The below code gives me a error:
Subscripted assignment dimension mismatch.
Error in ==> lookmcvmt at 18
M(:,:,j,i) = mcmvOUT2((k+1):(k+Nz), i:Nt:Nr);
Please help to solve.
load MCMVout1xzy
mcmvOUT2 = MCMVout1xzy;
whos
[Nr2 Nr] = size(mcmvOUT2);
Ny = 51;
Nx = 51;
Nz = 41;
Nt = 10;
M = zeros(Nz,Nx,Ny,Nt);
for j=1:Ny
for i=1:Nt
k = Nz*(j-1);
M(:,:,j,i) = mcmvOUT2((k+1):(k+Nz), i:Nt:Nr);
end
end
The error 'subscripted assignment dimension mismatch' means you are trying to assign a block of values into a space that is the wrong size.
This entity
mcmvOUT2((k+1):(k+Nz), i:Nt:Nr);
represents a matrix of values in 2 dimensions. Its size is defined by the two ranges specified by (k+1):(k+Nz) and i:Nt:Nr - you can check its size by typing
size(mcmvOUT2((k+1):(k+Nz), i:Nt:Nr))
The space you are trying to fit it into has to be exactly the same dimensions. The size of the range specified by
M(:,:,j,i)
is defined by the Nz and Nx arguments to the zeros call with which you preallocated the array.
We can't test this because the MCMVout1xzy file containing your data is not given, but you will be able to solve this yourself by using the size command and making sure all your dimensions match.
Because matlab uses column-wise indexing, and a lot of us are used to the row-wise paradigm of cartesian coordinate systems, getting the order of your indexes right can be confusing - this is the root of a lot of these kinds of error (for me anyway).
Things to check: your dimensions Nz etc. are correct and the order of your Nz etc. variables in the zeros call is correct.
Related
I am using a function in Matlab based on lp_solve. In my case, lp_solve is structured as follows:
A = rand (13336,3); %A is made of real numbers between 0 and 1. For this mwe, I thought 'rand' was fine
W = [0; 0; 1];
C = A(:,3);
B = 1E+09;
e = -1;
m= 13336;
xint = linspace(1,13336,13336);
xint = xint';
obj = lp_solve(A*W,C,B,e,zeros(m,1),ones(m,1),xint)
But when I run it, I get this error:
Error using mxlpsolve
invalid vector.
Error in lp_solve (line 46)
mxlpsolve('set_rh_vec', lp, b);
Error in mylpsolvefunction (line 32) %This is my function that uses lp_solve
obj = lp_solve(A*W,C,B,e,zeros(m,1),ones(m,1),xint);
I looked in the documentation, and it say, under the chapter "Matrices" that:
[...] if a dense matrix is provided, the dimension must exactly match the dimension that is expected by mxlpsolve. Matrices with too few or too much elements gives an 'invalid vector.' error. Sparse matrices can off course provide less elements (the non provided elements are seen as zero). However if too many elements are provided or an element with a too large index, again an 'invalid vector.' error is raised.
I did not understand what they mean when they say that the dimension "must exactly match the dimensions that is expected by mxlpsolve". Anyway, since they say that the error my also occur "if too many elements are provided", I tried to "cut" my inputs from 13336 elements to 50 (I am sure it works with 58 and I am quite sure it does also with 2000), but also this way I receive the same error. What may the problem be?
X= [P(1,:,:);
P(2,:,:);
P(3,:,:)];
y= P(4:end,:);
indTrain = randperm(4798);
indTrain = indTrain(1:3838);
trainX= X(indTrain,:);
trainy = y(indTrain);
indTest = 3839:4798;
indTest(indTrain) = [];
testX = X(indTest,:);
testy = y(indTest);
It shows error in trainX= X(indTrain,:); saying
Index exceeds matrix dimensions
can anyone pls clarify ? thanks.
by the way I am having a 4x4798 data which my first 3 rows serve as predictors, and last row (4th row) is my response. how would i correctly split the data into the first 3838 columns as my training set and remaining as testing set.
