my script creates a matrix and 2 vectors using several 'for' loops and as an example they are returned as follows:
K =
1.0e+006 *
1.2409 0.6250 0.8153 0.1250
0.6250 3.6591 -0.1250 3.5375
0.8153 -0.1250 1.2409 -0.6250
0.1250 3.5375 -0.6250 3.6591
F =
1.0e+006 *
0.1733
1.3533
-0.1066
1.3371
U =
u3
v3
u4
v4
As can be seen, the 'U' vector is a set of variables and I need to solve 'K*U=F' for variables contained in 'U'.
When I try to do that using linsolve or solve I get unexpected results and a message that the inverse of my matrix is close to singular.
HOWEVER, when I make another script and put in the SAME matrix and vectors BY HANDS it all works fine and I can't figure out what's wrong.
Is that somehow related to the way MATLAB stores matrices created by loop functions and I need to change the state of the matrix to something after the loop?
Also, when I put the matrix by hands it displays it without the 1.0e+006 multiplier in front of it:
K11 =
1240900 625000 815300 125000
625000 3659100 -125000 3537500
815300 -125000 1240900 -625000
125000 3537500 -625000 3659100
can that be related??
Thanks in advance.
Try the backslash operator:
U = K\F
See this reference.
From the previous discussion it's clear that your matrix is singular. This means that your equations are not linearly independent. When this happens there are two possibilities. Your system may be inconsistent (over-constrained), in which case no solutions exist. Or alternatively, it can also mean that your equations are under-constrained, in which case there is an infinite set of solutions.
To determine which case it is you can use rref to get the "row reduce echelon form" of the matrix. Do this as follows:
KF = [K,F]
rref(KF)
If the last row goes entirely to zeros then you're under-constrained and can extract a solution set (but not a unique solution) from your reduced matrix.
In this case however I get a row of [0 0 0 0 1], which makes the system over-constrained and hence without any solution.
Related
I have this program which calculates the realized covariance for each day in my sample but I have some troubles with storing the output in a matrix.
the program is as follows:
for i=1:66:(2071*66)
vec = realized_covariance(datapa(i:i+65),data(i:i+65),datapo(i:i+65),data(i:i+65),'wall','Fixed',fixedInterval,5)
mat(2,4142) = vec
end
Output:
vec =
1.0e-03 *
0.1353 -0.0283
-0.0283 0.0185
Subscripted assignment dimension mismatch.
I have tried various way to store the output in a matrix like defining a matrix on zeroes to store the output in or let the row dimension of the storing matrix be undefined, but nothing seems to do the job.
I would really appreciate an advice on how to tackle this challenge.
I have used a solution which does the job.
I defined a matrix and then filled in all my output one at the time using the following:
A = zeros(0,0) %before loop, only serve to define the storing matrix
A = [A; vec]%after the calculating function, inside the loop.
Actually mat(2,4142) is a single location in a matrix, you can't assign there four values.
You need to define the exact location inside mat every time you want to assign values into it. Try doing it like that:
mat=zeros(2,2142);
for k=1:66:(2071*66)
vec=realized_covariance(datapa(i:i+65),data(i:i+65),datapo(i:i+65),data(i:i+65),'wall','Fixed',fixedInterval,5)
mat(:,[(((k-1)/66)*2)+1 (((k-1)/66)*2)+2])=vec;
end
You're trying to store a 2 x 2 matrix into a single element. I.e. 4 elements on the right hand side, one on the left. That won't fit. See it like this: you have a garage besides your house where 1 car fits. You've got three friends coming over and they also want to park their car inside. That's a problem though, as you've got only space for one. So you have to buy a bigger garage: assign 4 elements on the left (e.g. mat(ii:ii+1,jj:jj+1) = [1 2;3 4]), or use a cell/structure array.
As Steve suggests in a comment below, you can use a 3D matrix quite easily:
counters = 1:66:(2071*66);
mat = zeros(2,2,numel(counters)); %// initialise output matrix
for ii=1:numel(counters)
vec = realized_covariance(datapa(counters(ii):counters(ii+65)),...
data(counters(ii):counters(ii+65)),datapo(counters(ii):counters(ii+65)),...
data(counters(ii):counters(ii+65)),'wall','Fixed',fixedInterval,5)
mat(:,:,ii) = vec; %// store in a 3D matrix
end
Now mat is 3D, with the first two coordinates being your regular output, i.e.e vec, and the last index is the iteration number. So to access the output of iteration 1032 you'd do mat(:,:,1032), possibly with a squeeze around that to make it 2D instead of 3D.
I'm attempting to run this simple diffusion case (I understand that it isn't ideal generally), and I'm doing fine with getting the inside of the solid, but need some help with the outer edges.
global M
size=100
M=zeros(size,size);
M(25,25)=50;
for diffusive_steps=1:500
oldM=M;
newM=zeros(size,size);
for i=2:size-1;
for j=2:size-1;
%we're considering the ij-th pixel
pixel_conc=oldM(i,j);
newM(i,j+1)=newM(i,j+1)+pixel_conc/4;
newM(i,j-1)=newM(i,j-1)+pixel_conc/4;
newM(i+1,j)=newM(i+1,j)+pixel_conc/4;
newM(i-1,j)=newM(i-1,j)+pixel_conc/4;
end
end
M=newM;
end
It's a pretty simple piece of code, and I know that. I'm not very good at using Octave yet (chemist by trade), so I'd appreciate any help!
