How to fix: Index Exceeds Matrix Dimensions
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Hi,
Im new to MATLAB and am trying to extract data from a sine wave. I've set the threshold to -1 SD and want all of the data below it to stack into one matrix. In other words, M has to be a stacked product of P. However, at the last part of the code it gives the error: Index Exceeds Matrix Dimensions. From what I understand, this means that length(P)+P(i)>length(noise). So the logical thing to do would be to make 'noise' bigger or P(i) smaller. However, I can't seem to get rid of the error. I was told not to index so far into P, but i have no idea how to do that. Hopefully someone can help me with this!
clear all
hold off
% parameters
srate=1000;
t=1:99/srate:10;
noiseAmplitude=2;
a=4;
f=4;
%signal
signal=a*sin(2*pi*f*t);
noise= signal + noiseAmplitude*randn(1,length(signal));
standdev=std(noise);
P=find(diff(noise<-standdev)==1);
for i=1:length(P)
M(i,:)=noise(P(i):P(i)+10); (<-- ERROR: index exceeds matrix dimensions)
end
plot(M)
0 个评论
采纳的回答
Alexandra Harkai
2017-9-14
P =
2 27 32 34 37 50 52 60 65 68 70 75 90
P+10
ans =
12 37 42 44 47 60 62 70 75 78 80 85 100
Therefore P+10 exceeds the size of noise, and is giving you the error.
How you go about fixing it depends what you want to achieve in those cases. This one would fill in the rightmost part of array M:
for i=1:length(P)
offset = min(10, size(noise,2)-P(i));
M(i,(end-offset):end)=noise(P(i):P(i)+offset);
end
更多回答(11 个)
Cam Salzberger
2017-9-14
I don't fully understand what you are trying to do here, but I can tell you how to examine the issue. With any kind of error, it's often easiest to set a breakpoint at the line the error occurs. With an error in a loop, however, it could be many iterations before you actually hit the parameter combination that will cause the error. In that case, it's probably easiest to just run this at the command line:
dbstop if error
Now MATLAB will automatically go into debug mode as soon as it encounters an error. Once in debug mode, you can examine the variables and see where the issue is. Of course, in your particular code, the exact iteration may change every time because of the randomized noise.
I suspect that you don't actually want to index into "P" using "10" as an offset to the next index value, when "P" already contains index values itself. But I don't fully understand your workflow, so I can't be sure.
-Cam
0 个评论
dawa lepcha
2020-2-3
Hi , i have an error in line 18,
kindly need helps
>> guidedfilter_color
Index exceeds matrix dimensions.
Error in guidedfilter_color (line 18)
mean_I_g = boxfilter(I(:, :, 2), r) ./ N;
function q = guidedfilter_color(I, p, r, eps)
% GUIDEDFILTER_COLOR O(1) time implementation of guided filter using a color image as the guidance.
%
% - guidance image: I (should be a color (RGB) image)
% - filtering input image: p (should be a gray-scale/single channel image)
% - local window radius: r
% - regularization parameter: eps
I = double(imread('.\img_smoothing\cat.bmp')) / 255;
p = I;
r = 4; % try r=2, 4, or 8
eps = 0.2^2; % try eps=0.1^2, 0.2^2, 0.4^2
[hei, wid] = size(p);
N = boxfilter(ones(hei, wid), r); % the size of each local patch; N=(2r+1)^2 except for boundary pixels.
mean_I_r = boxfilter(I(:, :, 1), r) ./ N;
mean_I_g = boxfilter(I(:, :, 2), r) ./ N;
mean_I_b = boxfilter(I(:, :, 3), r) ./ N;
mean_p = boxfilter(p, r) ./ N;
mean_Ip_r = boxfilter(I(:, :, 1).*p, r) ./ N;
mean_Ip_g = boxfilter(I(:, :, 2).*p, r) ./ N;
mean_Ip_b = boxfilter(I(:, :, 3).*p, r) ./ N;
% covariance of (I, p) in each local patch.
cov_Ip_r = mean_Ip_r - mean_I_r .* mean_p;
cov_Ip_g = mean_Ip_g - mean_I_g .* mean_p;
cov_Ip_b = mean_Ip_b - mean_I_b .* mean_p;
% variance of I in each local patch: the matrix Sigma in Eqn (14).
