Is it possible to find the difference beetwen two arrays of different size?
My problem is that I have two arrays, that scaled are pretty similar and I need the error in each point.
The data look like this:-
Yaw data is much bigger than Yaw Ref.
You could take a very naive approach and simply pad each element of the reference array. That is fairly simple to do:
n = length(yaw)/length(yaw_ref);
yaw_ref_pad = zeros(length(yaw), 1);
for j = 1:length(yaw_ref)-1
yaw_ref_pad((n*j):(n*(j+1)) = yaw_ref(j);
end
You could also do something more adaptive, which may or may not be what you want. This approach uses the derivatives to determine where the padded reference should switch. This might be considered a bit circular, since your system looks like an overdamped PID system and this uses the output to seed the input.
yaw_ref_pad = zeros(length(yaw), 1);
[x, peaks] = findpeaks(diff(yaw));
for j = 1:length(peaks)-1
yaw_ref_pad(peaks(j):peaks(j+1)) = yaw_ref(j);
end
Either way, after filling yaw_ref_pad, your result is simply
error = yaw_ref_pad - yaw;
Related
Following to the question published in How expressive can we be with arrays in Z3(Py)? An example, I expressed the following formula in Z3Py:
Exists i::Integer s.t. (0<=i<|arr|) & (avg(arr)+t<arr[i])
This means: whether there is a position i::0<i<|arr| in the array whose value a[i] is greater than the average of the array avg(arr) plus a given threshold t.
The solution in Z3Py:
t = Int('t')
avg_arr = Int('avg_arr')
len_arr = Int('len_arr')
arr = Array('arr', IntSort(), IntSort())
phi_1 = And(0 <= i, i< len_arr)
phi_2 = (t+avg_arr<arr[i])
phi = Exists(i, And(phi_1, phi_2))
s = Solver()
s.add(phi)
print(s.check())
print(s.model())
Note that, (1) the formula is satisfiable and (2) each time I execute it, I get a different model. For instance, I just got: [avg_a = 0, t = 7718, len_arr = 1, arr = K(Int, 7719)].
I have three questions now:
What does arr = K(Int, 7719)] mean? Does this mean the array contains one Int element with value 7719? In that case, what does the K mean?
Of course, this implementation is wrong in the sense that the average and length values are independent from the array itself. How can I implement simple avg and len functions?
Where is the i index in the model given by the solver?
Also, in which sense would this implementation be different using sequences instead of arrays?
(1) arr = K(Int, 7719) means that it's a constant array. That is, at every location it has the value 7719. Note that this is truly "at every location," i.e., at every integer value. There's no "size" of the array in SMTLib parlance. For that, use sequences.
(2) Indeed, your average/length etc are not related at all to the array. There are ways of modeling this using quantifiers, but I'd recommend staying away from that. They are brittle, hard to code and maintain, and furthermore any interesting theorem you want to prove will get an unknown as answer.
(3) The i you declared and the i you used as the existential is completely independent of each other. (Latter is just a trick so z3 can recognize it as a value.) But I guess you removed that now.
The proper way to model such problems is using sequences. (Although, you shouldn't expect much proof performance there either.) Start here: https://microsoft.github.io/z3guide/docs/theories/Sequences/ and see how much you can push it through. Functions like avg will need a recursive definition most likely, for that you can use RecAddDefinition, for an example see: https://stackoverflow.com/a/68457868/936310
Stack-overflow works the best when you try to code these yourself and ask very specific questions about how to proceed, as opposed to overarching questions. (But you already knew that!) Best of luck..
Working in MATLAB R2017a. I'm trying to optimise a piece of code I'm working on. It uses arrays to store field values on a grid.
In order to create a specific function in a field array I originally used the straight forward method of two for loops iterating over all the array elements. But i know for loops are slow so since then I came back and tried my best to remove them. However I could only manage to remove one of the loops; leaving me with this:
for n = 1:1:K
%%% define initial pertubation
t=n*dt;
% create array for source Ez field.
xtemps = (1:Ng)*dX;
for k = 1:Ng
ztemp = k*dX;
Ez0(k,:) = THzamp * (1/(1+exp(-(t-stepuppos)))) * exp(-((xtemps-...
THzstartx).^2)./(bx^2)) .* (t-((ztemp-THzstartz)/vg))*exp(-((t-((ztemp-...