If you have concerns about the border of your simulation you could pad your matrix with NaN values, and then remove the border after the simulation has completed. NaN stands for not a number and is often used to denote blank data. There are many MATLAB functions work in a useful way with these values.
e.g. finding the mean of an array which has blanks:
nanmean([0 nan 5 nan 10])
ans =
5
In your case, I would start by adding a border of NaNs to your M matrix. I'm using 'n' instead of 'size', since size is an important function in MATLAB, and using it as a variable can lead to confusing errors.
n=100;
blankM=zeros(n+2,n+2);
blankM([1,end],:) = nan;
blankM(:, [1,end]) = nan;
Now we can define 'M'. N.B that the first column and row will be NaNs so we need to add an offset (25+1):
M = blankM;
M(26,26)=50;
Run the simulation through,
m = size(blankM, 1);
n = size(blankM, 2);
for diffusive_steps=1:500
oldM = M;
newM = blankM;
for i=2:m-1;
for j=2:n-1;
pixel_conc=oldM(i,j);
newM(i,j+1)=newM(i,j+1)+pixel_conc/4;
newM(i,j-1)=newM(i,j-1)+pixel_conc/4;
newM(i+1,j)=newM(i+1,j)+pixel_conc/4;
newM(i-1,j)=newM(i-1,j)+pixel_conc/4;
end
end
M=newM;
end
and then extract the area of interest
finalResult = M(2:end-1, 2:end-1);
One simple change you might make is to add a boundary of ghost cells, or halo, around the domain of interest. Rather than mis-use the name size I've used a variable called sz. Replace:
M=zeros(sz,sz)
with
M=zeros(sz+2,sz+2)
and then compute your diffusion over the interior of this augmented matrix, ie over cells (2:sz+1,2:sz+1). When it comes to considering the results, discard or just ignore the halo.
Even simpler would be to simply take what you already have and ignore the cells in your existing matrix which are on the N,S,E,W edges.
This technique is widely used in problems such as, and similar to, yours and avoids the need to write code which deals with the computations on cells which don't have a full complement of neighbours. Setting the appropriate value for the contents of the halo cells is a problem-dependent matter, 0 isn't always the right value.
I have a matrix of numbers for one of the variables in an fsolve equation so when I run matlab I am hoping to get back a matrix but instead get a scalar. I even tried a for loop but this gave me an error about size so that is not the solution. I am including the code to get some feedback as to what i am doing wrong.
z=0.1;
bubba =[1 1.5 2];
bubba = bubba';
joe = 0:0.1:1.5;
joe = repmat(joe,3,1);
bubba = repmat(bubba,1,length(joe));
for x=1:1:16
eqn0 = #(psi0) (joe.-bubba.*(sqrt((psi0+z))));
result0(x) = fsolve(eqn0,0.1,options);
end
note I need the joe variable later for plotting so I clipped that part of the code.
Based on your earlier comments, let me take a shot at a solution... still not sure this is what you want:
bubba =[1 1.5 2];
joe = 0:0.1:1.5;
for xi = 1:numel(joe)
for xj = 1:numel(bubba)
eqn0 = #(psi0) (joe(xi).-bubba(xj).*(sqrt((psi0+z))));
result(xi,xj) = fsolve(eqn0,0.1,options);
end
end
It is pedestrian; but is it what you want? I can't access matlab right now, otherwise I might come up with something more efficient.
To elaborate on my comment:
psi0 is the independent variable in your solver. You set the dimension of it to [1 1] when you use a scalar as the second argument of fsolve(eqn0, 0.1, options); - this tells Matlab to optimize the scalar psi0, starting at a value of 0.1. The result will be a scalar - the value that minimizes the function
0.1 * sqrt(psi0 + 0.1)
since you had set z=0.1
You should get a value of -0.1 returned for every iteration of your loop, since you never changed anything. There is not enough information right now to figure out which factor you would like to be a matrix - especially since your expression for eqn0 involves a matrix multiplication, it's hard to know what you expect the dimensionality of the result to be.
I hope that you will use this initial answer as a springboard to modify your question so it can be answered properly!?
I am loading a sparse matrix in MATLAB using the command:
A = spconvert(load('mymatrix.txt'));
I know that the dimension of my matrix is 1222 x 1222, but the matrix is loaded as 1220 x 1221. I know that it is impossible for MATLAB to infer the real size of my matrix, when it is saved sparse.
A possible solution for making A the right size, is to include a line in mymatrix.txt with the contents "1222 1222 0". But I have hundreds of matrices, and I do not want to do this in all of them.
How can I make MATLAB change the size of the matrix to a 1222 x 1222?
