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?
Related
Matlab (2015a) is behaving weirdly: a number of builtin functions are not responding as expected. For instance, typing
ttest([1 2], [1 2])
results in
Error using size
Dimension argument must be a positive integer scalar within indexing range.
Error in nanstd (line 59)
tile(dim) = size(x,dim);
Error in ttest (line 132)
sdpop = nanstd(x,[],dim);
If I do a which for each of these functions:
which size
which nanstd
which ttest
I get, respetively:
built-in (C:\Program Files\MATLAB\R2015a\toolbox\matlab\elmat\size)
C:\Program Files\MATLAB\R2015a\toolbox\stats\eml\nanstd.m
C:\Program Files\MATLAB\R2015a\toolbox\stats\stats\ttest.m
Each of these files looks fine, except that size.m has each one of its rows commented out.
What could be the problem here?
Perhaps related to your problem:
ttest for R2013a makes the following call:
sdpop = nanstd(x,[],dim);
The helpfile for R2013a version of nanstd states:
Y = nanstd(X,FLAG,DIM) takes the standard deviation along dimension DIM of X.
On the other hand, nanstd in the 2005 nansuite package downloaded off Mathworks file exchange states:
FORMAT: Y = nanstd(X,DIM,FLAG)
Notice how DIM and FLAG are reversed!
If I call R2013a's ttest such that it makes a call to the old, 2005 nansuite function nanstd, Matlab generates an error similar to yours:
Error using size
Dimension argument must be a positive integer scalar within indexing range.
Error in nanmean (line 46)
count = size(x,dim) - sum(nans,dim);
Error in nanstd (line 54)
avg = nanmean(x,dim);
Error in ttest (line 132)
sdpop = nanstd(x,[],dim);
If [] is passed as DIM instead of FLAG, then nanstd's call to size(x, DIM) triggers an error because [] is not a positive integer scalar. If something like this is the cause, the broader question is, what's going on with your Matlab install or setup or downloads or whatever such that you're calling archaic code? Or why is that archaic code even around? I don't know at what point in Matlab's release history that nanstd(x, FLAG, DIM) became supported (instead of simply nanstd(x, DIM))?
Archive: below is my old answer which misdiagnosed your problem
Both of your sample vectors x and y are the same (i.e. [1,2]). The estimated variance of the difference is 0, and all your stats are going to blow up with NaN.
Do the stats step by step, and it will be clear what's going on.
x = [1; 2]; % Data you used in the example.
y = [1; 2]; % Data you used in the example.
z = x - y; % Your call to ttest tests whether this vector is different from zero at a statistically significant level.
Now we do all the stats on z
r.n = length(z);
r.mu = mean(z);
r.standard_error = sqrt(var(z,1) / (r.n-1)); % For your data, this will be zero since z is constant!
r.t = r.mu ./ r.standard_error; % For your data, this will be inf because dividing by zero!
r.df = r.n - 1;
r.pvals(r.t >= 0) = 2 * (1 - tcdf(r.t(r.t>=0), r.df)); % For your data, tcdf returns NaN and this all fails...
r.pvals(r.t < 0) = 2 * tcdf(r.t(r.t<0), r.df);
etc...
This should match a call to
[h, p, ci, stats] = ttest(x-y);
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.
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.
keep getting an indexing error " ()-indexing must appear last in an index expression"
bascially im trying to create a matrix x made of randn(n,1000000) where every jth row is multiplied by matrix NV(i,j).
%monte carlo simulation
function [y1,y2,y3,y4]= ed1(SNRL,SNRS,SNRH,n) %ed is the energy detection
g1= SNRL:SNRS:SNRH;
g=10.^(g1/10);
beta=0.8; % is the probability pfa, it cannot be more than 1
pf1=zeros(1,length(g));
pd1=zeros(1,length(g));
pf2=zeros(1,length(g));
pd2=zeros(1,length(g));
x=zeros(n,length(g));
y=zeros(n,length(g));
for i=1 : length(g)
for j=1:n
NV=(i,j);
x(j,:) = randn(n,1000000)*sqrt(NV(i,j));
y(j,:)=randn(n,1000000*(j),:)*sqrt(g(i))+x(j,:);
end
%Tgam is the threshold of gamma distribution
Tgam = gaminv((1-beta),n/2,(2/n)*(1+g(i))); %probab of flase detection
pf1(i)= gamcdf(Tgam,n/2,(2/n)*(1+g(i))); %ho
pd1 (i) = gamcdf(Tgam,n/2,2/n); %h1 % prob of detection
pf2(i)= length (find(sum(y.^2)/n<Tgam))/1000000;
pd2 (i) = length (find(sum(x.^2)/n<Tgam))/1000000;
y1=pf1; y2=pd1; y3=pf2; y4=pd2;
end
You have three things giving you errors:
1) NV=(i,j) This syntax will give you an error message. If you are not getting an error message about this line, you have an error in the line calling this code and should post that too. Try NV=zeros(length(g),n); for a temporary fix until you know what NV should be.
2) randn(n,1000000*(j),:) is also bad syntax. Do you mean randn(n,1000000*(j))?
3) x(j,:) = randn(1,10000)*sqrt(NV(i,j));: x as written in your code is not the right size unless g is magically of length 10000.
Hopefully checking these 3 things will give you an idea of what is giving your specific error message (which is not the same as any of the messages these errors gave me.)