I have a dataset with 274 samples (9 months) of the daily energy (Watts.hour) used on a residential household. I'm not sure if i'm applying the lpc function correctly.
My code is the following:
filename='9-months.csv';
energy = csvread(filename);
C=zeros(5,1);
counter=0;
N=3;
for n=274:-1:31
w2=energy(1:n-1,1);
a=lpc(w2,N);
energy_estimated=0;
for X = 1:N
energy_estimated = energy_estimated + (-a(X+1)*energy(n-X));
end
w_real=energy(n);
error2=abs(w_real-energy_estimated);
counter=counter+1;
C(counter,1)=error2;
end
mean_error=round(mean(C));
Being "n" the sample on analysis, I will use the energy array's values, from 1 to n-1, to calculate the lpc coefficientes (with N=3).
After that, it will apply the calculated coefficients on the "for" cycle presented, in order to calculate the estimated energy.
Finally, error2 outputs the error between the real energy and estimated value.
On the example presented ( http://www.mathworks.com/help/signal/ref/lpc.html ) some filters are used. Do I need to apply any filter to it? Is my methodology correct?
Thank you very much in advance!
The lpc seems to be used correctly, but there are a few other things about your code. I am adressign the part at he "for n" :
for n=31:274 %for me it would seem more logically to go forward in time
w2=energy(1:n-1,1);
a=lpc(w2,N);
energy_estimate=filter([0 -a(2:end)],1,w2);
energy_estimate=energy_estimate(end);
estimates(n)=energy_estimate;
end
error=energy(31:274)-estimates(31:274)';
meanerror=mean(error); %you dont really round mean errors
filter is exactly what you are trying to do with the X=1:N loop. but this will perform the calculation for the entire w2 vector. If you just want the last value take the (end) command as well.
Now there is no reason to calculate the error for every single value and then add them to a vector you can do that faster after the calculation.
Now if your trying to estimate future values with a lpc it could work like that, but you are implying that every value is only dependend on the last 3 values. Have you tried something like a polynominal approach? i would think that this would be closer to reality.
Related
I have the following code from my predecessor. I am unable to figure out what is the math that is happening here and how is the values avgCov and stdCov are different, and what they signify.
Cprofile_f is a curve similar to gaussian curve, like a peak. Cprofile_f is an array of known size (5700).
b1, d1 are index values. Usually, b1 is 2000, d1 is 4300.
avgCov=sum(Cprofile_f(b1:d1))/(d1-b1)
stdCov=0;
for ii=b1:d1
stdCov =stdCov + sqrt((avgCov - Cprofile_f(ii))^2);
end
stdCov =1- stdCov/(d1-b1)/avgCov
Trying to figure out, what stdCov mean here.
Looks like it's computing the average (avgCov) and standard deviation (stdCov, sort of) in order to compute 1 minus the coefficient of variation (stored in stdCov).
https://en.wikipedia.org/wiki/Coefficient_of_variation
I have this piece of code
N=10^4;
for i = 1:N
[E,X,T] = fffun(); % Stochastic simulation. Returns every time three different vectors (whose length is 10^3).
X_(i,:)=X;
T_(i,:)=T;
GRID=[GRID T];
end
GRID=unique(GRID);
% Second part
for i=1:N
for j=1:(kmax)
f=find(GRID==T_(i,j) | GRID==T_(i,j+1));
s=f(1);
e=f(2)-1;
counter(X_(i,j), s:e)=counter(X_(i,j), s:e)+1;
end
end
The code performs N different simulations of a stochastic process (which consists of 10^3 events, occurring at discrete moments (T vector) that depends on the specific simulation.
Now (second part) I want to know, as a function of time istant, how many simulations are in a particular state (X assumes value between 1 and 10). The idea I had: create a grid vector with all the moments at which something happens in any simulation. Then, looping over the simulations, loop over the timesteps in which something happens and incrementing all the counter indeces that corresponds to this particular slice of time.
However this second part is very heavy (I mean days of processing on a standard quad-core CPU). And it shouldn't.