I found the following solution to the problem, which is simple and short, but not as elegant as I hoped:
A = spconvert(load('mymatrix.txt'));
if size(A,1) ~= pSize || size(A,2) ~= pSize
A(pSize,pSize) = 0;
end
where pSize is the preferred size of the matrix. So I load the matrix, and if the dimensions are not as I wanted, I insert a 0-element in the lower right corner.
Sorry, this post is more a pair of clarifying questions than it is an answer.
First, is the issue with the 'load' command or with 'spconvert'? As in, if you do
B = load('mymatrix.txt')
is B the size you expect? If not, then you can use 'textread' or 'fread' to write a function that creates the matrix of the right size before inputting into 'spconvert'.
Second, you say that you are loading several matrices. Is the issue consistent among all the matrices you are loading. As in, does the matrix always end up being two rows less and one column less than you expect?
I had the same problem, and this is the solution I came across:
nRows = 1222;
nCols = 1222;
A = spconvert(load('mymatrix.txt'));
[i,j,s] = find(A);
A = sparse(i,j,s,nRows,nCols);
It's an adaptation of one of the examples here.
I have cell array each containing a sequence of values as a row vector. The sequences contain some missing values represented by NaN.
I would like to replace all NaNs using some sort of interpolation method, how can I can do this in MATLAB? I am also open to other suggestions on how to deal with these missing values.
Consider this sample data to illustrate the problem:
seq = {randn(1,10); randn(1,7); randn(1,8)};
for i=1:numel(seq)
%# simulate some missing values
ind = rand( size(seq{i}) ) < 0.2;
seq{i}(ind) = nan;
end
The resulting sequences:
seq{1}
ans =
-0.50782 -0.32058 NaN -3.0292 -0.45701 1.2424 NaN 0.93373 NaN -0.029006
seq{2}
ans =
0.18245 -1.5651 -0.084539 1.6039 0.098348 0.041374 -0.73417
seq{3}
ans =
NaN NaN 0.42639 -0.37281 -0.23645 2.0237 -2.2584 2.2294
Edit:
Based on the responses, I think there's been a confusion: obviously I'm not working with random data, the code shown above is simply an example of how the data is structured.
The actual data is some form of processed signals. The problem is that during the analysis, my solution would fail if the sequences contain missing values, hence the need for filtering/interpolation (I already considered using the mean of each sequence to fill the blanks, but I am hoping for something more powerful)
Well, if you're working with time-series data then you can use Matlab's built in interpolation function.
Something like this should work for your situation, but you'll need to tailor it a little ... ie. if you don't have equal spaced sampling you'll need to modify the times line.
nseq = cell(size(seq))
for i = 1:numel(seq)
times = 1:length(seq{i});
mask = ~isnan(seq{i});
nseq{i} = seq{i};
nseq{i}(~mask) = interp1(times(mask), seq{i}(mask), times(~mask));
end
You'll need to play around with the options of interp1 to figure out which ones work best for your situation.
I would use inpaint_nans, a tool designed to replace nan elements in 1-d or 2-d matrices by interpolation.
seq{1} = [-0.50782 -0.32058 NaN -3.0292 -0.45701 1.2424 NaN 0.93373 NaN -0.029006];
seq{2} = [0.18245 -1.5651 -0.084539 1.6039 0.098348 0.041374 -0.73417];
seq{3} = [NaN NaN 0.42639 -0.37281 -0.23645 2.0237];
for i = 1:3
seq{i} = inpaint_nans(seq{i});
end
seq{:}
ans =
-0.50782 -0.32058 -2.0724 -3.0292 -0.45701 1.2424 1.4528 0.93373 0.44482 -0.029006
ans =
0.18245 -1.5651 -0.084539 1.6039 0.098348 0.041374 -0.73417
ans =
2.0248 1.2256 0.42639 -0.37281 -0.23645 2.0237
If you have access to the System Identification Toolbox, you can use the MISDATA function to estimate missing values. According to the documentation:
This command linearly interpolates
missing values to estimate the first
model. Then, it uses this model to
estimate the missing data as
parameters by minimizing the output
prediction errors obtained from the
reconstructed data.
Basically the algorithm alternates between estimating missing data and estimating models, in a way similar to the Expectation Maximization (EM) algorithm.
The model estimated can be any of the linear models idmodel (AR/ARX/..), or if non given, uses a default-order state-space model.
Here's how to apply it to your data:
for i=1:numel(seq)
dat = misdata( iddata(seq{i}(:)) );
seq{i} = dat.OutputData;
end
Use griddedInterpolant
There also some other functions like interp1. For curved plots spline is the the best method to find missing data.
As JudoWill says, you need to assume some sort of relationship between your data.
One trivial option would be to compute the mean of your total series, and use those for missing data. Another trivial option would be to take the mean of the n previous and n next values.
But be very careful with this: if you're missing data, you're generally better to deal with those missing data, than to make up some fake data that could screw up your analysis.
Consider the following example
X=some Nx1 array
Y=F(X) with some NaNs in it
then use
X1=X(find(~isnan(Y)));
Y1=Y(find(~isnan(Y)));
Now interpolate over X1 and Y1 to compute all values at all X.