I'm pretty new to Simulink and I was wondering if the following thing was possible somehow:
I have a signal of let's say 10000 data points.
On this signal I want to run a certain algorithm, however said algorithm needs exactly 1000 samples to work properly.
Now with normal matlab functions this is no problem. You cut the signal in 10 pieces, perform the algorithm for each one, stitch the processed parts back together and you get your result.
In Simulink however this creates sort of a problem, since (to my understanding right now) Simulinks blocks work sample per sample (one sample in, one sample out). So I don't have the necessary data to perform the algortihm within a block.
Is there any way to increase the number of processed samples per block?
Reshape 10,000 data points with 1000 samples and create Column wise data
lets say
data = [1 2 3 4 5 6], convert
data = [1 4
2 5
3 6]
Now, define sample time for lets say 1 sec, use fromworkspace blocks. In this example 2 (No of column), with each workspace block format a array as [t data(:,1)],[t data(:,2)] where t = [1 2 3], multiples of sampletime.
On Simulink Model, set running time equal to 3 sec as there are 3 data points and store output via to workspace block
Related
I am wondering if anyone can explain the interpretation of the size (number of feature) in a time series? For example consider a simple script in Matlab
X= randn(2,5,2)
X(:,:,1) =
-0.5530 0.4291 0.3937 -1.2534 0.2811
-1.4926 -0.7019 -0.8305 -1.4034 1.9545
X(:,:,2) =
0.2004 0.1438 2.3655 -0.1589 0.7140
0.4905 0.2301 -0.7813 -0.6737 0.2552
Assume X is a time series with the following output
This generates 2 vectors of length 5 each has 2 rows. Can anyone tell me what is exactly the meaning of first 2 and 5?
In some websites it says a creating 5 vectors of length 5 and size 2. What does size mean here?
Is 2 like number of features and 5 is like number of time series. The reason for this confusion is because I do not understand how to interpret following sentence:
"Generate 2 vector-valued sequences of length 5; each vector has size
2."
What do size 2 and length 5 mean here?
This entirely depends on your data, and how you want to store this. If you have some 2D data over time, I find it convenient to have a data matrix with in the 1st and 2nd dimension the 2D data per time step, and in the 3rd dimension time.
Say I have a movie of 1920 by 1080 pixels with 100 frames, I'd store this as mov = rand(1080,1920,100) (1080 and 1920 swapped because of row, col order of indexing). So now mov(:,:,1) would give me the first frame etc.
BTW, your X is a normal array, not to be confused with the timeseries object.
I have two for loops in MATLAB.
One of the for loops leads to different variables being inserted into the model, which are 43 and then I have 5 horizons.
So I estimate the model 215 times.
My problem is I want to store this in 215x5 matrix, the reason I have x5 is that I am estimating 5 variables, 4 are fixed and the other one comes from the for loop.
I have tried to do this in two ways,
Firstly, I create a variable called out,
out=zeros(215,5);
The first for loop is,
for i=[1,2,3,4,5];
The second for loop is,
for ii=18:60;
The 18:60 is how I define my variables using XLS read, e.g. they are inserted into the model as (data:,ii).
I have tried to store the data in two ways, I want to store OLS which contains the five estimates
First,
out(i,:)=OLS;
This method creates a 5 x 5 matrix, with the estimates for one of the (18:60), at each of the horizons.
Second,
out(ii,:)=OLS;
This stores the variables for each of the variables (18:60), at just one horizon.
I want to have a matrix which stores all of the estimates OLS, at each of the horizons, for each of my (18:60).
Minimal example
clear;
for i=[1,2,3,4,5];
K=i;
for ii=18:60
x=[1,2,3,i,ii];
out(i,:)=x;
end
end
So the variable out will store 1 2 3 5 60
I want the variable out to store all of the combinations
i.e.
1 2 3 1 1
1 2 3 1 2
...