Are there any ideas (maybe about comparing vectors in a more efficient way) to cut the CPU time?
This is a standalone 'second_part'
N=5000;
counter=zeros(11,length(GRID));
for i=1:N
disp(['Counting sim #' num2str(i)]);
for j=1:(kmax)
f=find(GRID==T_(i,j) | GRID==T_(i,j+1),2);
s=f(1);
e=f(2)-1;
counter(X_(i,j), s:e)=counter(X_(i,j), s:e)+1;
end
end
counter=counter/N;
stop=find(GRID==Tmin);
stop=stop-1;
plot(counter(:,(stop-500):stop)')
with associated dummy data ( filedropper.com/data_38 ). In the real context the matrix has 2x rows and 10x columns.
Here is what I understand:
T_ is a matrix of time steps from N simulations.
X_ is a matrix of simulation state at T_ in those simulations.
so if you do:
[ut,~,ic]= unique(T_(:));
you get ic which is a vector of indices for all unique elements in T_. Then you can write:
counter = accumarray([ic X_(:)],1);
and get counter with no. of rows as your unique timesteps, and no. of columns as the unique states in X_ (which are all, and must be, integers). Now you can say that for each timestep ut(k) the number of time that the simulation was in state m is counter(k,m).
In your data, the only combination of m and k that has a value greater than 1 is (1,1).
Edit:
From the comments below, I understand that you record all state changes, and the time steps when they occur. Then every time a simulation change a state you want to collect all the states from all simulations and count how many states are from each type.
The main problem here is that your time is continuous, so basically each element in T_ is unique, and you have over a million time steps to loop over. Fully vectorizing such a process will need about 80GB of memory which will probably stuck your computer.
So I looked for a combination of vectorizing and looping through the time steps. We start by finding all unique intervals, and preallocating counter:
ut = unique(T_(:));
stt = 11; % no. of states
counter = zeros(stt,numel(ut));r = 1:size(T_,1);
r = 1:size(T_,1); % we will need that also later
Then we loop over all element in ut, and each time look for the relevant timestep in T_ in all simulations in a vectorized way. And finally we use histcounts to count all the states:
for k = 1:numel(ut)
temp = T_<=ut(k); % mark all time steps before ut(k)
s = cumsum(temp,2); % count the columns
col_ind = s(:,end); % fins the column index for each simulation
% convert the coulmns to linear indices:
linind = sub2ind(size(T_),r,col_ind.');
% count the states:
counter(:,k) = histcounts(X_(linind),1:stt+1);
end
This takes about 4 seconds at my computer for 1000 simulations, so it adds to a little more than one hour for the whole process. Not very quick...
You can try also one or two of the tweaks below to squeeze run time a little bit more:
As you can read here, accumarray seems to work faster in small arrays then histcouns. So may want to switch to it.
Also, computing linear indices directly is a quicker method than sub2ind, so you may want to try that.
implementing these suggestions in the loop above, we get:
R = size(T_,1);
r = (1:R).';
for k = 1:K
temp = T_<=ut(k); % mark all time steps before ut(k)
s = cumsum(temp,2); % count the columns
col_ind = s(:,end); % fins the column index for each simulation
% convert the coulmns to linear indices:
linind = R*(col_ind-1)+r;
% count the states:
counter(:,k) = accumarray(X_(linind),1,[stt 1]);
end
In my computer switching to accumarray and or removing sub2ind gain a slight improvement but it was not consistent (using timeit for testing on 100 or 1K elements in ut), so you better test it yourself. However, this still remains very long.
One thing that you may want to consider is trying to discretize your timesteps, so you will have much less unique elements to loop over. In your data about 8% of the time intervals a smaller than 1. If you can assume that this is short enough to be treated as one time step, then you could round your T_ and get only ~12.5K unique elements, which take about a minute to loop over. You can do the same for 0.1 intervals (which are less than 1% of the time intervals), and get 122K elements to loop over, what will take about 8 hours...
