My motive here is to calculate parameters of a model using Least Square Estimation method.
Variable W stores the 'time' and CF stores the no. of faults occurred. I tried testing it on different Software Reliability Models (more like functions, as I've define below in 'func' variable), and only in some of them this error occurred.
W=xlsread(file,'A:A');
CF=xlsread(file,'B:B');
a(1)=100;
a(2)=1;
dy = CF*0.04;
func = inline('a(1).*(1-exp(-a(2).*W))','a','W');
Nrepeat=100;
sd = 0.3;
zfactor = 2;
outcut=10;
pList=zeros(Nrepeat,3);
for rep =1:Nrepeat
rep
[a,r,j] = nlinfit(W,CF,func,a);
%copy fit to list
pList(rep,:) = a'; <--- ERROR
ERROR :
Subscripted assignment dimension mismatch.
Error in lse_go (line 59)
pList(rep,:) = a';
I have tried using a.'; , a,' , a' but still it shows the same error. How do I remove this error?
Related
I so far have the following
y = log(x);
% Ask user for input values for h and M
% M denotes the number of steps of the algorithm.
h = input('Input value h: ');
M = input('Input value M: ');
%Initialize an MxM matrix
D = zeros(M);
phi = (1/(2*h)) * (y(x+h) - y(x-h));
print(phi);
I obtain the error
Error using symengine (line 58) Index exceeds matrix dimensions.
Error in sym/subsref (line 696)
B = mupadmex('symobj::subsref',A.s,inds{:});
Error in RE (line 12) phi = (1/(2*h)) * (y(x+h) - y(x-h));
First, I believe I should be getting an error message about x not being defined. Second, I have no idea what the matrix dimension error is about. Third, and most importantly, how can I declare the function phi so that it becomes what I wrote?
First, I believe I should be getting an error message about x not being defined.
I'm guessing that x is defined, or you would get that error upon the line defining phi. To check whether x is defined, type "who" or "whos".
Second, I have no idea what the matrix dimension error is about.
This is most likely because y is a scalar, x + h is equal to some nonzero integer that is not 1, and you're trying to access y(x + h). For your own edification try setting y equal to a scalar (e.g. y = 5;) and seeing what errors are produced by indexing it in various legitimate and non-legitimate ways (e.g. y(1), y(0), y(3), y(-1), y(1.5)).
Third, and most importantly, how can I declare the function phi so that it becomes what I wrote?
Based on the context it looks like you want y to be defined as a function of x instead of a scalar. In other words:
y = #(x)log(x);
phi = (1/(2*h)) * (y(x+h) - y(x-h));
The code runs without error when you change the definitions to the above.
One other error you will run into: the print command is not what you're looking for - this prints a figure to a file. You're probably looking for:
disp(phi);
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?
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.
i have my objective function as
function val = fitness( X )
val = 10*X(7)+20*X(8)+50*X(9)+10*X(10)+20*X(11)+50*X(12);
end
and i am trying to call ga as
ga(#fitness,12,A,b,[],[],lb,[],[],IntCon)
A = 9X9 matrix
b = 9X1 matrix
lb = 9X1 Zero matrix
IntCon = [1:12]
i am getting the following error message
Error using preProcessLinearConstr (line 48)
The number of columns in A must be the same as the length of X0.
Error in gacommon (line 100)
[Iterate.x,Aineq,bineq,Aeq,beq,lb,ub,msg,exitFlag] = ...
Error in ga (line 319)
[x,fval,exitFlag,output,population,scores,FitnessFcn,nvars,Aineq,bineq,Aeq,beq,lb,ub,
please provide an example using ga function solving mixed integer problem.
The issue is that Ab is of size 9x1, while fitness() expects the size to be at least 12x1.
E.g. the following has no error:
A = ones(12,12);
b = ones(12,1);
lb = zeros(12,1);
IntCon = [1:12];
ga(#fitness,12,A,b,[],[],lb,[],[],IntCon)
For more information, see Mixed Integer Optimization.