1 2 3 5 60
Thanks
The simplest solution is to use a 3D matrix:
for jj=[1,2,3,4,5];
K=jj;
for ii=18:60
x=[1,2,3,jj,ii];
out(ii-17,jj,:)=x;
end
end
If you now reshape the out matrix you get the same result as the first block in etmuse's answer:
out = reshape(out,[],size(out,3));
(Note I replaced i by jj. i and ii are too similar to use both, it leads to confusion. It is better to use different letters for loop indices. Also, i is OK to use, but it also is the built-in imaginary number sqrt(-1). So I prefer to use ii over i.)
As you've discovered, using just one of your loop variables to index the output results in most of the results being overwritten, leaving only the results from the final iteration of the relevant loop.
There are 2 ways to create an indexing variable.
1- You can use an independent variable, initialised before the loops and incremented at the end of the internal loop.
kk=1;
for i=1:5
for ii=18:60
%calculate OLC
out(kk,:)=OLC;
kk = kk+1;
end
end
2- Use a calculation of i and ii
kk = i + 5*(ii-18)
(Use in loop as before, without increment)
I'm using a kd-tree to perform quick nearest neighbor search queries. I'm using the following piece of code to generate the kd-tree and perform queries on it:
% 3 dimensional vertex data
x = [1 2 2 1 2 5 6 3 4;
3 2 3 2 2 7 6 5 2;
1 2 9 9 7 5 8 9 3]';
% create the kd-tree
kdtree = createns(x, 'NSMethod', 'kdtree');
% perform a nearest neighbor search
nearestNeighborIndex = knnsearch(kdtree, [1 1 1]);
This works well enough for when the data is static. However, every once in a while, I need to translate every vertex on the kd-tree. I know that changing the whole data means I need to re-generate the whole tree to perform a nearest neighbor search again. Having a couple of thousand vertices for each kd-tree, re-generating the whole tree from scratch seems to me like an overkill as it takes a significant amount of time. Is there a way to translate the kd-tree without re-generating it from scratch? I tried accessing and changing the X property (which holds the actual vertex data) of the kd-tree but it seems to be read-only, and it probably wouldn't have worked even if it wasn't since there is a lot more going on behind the curtains.
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 a newbie to Matlab and just stumped how to do a simple task that can be easily performed in excel. I'm simply trying to get the percent change between cells in a matrix. I would like to create a for loop for this task. The data is setup in the following format:
DAY1 DAY2 DAY3...DAY 100
SUBJECT RESULTS
I could only perform getting the percent change between two data points. How would I conduct it if across multiple days and multiple subjects? And please provide explanation
Thanks a bunch
FOR EXAMPLE, FOR DAY 1 SUBJECT1(RESULT=1), SUBJECT2(RESULT=4), SUBJECT3(RESULT=5), DAY 2 SUBJECT1(RESULT=2), SUBJECT2(RESULT=8), SUBJECT3(RESULT=10), DAY 3 SUBJECT1(RESULT=1), SUBJECT2(RESULT=4), SUBJECT3(RESULT=5).
I WANT THE PERCENT CHANGE SO OUTPUT WILL BE DAY 2 SUBJECT1(RESULT=100%), SUBJECT2(RESULT=100%), SUBJECT3(RESULT=100%). DAY3 SUBJECT1(RESULT=50%), SUBJECT2(RESULT=50%), SUBJECT3(RESULT=50%)
updated:
Hi thanks for responding guys. sorry for the confusion. zebediah49 is pretty close to what I'm looking for. My data is for example a 10 x 10 double. I merely wanted to get the percentage change from column to column. For example, if I want the percentage change from rows 1 through 10 on all columns (from columns 2:10). I would like the code to function for any matrix dimension (e.g., 1000 x 1000 double) zebediah49 could you explain the code you posted? thanks
updated2:
zebediah49,
(data(1:end,100)- data(1:end,99))./data(1:end,99)
output=[data(:,2:end)-data(:,1:end-1)]./data(:,1:end-1)*100;
Observing the code above, How would I go about modifying it so that column 100 is used as the index against all of the other columns(1-99)? If I change the code to the following:
(data(1:end,100)- data(1:end,:))./data(1:end,:)
matlab is unable because of exceeding matrix dimensions. How would I go about implementing that?
