could be the rest of the values but I'm not sure.
Final question. The other code does not 'scale' the HRF to have an amplitude of 1. Will that matter, ultimately?
The canonical HRF you choose is dependent upon where in the brain the BOLD signal is coming from. It would be inappropriate to choose just any HRF. Your best source of a model is going to come from a lit review. I've linked a paper discussing the merits of multiple HRF models. The methods section brings up some salient points.
I have a univariate time series data. I want to do a multistep prediction.
I came across this question which explains time series one step prediction.
but I am interested in multistep ahead prediction.
e.g typical univariate time series data looks like
time value
---- ------
t1 a1
t2 a2
..........
..........
t100 a100.
Suppose, I want 3 step ahead prediction.
Can I frame my problem like
TrainX TrainY
[a1,a2,a3,a4,a5,a6] -> [a7,a8,a9]
[a2,a3,a4,a5,a6,a7] -> [a8,a9,a10]
[a3,a4,a5,a6,a7,a8] -> [a9,a10,a11]
.................. ...........
.................. ...........
I am using keras and tensorflow as backend
First layer has 50 neurons and expects 6 inputs.
hidden layer has 30 neurons
output layer has 3 neurons i.e (outputs three time series values)
model = Sequential()
model.add(Dense(50, input_dim=6, activation='relu',kernel_regularizer=regularizers.l2(0.01)))
model.add(Dense(30, activation='relu',kernel_regularizer=regularizers.l2(0.01)))
model.add(Dense(3))
model.compile(loss='mean_squared_error', optimizer='adam')
model.fit(TrainX, TrainY, epochs=300, batch_size=16)
My model will be able to predict a107,a108,a109
,when my input is a101,a102,a103,a104,a105,a106
Is this a valid model ? Am I missing some thing?
That model might do it, but you should probably benefit from using LSTM layers (recurrent networks for sequences).
#TrainX.shape = (total of samples, time steps, features per step)
#TrainX.shape = (total of samples, 6, 1)
model.add(LSTM(50,input_shape=(6,1),return_sequences=True, ....))
model.add(LSTM(30,return_sequences=True, ....))
model.add(LSTM(3,return_sequences=False, ....))
You may be missing an activation function that limits the result to the possible range of the value you want to predict.
Often we work with values from 0 to 1 (activation='sigmoid') or from -1 to 1 (activation='tanh').
This would also require that the input be limited to these values, since inputs and outputs are the same.
I want to fit a data sets with Gaussian mixture model, the data sets contains about 120k samples and each sample has about 130 dimensions. When I use matlab to do it, so I run scripts (with cluster number 1000):
gm = fitgmdist(data, 1000, 'Options', statset('Display', 'iter'), 'RegularizationValue', 0.01);
I get the following outputs:
iter log-likelihood
1 -6.66298e+07
2 -1.87763e+07
3 -5.00384e+06
4 -1.11863e+06
5 299767
6 985834
7 1.39525e+06
8 1.70956e+06
9 1.94637e+06
The log likelihood is bigger than 0! I think it's unreasonable, and don't know why.
Could somebody help me?
First of all, it is not a problem of how large your dataset is.
Here is some code that produces similar results with a quite small dataset:
options = statset('Display', 'iter');
x = ones(5,2) + (rand(5,2)-0.5)/1000;
fitgmdist(x,1,'Options',options);
this produces
iter log-likelihood
1 64.4731
2 73.4987
3 73.4987
Of course you know that the log function (the natural logarithm) has a range from -inf to +inf. I guess your problem is that you think the input to the log (i.e. the aposteriori function) should be bounded by [0,1]. Well, the aposteriori function is a pdf function, which means that its value can be very large for very dense dataset.
PDFs must be positive (which is why we can use the log on them) and must integrate to 1. But they are not bounded by [0,1].
You can verify this by reducing the density in the above code
x = ones(5,2) + (rand(5,2)-0.5)/1;
fitgmdist(x,1,'Options',options);
this produces
iter log-likelihood
1 -8.99083
2 -3.06465
3 -3.06465
So, I would rather assume that your dataset contains several duplicate (or very close) values.
I'm trying to simulate an optical network algorithm in MATLAB for a homework project. Most of it is already done, but I have an issue with the diagrams I'm getting.
In the simulation I'm generating exponential traffic, however, for low lambda values (0.1) I'm getting very high packet drop rates (99%). I wrote a sample here which is very close to the testbench I'm running on my simulator.
% Run the simulation 10 times, with different lambda values
l = [1 2 3 4 5 6 7 8 9 10];
for i=l(1):l(end)
X = rand();
% In the 'real' simulation the following line defines the time
% when the next packet generation event will occur. Suppose that
% i is the current time
t_poiss = i + ceil((-log(X)/(i/10)));
distr(i)=t_poiss;
end
figure, plot(distr)
axis square
grid on;
title('Exponential test:')
The resulting image is
The diagram I'm getting in this sample is IDENTICAL to the diagram I'm getting for the drop rate/λ. So I would like to ask if I'm doing something wrong or if I miss something? Is this the right thing to expect?
So the problem is coming from might be a numerical problem. Since you are generating a random number for X, the number might be incredibly small - say, close to zero. If you have a number close to zero numerically, log(X) is going to be HUGE. So your calculation of t_poiss will be huge. I would suggest doing something like X = rand() + 1 to make sure that X is never close to zero.