Thanks..!!
Fix your error
To fix the indexing error you need to select the column indices of X, rather than the row indices:
trainX = X(:, indTrain );
Some words of advice
It seems like your P matrix is 4-by-4798 and it is two dimensional. Therefore, writing P(1,:,:) does select the first row, but it gives the impression as if P is three dimensional because of the extra : at the end. Don't do that. It's bad habit and makes your code harder to read/understand/debug.
X = P(1:3,:); % select all three rows at once
y = P(4,:); % no need for 4:end here - again, gives wrong impression as if you expect more than a single label per x.
Moreover, I do not understand what you are trying to accomplish with indTest(indTrain)=[]? Are you trying to ascertain that the train and test set are mutually exclusive?
This line will most likely cause an error since the size of your test set is 960 and indTrain contains 1:3838 (randomly permuted) so you will get "index exceeds..." error again.
You already defined your indTrain and indTest as mutually exclusive, no need for another operation. If you want to be extra careful you can use setdiff
indTest = setdiff( indTest, indTrain );
having a problem with my "new love", matlab: I wrote a function to calculate an integral using the trapz-method: `
function [L]=bogenlaenge_innen(schwingungen)
R = 1500; %Ablegeradius
OA = 1; %Amplitude
S = schwingungen; %Schwingungszahl
B = 3.175; %Tapebreite
phi = 0:2.*pi./10000:2.*pi;
BL = sqrt((R-B).^2+2.*(R-B).*OA.*sin(S.*phi)+OA.^2.*(sin(S.*phi)).^2+OA.^2.*S.^2.*(cos(S.*phi)).^2);
L = trapz(phi,BL)`
this works fine when i start it with one specific number out of the command-window. Now I want to plot the values of "L" for several S.
I did the following in a new *.m-file:
W = (0:1:1500);
T = bogenlaenge_innen(W);
plot(W,T)
And there it is:
Error using .*
Matrix dimensions must agree.
What is wrong? Is it just a dot somewhere? I am using matlab for the second day now, so please be patient.... ;) Thank you so much in advance!
PS: just ignore the german part of the code, it does not really matter :)
In your code, the arrays S and phi in the expression sin(S.*phi) should have same size or one of them should be a constant in order the code works
The error is most likely because you have made it so that the number of elements in schwingungen, i.e. W in your code, must be equal to the number of elements in phi. Since size(W) gives you a different result from size(0:2.*pi./10000:2.*pi), you get the error.
The way .* works is that is multiplies each corresponding elements of two matrices provided that they either have the same dimensions or that one of them is a scalar. So your code will work when schwingungen is a scalar, but not when it's a vector as chances are it has a different number of elements from the way you hard coded phi.
The simplest course of action (not necessarily the most Matlabesque though) for you is to loop through the different values of S:
W = (0:1:1500);
T = zeros(size(W); %Preallocate for speed)
for ii = 1:length(W)
T(ii) = bogenlaenge_innen(W(ii));
end
plot(W,T)
In your function you define phi as a vector of 10001 elements.
In this same function you do S.*phi, so if S is not the same length as phi, you will get the "dimensions must agree" error.
In your call to the function you are doing it with a vector of length 1501, so there is your error.
Regards
I am trying to model a vector data containing 200 sample points denoting a measurement.I want to see "goodness of fit" and after reading I found that this can be done by predicting the next set of values(I am not that confident though if this is the correct way).I am stuck at this since the following code gives an error and I am just unable to solve it.Can somebody please help in removing the error
Error using *
Inner matrix dimensions must agree.
Error in data_predict (line 27)
ypred(j) = ar_coeff' * y{i}(j-1:-1:j-p);
Also,can somebody tell me how to do the same thing i.e get the coefficients using nonlinear AR modelling,moving average and ARMA since using the command nlarx() did not return any model coefficients?