Of course, all the timing above are rough estimates using the same algorithm. If you do choose to round the times there may be even better ways to solve this.
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.
Problem
I have two arrays (Xa and Xb) that contain measurements of the same physical signal, but they are taken at different sample rates. Lastly, physical logging of Xa data starts at a different time, than that of Xb. The logging of data also stops at different time.
i.e.
(The following is just a summary of important statements, not code.)
sampleRatea > sampleRateb % Resolution of Xa is greater than that of Xb
t0a ~= t0b % Start times are not equal
t1a ~= t1b % End times are not equal
Objective
Find the necessary shift in indices that will best line up these sets of data.
Approach
Use fmincon to find the index that minimizes the mean squared error (MSE) between versions Xa and Xb that are edited to have the same sample rate (perhaps using the interpolation function).
I have tried to do this but it always seems that I have too many degrees of freedom. Is there anyone who can shed some light on a process that might facilitate this process?
Assuming you have two samples with constant frequencies, the problem reduces to something quite simple:
Find scale, location such that:
Xa , at timestamps corresponding to its index, makes the best match with Xb at timstamps corresponding to location + scale * its index.
If you agree with this you can see that only two degrees of freedom are left, if you know the ratio of sample rates it even reduces to just 1 degree of freedom.
I believe that now the hard part is done, but some work still remains:
Judge how good two samples with timestamps and values match
Find the optimal combination of your location and scale parameter
Note that, assuming you complete these 2 steps properly, the solution should be optimal for finding the optimal timestamps. As you are looking for a shift in (integer) indices, translating these timestamps back to indices may not be result in the real optimum but it should be pretty close.
Here is a quick-and-dirty solution that should be enough to get you started. Given your input signals Xa and Xb sampled at sampleRatea and sampleRateb respectively:
g = gcd(sampleRatea,sampleRateb);
Ya = interp(Xa,sampleRateb/g);
Yb = interp(Xb,sampleRatea/g);
Yfs = sampleRatea*sampleRateb/g;
[acor,lag] = xcorr(Ya,Yb);
time_shift = lag(acor == max(acor))/Yfs;
The variable time_shift will tell you the time elapsed between the start of A and the start of B. If B starts first, the result will be negative.
If your sampling rates are relatively prime, this will be horribly inefficient. If one is an integer multiple of the other, or they have a relatively large GCD, it will be much better.
Having read carefully the previous question
Random numbers that add to 100: Matlab
I am struggling to solve a similar but slightly more complex problem.
I would like to create an array of n elements that sums to 1, however I want an added constraint that the minimum increment (or if you like number of significant figures) for each element is fixed.
For example if I want 10 numbers that sum to 1 without any constraint the following works perfectly:
num_stocks=10;
num_simulations=100000;
temp = [zeros(num_simulations,1),sort(rand(num_simulations,num_stocks-1),2),ones(num_simulations,1)];
weights = diff(temp,[],2);
I foolishly thought that by scaling this I could add the constraint as follows
num_stocks=10;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp2 = [zeros(num_simulations,1),sort(round(rand(num_simulations,num_stocks-1)*scaling)/scaling,2),ones(num_simulations,1)];
weights2 = diff(temp2,[],2);
However though this works for small values of n & small values of increment, if for example n=1,000 & the increment is 0.1% then over a large number of trials the first and last numbers have a mean which is consistently below 0.1%.
I am sure there is a logical explanation/solution to this but I have been tearing my hair out to try & find it & wondered anybody would be so kind as to point me in the right direction. To put the problem into context create random stock portfolios (hence the sum to 1).
Thanks in advance
Thank you for the responses so far, just to clarify (as I think my initial question was perhaps badly phrased), it is the weights that have a fixed increment of 0.1% so 0%, 0.1%, 0.2% etc.
I did try using integers initially
num_stocks=1000;
min_increment=0.001;
num_simulations=100000;
scaling=1/min_increment;
temp = [zeros(num_simulations,1),sort(randi([0 scaling],num_simulations,num_stocks-1),2),ones(num_simulations,1)*scaling];
weights = (diff(temp,[],2)/scaling);
test=mean(weights);
but this was worse, the mean for the 1st & last weights is well below 0.1%.....