UPDATE 3
zebediah49,
Worked perfectly!!! Originally I created a new variable for the index and repmat the index to match the matrices which was not a good idea. It took forever to replicate when dealing with large numbers.
Thanks for you contribution once again.
Thanks Chris for your contribution too!!! I was looking more on how to address and manipulate arrays within a matrix.
It's matlab; you don't actually want a loop.
output=input(2:end,:)./input(1:end-1,:)*100;
will probably do roughly what you want. Since you didn't give anything about your matlab structure, you may have to change index order, etc. in order to make it work.
If it's not obvious, that line defines output as a matrix consisting of the input matrix, divided by the input matrix shifted right by one element. The ./ operator is important, because it means that you will divide each element by its corresponding one, as opposed to doing matrix division.
EDIT: further explanation was requested:
I assumed you wanted % change of the form 1->1->2->3->1 to be 100%, 200%, 150%, 33%.
The other form can be obtained by subtracting 100%.
input(2:end,:) will grab a sub-matrix, where the first row is cut off. (I put the time along the first dimension... if you want it the other way it would be input(:,2:end).
Matlab is 1-indexed, and lets you use the special value end to refer to the las element.
Thus, end-1 is the second-last.
The point here is that element (i) of this matrix is element (i+1) of the original.
input(1:end-1,:), like the above, will also grab a sub-matrix, except that that it's missing the last column.
I then divide element (i) by element (i+1). Because of how I picked out the sub-matrices, they now line up.
As a semi-graphical demonstration, using my above numbers:
input: [1 1 2 3 1]
input(2,end): [1 2 3 1]
input(1,end-1): [1 1 2 3]
When I do the division, it's first/first, second/second, etc.
input(2:end,:)./input(1:end-1,:):
[1 2 3 1 ]
./ [1 1 2 3 ]
---------------------
== [1.0 2.0 1.5 0.3]
The extra index set to (:) means that it will do that procedure across all of the other dimension.
EDIT2: Revised question: How do I exclude a row, and keep it as an index.
You say you tried something to the effect of (data(1:end,100)- data(1:end,:))./data(1:end,:). Matlab will not like this, because the element-by-element operators need them to be the same size. If you wanted it to only work on the 100th column, setting the second index to be 100 instead of : would do that.
I would, instead, suggest setting the first to be the index, and the rest to be data.
Thus, the data is processed by cutting off the first:
output=[data(2:end,2:end)-data(2:end,1:end-1)]./data(2:end,1:end-1)*100;
OR, (if you neglect the start, matlab assumes 1; neglect the end and it assumes end, making (:) shorthand for (1:end).
output=[data(2:,2:end)-data(2:,1:end-1)]./data(2:,1:end-1)*100;
However, you will probably still want the indices back, in which case you will need to append that subarray back:
output=[data(1,1:end-1) data(2:,2:end)-data(2:,1:end-1)]./data(2:,1:end-1)*100];
This is probably not how you should be doing it though-- keep data in one matrix, and time or whatever else in a separate array. That makes it much easier to do stuff like this to data, without having to worry about excluding time. It's especially nice when graphing.
Oh, and one more thing:
(data(:,2:end)-data(:,1:end-1))./data(:,1:end-1)*100;
is identically equivalent to
data(:,2:end)./data(:,1:end-1)*100-100;
Assuming zebediah49 guessed right in the comment above and you want
1 4 5
2 8 10
1 4 5
to turn into
1 1 1
-.5 -.5 -.5
then try this:
data = [1,4,5; 2,8,10; 1,4,5];
changes_absolute = diff(data);
changes_absolute./data(1:end-1,:)
ans =
1.0000 1.0000 1.0000
-0.5000 -0.5000 -0.5000
You don't need the intermediate variable, you can directly write diff(data)./data(1:end,:). I just thought the above might be easier to read. Getting from that result to percentage numbers is left as an exercise to the reader. :-)
Oh, and if you really want 50%, not -50%, just use abs around the final line.