CODE
if ~iscell(y); y = {y}; end
model = ar(y, 2, 'yw');
%prediction
yresiduals=[];
nsegments=length(y);
ar_coeffs = model.a;
ar_coeff=[ar_coeffs(2) ar_coeffs(3)]
for i=1:nsegments
pred = zeros(length(y{i}),1);
for j=p+1:length(y{i})
ypred(j) = ar_coeff(:)' * y{i}(j-1:-1:j-p);
end
yresiduals = [yresiduals; y{i}(p+1:end) - ypred(p+1:end)];
end
In matlab, * is the matrix product between two matrices. That means that the number of columns in the first matrix must equal the number of rows in the second matrix. You might have intended to use .* element by element multiplication. EDIT: For element by element multiplication, the matrices must be the same size. Check the size of your matrices. If they don't fit either of these conditions, something needs to change.
I am currently using the Toolbox Graph on the Matlab File Exchange to calculate curvature on 3D surfaces and find them very helpful (http://www.mathworks.com/matlabcentral/fileexchange/5355). However, the following error message is issued in “compute_curvature” for certain surface descriptions and the code fails to run completely:
> Error in ==> compute_curvature_mod at 75
> dp = sum( normal(:,E(:,1)) .* normal(:,E(:,2)), 1 );
> ??? Index exceeds matrix dimensions.
This happens only sporadically, but there is no obvious reason why the toolbox works perfectly fine for some surfaces and not for others (of a similar topology). I also noticed that someone had asked about this bug back in November 2009 on File Exchange, but that the question had gone unanswered. The post states
"compute_curvature will generate an error on line 75 ("dp = sum(
normal(:,E(:,1)) .* normal(:,E(:,2)), 1 );") for SOME surfaces. The
error stems from E containing indices that are out of range which is
caused by line 48 ("A = sparse(double(i),double(j),s,n,n);") where A's
values eventually entirely make up the E matrix. The problem occurs
when the i and j vectors create the same ordered pair twice in which
case the sparse function adds the two s vector elements together for
that matrix location resulting in a value that is too large to be used
as an index on line 75. For example, if i = [1 1] and j = [2 2] and s
= [3 4] then A(1,2) will equal 3 + 4 = 7.
The i and j vectors are created here:
i = [face(1,:) face(2,:) face(3,:)];
j = [face(2,:) face(3,:) face(1,:)];
Just wanted to add that the error I mentioned is caused by the
flipping of the sign of the surface normal of just one face by
rearranging the order of the vertices in the face matrix"
I have tried debugging the code myself but have not had any luck. I am wondering if anyone here has solved the problem or could give me insight – I need the code to be sufficiently general-purpose in order to calculate curvature for a variety of surfaces, not just for a select few.
The November 2009 bug report on File Exchange traces the problem back to the behavior of sparse:
S = SPARSE(i,j,s,m,n,nzmax) uses the rows of [i,j,s] to generate an
m-by-n sparse matrix with space allocated for nzmax nonzeros. The
two integer index vectors, i and j, and the real or complex entries
vector, s, all have the same length, nnz, which is the number of
nonzeros in the resulting sparse matrix S . Any elements of s
which have duplicate values of i and j are added together.
The lines of code where the problem originates are here:
i = [face(1,:) face(2,:) face(3,:)];
j = [face(2,:) face(3,:) face(1,:)];
s = [1:m 1:m 1:m];
A = sparse(i,j,s,n,n);
Based on this information removal of the repeat indices, presumably using unique or similar, might solve the problem:
[B,I,J] = unique([i.' j.'],'rows');
i = B(:,1).';
j = B(:,2).';
s = s(I);
The full solution may look something like this:
i = [face(1,:) face(2,:) face(3,:)];
j = [face(2,:) face(3,:) face(1,:)];
s = [1:m 1:m 1:m];
[B,I,J] = unique([i.' j.'],'rows');
i = B(:,1).';
j = B(:,2).';
s = s(I);
A = sparse(i,j,s,n,n);
Since I do not have a detailed understanding of the algorithm it is hard to tell whether the removal of entries will have a negative effect.