Edit to reflect excellent answer by Floris & clarify
The original code I was using to solve this problem (before finding this forum) was
function x = monkey_weights_original(simulations,stocks)
stockmatrix=1:stocks;
base_weight=1/stocks;
r=randi(stocks,stocks,simulations);
x=histc(r,stockmatrix)*base_weight;
end
This runs very fast, which was important considering I want to run a total of 10,000,000 simulations, 10,000 simulations on 1,000 stocks takes just over 2 seconds with a single core & I am running the whole code on an 8 core machine using the parallel toolbox.
It also gives exactly the distribution I was looking for in terms of means, and I think that it is just as likely to get a portfolio that is 100% in 1 stock as it is to geta portfolio that is 0.1% in every stock (though I'm happy to be corrected).
My issue issue is that although it works for 1,000 stocks & an increment of 0.1% and I guess it works for 100 stocks & an increment of 1%, as the number of stocks decreases then each pick becomes a very large percentage (in the extreme with 2 stocks you will always get a 50/50 portfolio).
In effect I think this solution is like the binomial solution Floris suggests (but more limited)
However my question has arrisen because I would like to make my approach more flexible & have the possibility of say 3 stocks & an increment of 1% which my current code will not handle correctly, hence how I stumbled accross the original question on stackoverflow
Floris's recursive approach will get to the right answer, but the speed will be a major issue considering the scale of the problem.
An example of the original research is here
http://www.huffingtonpost.com/2013/04/05/monkeys-stocks-study_n_3021285.html
I am currently working on extending it with more flexibility on portfolio weights & numbers of stock in the index, but it appears my programming & probability theory ability are a limiting factor.......
One problem I can see is that your formula allows for numbers to be zero - when the rounding operation results in two consecutive numbers to be the same after sorting. Not sure if you consider that a problem - but I suggest you think about it (it would mean your model portfolio has fewer than N stocks in it since the contribution of one of the stocks would be zero).
The other thing to note is that the probability of getting the extreme values in your distribution is half of what you want them to be: If you have uniformly distributed numbers from 0 to 1000, and you round them, the numbers that round to 0 were in the interval [0 0.5>; the ones that round to 1 came from [0.5 1.5> - twice as big. The last number (rounding to 1000) is again from a smaller interval: [999.5 1000]. Thus you will not get the first and last number as often as you think. If instead of round you use floor I think you will get the answer you expect.
EDIT
I thought about this some more, and came up with a slow but (I think) accurate method for doing this. The basic idea is this:
Think in terms of integers; rather than dividing the interval 0 - 1 in steps of 0.001, divide the interval 0 - 1000 in integer steps
If we try to divide N into m intervals, the mean size of a step should be N / m; but being integer, we would expect the intervals to be binomially distributed
This suggests an algorithm in which we choose the first interval as a binomially distributed variate with mean (N/m) - call the first value v1; then divide the remaining interval N - v1 into m-1 steps; we can do so recursively.
The following code implements this:
% random integers adding up to a definite sum
function r = randomInt(n, limit)
% returns an array of n random integers
% whose sum is limit
% calls itself recursively; slow but accurate
if n>1
v = binomialRandom(limit, 1 / n);
r = [v randomInt(n-1, limit - v)];
else
r = limit;
end
function b = binomialRandom(N, p)
b = sum(rand(1,N)<p); % slow but direct
To get 10000 instances, you run this as follows:
tic
portfolio = zeros(10000, 10);
for ii = 1:10000
portfolio(ii,:) = randomInt(10, 1000);
end
toc
This ran in 3.8 seconds on a modest machine (single thread) - of course the method for obtaining a binomially distributed random variate is the thing slowing it down; there are statistical toolboxes with more efficient functions but I don't have one. If you increase the granularity (for example, by setting limit=10000) it will slow down more since you increase the number of random number samples that are generated; with limit = 10000 the above loop took 13.3 seconds to complete.