THzstartz)/vg))^2)/(bt^2));
end
The important bit here is the last 5 lines, but I figured the stuff before might be important for context. I've removed the for loop looping over the x coordinates. I want to vectorize the z/k for loop but I can't figure out how to distinguish between the dimensions with the array oporators.
Edit: THzamp, stepuppos, bx, bt, THzstartz, THzstartx are all just scalars, they control the function (Ez0) I'm trying to create. dX and t are also just scalars. Ez0 is a square array of size Ng.
What I want to achieve is to remove the for loop that loops over k, so that that the values of ztemp are defined in a vector (like xtemps already is), rather than individually in the loop. However, I don't know how I'd write the definition of Ez0 in that case.
First time posting here, if I'm doing it wrong let me know. If you need more info just ask.
It isn't clear if n is used in the other headers and as stated in the comments your sizes aren't properly defined so you'll have to ensure the sizes are correct.
However, you can give this vectorize code a try.
n = 1:K
%%% define initial pertubation
t=n*dt;
% create array for source Ez field.
xtemps = (1:Ng)*dX;
for k = 1:Ng
ztemp = k*dX;
Ez0(k,:) = THzamp .* (1./(1+exp(-(t-stepuppos)))) .* exp(-((xtemps-...
THzstartx).^2)./(bx^2)) .* (t-((ztemp-THzstartz)/vg)).*exp(-((t-((ztemp-...
THzstartz)/vg)).^2)/(bt.^2));
end
So now t has the size K you'll need to ensure stepupposand (ztemp-THzstartz)/vg) have the same size K. Also you can take a look at vectors vs array operators here.
I have a time series data or considering a real valued data of length N. I want to create sub-blocks of length k, which is the window length. The value of k can be arbitrarily chosen. This creates problem since the window size is the same across the data. I want to store each subblock in an array. But I am stuck in creating sub-blocks of the data and to include a check so that the (mod(N, nseg)) nseg must be divisible by the data length.
N = 512; %length of the time series
data = rand(N,1);
window_length = 30; %k
Nseg = floor(N/window_length) %Number of segments or blocks
Modified_Data = [mean(reshape(data,window_length,Nseg))]; %Throws error
If you have the Image Processing toolbox you could use im2col to slide a specific block size over the entire time series. Each column of the output represents the data from one of those blocks.
values = im2col(data, [window_length 1], 'distinct');
Since it looks like you just want the mean over each block, you could also use blockproc to do this.
means = blockproc(data, [window_length, 1], #(x)mean(x.data));
If you do not have the Image Processing Toolbox, you can instead use accumarray to perform this task.
means = accumarray(floor((0:(N-1)).'/window_length) + 1, data, [], #mean);
If you want to discard any data that extends beyond a number which is divisible by window_length, you can do this with something like the following:
data = data(1:(numel(data) - mod(numel(data), window_length)));
If you want overlapping data, you'll either want to use straight-up convolution (the preferred method)
means = conv(data(:), ones(5, 1)/5, 'same');
Or you can create overlapping blocks with im2col by omitting the last input.
values = im2col(data, [window_length 1]);
means = mean(values 1);
If you have R2016a+, consider using the built-in movmean function:
N = 512; %length of the time series
data = rand(N,1);
window_length = 30; %k
Modified_Data = movmean(data, window_length);
See the documentation for further details and other options.
If I understand your question correctly, it's pretty straightforward:
filter(ones(N,1)/N,1,signal)
If you think about it filtering with [1/N 1/N 1/N...1/N] is exactly calculating the localized mean...
I have an ODE system of two equations, but want to minimize it with using just one equation with the result of the other.
1)
t=linspace(0,2,3);
syms x(t) y(t);
inits='x(0)=2,y(0)=0';
[x,y]=dsolve('Dx=y','Dy=(y*2)-x', inits)
x = 2*exp(t) - 2*t*exp(t);
y = -2*t*exp(t)
xx=eval(vectorize(x));
xx = 2.0000; 0; -14.7781
yy=eval(vectorize(y));
yy = 0; -5.4366; -29.5562
After I had got the results, I tried to solve it just with one equation and use xx array in Dy equation.
2)
inits='y(0)=0';
[y]=dsolve('Dy=(y*2)-xx', inits);
y = xx/2 - (xx*exp(2*t))/2
yy=eval(vectorize(y));
yy = 0; 0; 396.0397
The values are not the same as it was in the first example. How to get the same result using array?