As a test, I found mean(portfolio)' and std(portfolio)' as follows (with limit=1000):
100.20 9.446
99.90 9.547
100.09 9.456
100.00 9.548
100.01 9.356
100.00 9.484
99.69 9.639
100.06 9.493
99.94 9.599
100.11 9.453
This looks like a pretty convincing "flat" distribution to me. We would expect the numbers to be binomially distributed with a mean of 100, and standard deviation of sqrt(p*(1-p)*n). In this case, p=0.1 so we expect s = 9.4868. The values I actually got were again quite close.
I realize that this is inefficient for large values of limit, and I made no attempt at efficiency. I find that clarity trumps speed when you develop something new. But for instance you could pre-compute the cumulative binomial distributions for p=1./(1:10), then do a random lookup; but if you are just going to do this once, for 100,000 instances, it will run in under a minute; unless you intend to do it many times, I wouldn't bother. But if anyone wants to improve this code I'd be happy to hear from them.
Eventually I have solved this problem!
I found a paper by 2 academics at John Hopkins University "Sampling Uniformly From The Unit Simplex"
http://www.cs.cmu.edu/~nasmith/papers/smith+tromble.tr04.pdf
In the paper they outline how naive algorthms don't work, in a way very similar to woodchips answer to the Random numbers that add to 100 question. They then go on to show that the method suggested by David Schwartz can also be slightly biased and propose a modified algorithm which appear to work.
If you want x numbers that sum to y
Sample uniformly x-1 random numbers from the range 1 to x+y-1 without replacement
Sort them
Add a zero at the beginning & x+y at the end
difference them & subtract 1 from each value
If you want to scale them as I do, then divide by y
It took me a while to realise why this works when the original approach didn't and it come down to the probability of getting a zero weight (as highlighted by Floris in his answer). To get a zero weight in the original version for all but the 1st or last weights your random numbers had to have 2 values the same but for the 1st & last ones then a random number of zero or the maximum number would result in a zero weight which is more likely.
In the revised algorithm, zero & the maximum number are not in the set of random choices & a zero weight occurs only if you select two consecutive numbers which is equally likely for every position.
I coded it up in Matlab as follows
function weights = unbiased_monkey_weights(num_simulations,num_stocks,min_increment)
scaling=1/min_increment;
sample=NaN(num_simulations,num_stocks-1);
for i=1:num_simulations
allcomb=randperm(scaling+num_stocks-1);
sample(i,:)=allcomb(1:num_stocks-1);
end
temp = [zeros(num_simulations,1),sort(sample,2),ones(num_simulations,1)*(scaling+num_stocks)];
weights = (diff(temp,[],2)-1)/scaling;
end
Obviously the loop is a bit clunky and as I'm using the 2009 version the randperm function only allows you to generate permutations of the whole set, however despite this I can run 10,000 simulations for 1,000 numbers in 5 seconds on my clunky laptop which is fast enough.
The mean weights are now correct & as a quick test I replicated woodchips generating 3 numbers that sum to 1 with the minimum increment being 0.01% & it also look right
Thank you all for your help and I hope this solution is useful to somebody else in the future
The simple answer is to use the schemes that work well with NO minimum increment, then transform the problem. As always, be careful. Some methods do NOT yield uniform sets of numbers.
Thus, suppose I want 11 numbers that sum to 100, with a constraint of a minimum increment of 5. I would first find 11 numbers that sum to 45, with no lower bound on the samples (other than zero.) I could use a tool from the file exchange for this. Simplest is to simply sample 10 numbers in the interval [0,45]. Sort them, then find the differences.
X = diff([0,sort(rand(1,10)),1]*45);
The vector X is a sample of numbers that sums to 45. But the vector Y sums to 100, with a minimum value of 5.
Y = X + 5;
Of course, this is trivially vectorized if you wish to find multiple sets of numbers with the given constraint.