One problem seems to be that variable xx is not symbolic, so the symbolic solver appears to be considering it as a constant.
A bigger problem is that you really haven't identified how exactly you want matlab to treat the xx values as a continuous function, when it's merely a vector of three points! The fact that you are even expecting the output to be the same for the second case indicates some kind of misunderstanding to me.
But to make this definite, lets assume that you want it to treat xx as a ZOH (zero order held) continuous signal. To handle this symbolically I believe you would need to construct the ZOH signal explicitly using Heaviside functions.
Alternative you could solve it numerically using ode45 for example
t = [0,1,2];
xx = [2, 0, -14.7781];
dydy = #(t,y) 2*y - xx(1+trunc(t));
y = ode45(dydt, [0,2], 0);
This will return yy values of [0, -6.39, -47.21] at the t values of [0, 1, 2] respectively.
This corresponds well with the theoretical values (calculated by hand) of [0, 1-e^2, e^2-e^4] for the ZOH system.
As you can see the above answer is much more in line with your original solution of yy = [0, -5.4366, -29.5562]. Naturally however the two systems differ, as the first one was fed with a continuous time exponential signal whereas the second system was fed with a very coarsely sampled approximation!
You make the two more similar by sampling at a faster rate (finer time interval), and also by interpolating the intersample points with something better than a ZOH.
Update:
Thank you. Maybe can you help me with creating ZOH continuous signal? How to do that?
In the above example I created a ZOH in my derivative function (dydx) by using the three given points in the xx vector and accessing these using "xx(1+trunc(t))". This uses trunc (truncate) to explicitly hold the input constant during the inter-sample (non integer) times.
Seeing as your ODE is linear, you could also use the matlab function "lsim()" which allows you to directly specify the time vector and input vector, and also to directly specify the type of input interpolation (including ZOH, which is actually the default).
For example:
t=[0,1,2]
x=[2,0,-2*e^2]
num=-1
den=[1,-2]
mytf = tf(num,den)
y = lsim(mytf,x,t,0,'zoh');
As with my previous ode45 numerical solution, this gives the identical solution of,
y = [0.00000; -6.38906; -47.20909]
Update (again)
Thank you. Maybe can you help me with creating ZOH continuous signal? How to do that?
Re the symbolic solver. I don't have access to the Matlab symbolic library, but if you really want to use the symbolic solver, then as I explained previous, you can construct a continuous time ZOH signal using the heaviside (unit step) function. Something like the following should do it:
syms xzoh(t)
xzoh = xx(1)*heaviside(t) + (xx(2)-xx(1))*heaviside(t-1) + (xx(3)-xx(2))*heaviside(t-2)
Here's the code I am using now, where decimal1 is an array of decimal values, and B is the number of bits in binary for each value:
for (i = 0:1:length(decimal1)-1)
out = dec2binvec(decimal1(i+1),B);
for (j = 0:B-1)
bit_stream(B*i+j+1) = out(B-j);
end
end
The code works, but it takes a long time if the length of the decimal array is large. Is there a more efficient way to do this?
bitstream = zeros(nelem * B,1);
for i = 1:nelem
bitstream((i-1)*B+1:i*B) = fliplr(dec2binvec(decimal1(i),B));
end
I think that should be correct and a lot faster (hope so :) ).
edit:
I think your main problem is that you probably don't preallocate the bit_stream matrix.
I tested both codes for speed and I see that yours is faster than mine (not very much tho), if we both preallocate bitstream, even though I (kinda) vectorized my code.
If we DONT preallocate the bitstream my code is A LOT faster. That happens because your code reallocates the matrix more often than mine.
So, if you know the B upfront, use your code, else use mine (of course both have to be modified a little bit to determine the length at runtime, which is no problem since dec2binvec can be called without the B parameter).
The function DEC2BINVEC from the Data Acquisition Toolbox is very similar to the built-in function DEC2BIN, so some of the alternatives discussed in this question may be of use to you. Here's one option to try, using the function BITGET:
decimal1 = ...; %# Your array of decimal values
B = ...; %# The number of bits to get for each value
nValues = numel(decimal1); %# Number of values in decimal1
bit_stream = zeros(1,nValues*B); %# Initialize bit stream
for iBit = 1:B %# Loop over the bits
bit_stream(iBit:B:end) = bitget(decimal1,B-iBit+1); %# Get the bit values
end
This should give the same results as your sample code, but should be significantly faster.