% Note the variance in each local patch is a 3x3 symmetric matrix:
% rr, rg, rb
% Sigma = rg, gg, gb
% rb, gb, bb
var_I_rr = boxfilter(I(:, :, 1).*I(:, :, 1), r) ./ N - mean_I_r .* mean_I_r;
var_I_rg = boxfilter(I(:, :, 1).*I(:, :, 2), r) ./ N - mean_I_r .* mean_I_g;
var_I_rb = boxfilter(I(:, :, 1).*I(:, :, 3), r) ./ N - mean_I_r .* mean_I_b;
var_I_gg = boxfilter(I(:, :, 2).*I(:, :, 2), r) ./ N - mean_I_g .* mean_I_g;
var_I_gb = boxfilter(I(:, :, 2).*I(:, :, 3), r) ./ N - mean_I_g .* mean_I_b;
var_I_bb = boxfilter(I(:, :, 3).*I(:, :, 3), r) ./ N - mean_I_b .* mean_I_b;
a = zeros(hei, wid, 3);
for y=1:hei
for x=1:wid
Sigma = [var_I_rr(y, x), var_I_rg(y, x), var_I_rb(y, x);
var_I_rg(y, x), var_I_gg(y, x), var_I_gb(y, x);
var_I_rb(y, x), var_I_gb(y, x), var_I_bb(y, x)];
%Sigma = Sigma + eps * eye(3);
cov_Ip = [cov_Ip_r(y, x), cov_Ip_g(y, x), cov_Ip_b(y, x)];
a(y, x, :) = cov_Ip * inv(Sigma + eps * eye(3)); % Eqn. (14) in the paper;
end
end
b = mean_p - a(:, :, 1) .* mean_I_r - a(:, :, 2) .* mean_I_g - a(:, :, 3) .* mean_I_b; % Eqn. (15) in the paper;
q = (boxfilter(a(:, :, 1), r).* I(:, :, 1)...
+ boxfilter(a(:, :, 2), r).* I(:, :, 2)...
+ boxfilter(a(:, :, 3), r).* I(:, :, 3)...
+ boxfilter(b, r)) ./ N; % Eqn. (16) in the paper;
figure();imshow(q)
end
0 个评论
Walter Roberson
2020-2-3
Bmp files are not always rgb.
And when they are rgb then
[hei, wid] = size(p);
is always wrong.
[hei, wid, panes] = size(p);
0 个评论
Getnet Belie
2020-6-14
编辑:Walter Roberson
2020-6-14
Index exceeds matrix dimensions.
Error in feasmbll (line 21)
kk(ii,jj)=kk(ii,jj)+k(i,j)
function [kk]=feasmbll(kk,k,index)
%..................
%purpose;
%assembly of element matrices into the system matricx
%
% Synopsis;
%[kk]=feasmbll(kk,k,index)
%
%Variable Description;
%kk-system matrix
%k-element matrix
%index-d.o.f.vector associated with an element
%.................
%
edof=length(index);
%
for i=1:edof
ii=index(i);
for j=1:edof
jj=index(j);
26. kk(ii,jj)=kk(ii,jj)+k(i,j)
end
end
end
could you help me
kindly
1 个评论
Josh Meyer
2017-9-14
编辑:Josh Meyer
2017-9-14
The issue is with the line:
M(i,:)=noise(P(i):P(i)+10);
You want to take the index P(i) and grab the next 10 elements after that point, then put them into a row in M. This works fine for the first several iterations where P is picking out relatively small indices. However, some of the elements of P are picking out array elements near the end of noise, so there aren't 10 elements after those points to grab.
To put it more concretely, here is what P contains when I ran the code:
>> P
P =
2 12 14 20 27 30 32 40 55 65 70 75 78 80 88
So, when you get to the last element of P, 88, you will want to grab elements 88-98 from noise and put them into M. However, noise only has 91 elements:
>> size(noise)
ans =
1 91
So to fix this error you need to replace the +10 with something else. You can use min(diff(P)) to see the minimum distance between consecutive indices in P, or min(abs(P-length(noise))) to see how close the last element in P is to the end of noise, or maybe you could just loop to length(P)-1 or length(P)-2, avoiding the last few large indices in P.
0 个评论
yuvarani divakaramoorthy
2018-9-19
for i=1:s/k sum=0; for j=1:k sum=m((i-1)*k+j)+sum; <--(ERROR: index exceeds matrix dimensions) end M(i)=sum/k;
what can be done here?
0 个评论
rubina naz
2018-10-22
编辑:Walter Roberson
2020-6-14
Rmin = 60;
Rmax = 100;
[center, radius] = imfindcircles(RGB,[Rmin Rmax],'Sensitivity',0.9); % Display the circle
viscircles(center,radius); % Display the calculated center
hold on;
plot(center(:,1),center(:,2),'yx','LineWidth',2);%index axceed matrix problems
hold off;
0 个评论
MathWorks Support Team
2018-11-28
This error is returned when MATLAB tries to access elements of an array that do not exist. In this case, since “noise” only has 91 elements, MATLAB errors when the loop reaches an element of “P” such that “P(i) + 10” exceeds 91. Starting in R2018b, you will see the following error instead:
Index exceeds the number of array elements (91).
To fix this error, you will need to replace the index “P(i)+10” with something that doesn’t exceed 91, or end your loop earlier. For example, you could stop the loop at “length(P)-2”, and concatenate the remaining elements of “M” outside of the loop.
0 个评论
Abdul Basit
2020-6-15
clc;
clear
close all;
%% Phase 1 - Define Objective Function
%% Phase 2 - PSO Parameters
LB=[0 0 0];
UB=[10 10 10];
m=3;
n=500;
wmax=0.9;
wmin=0.4;
c1=2.05;
c2=2.05;
Maxiter=100;
% Loop for maximum run start
for run=1:20
%% Phase 3 - Initialization of Position & Velocity
for i=1:n
for j=1:m
pos(i,j)=(LB(i,j)+rand()).*(UB(i,j)-LB(i,j)); <===Index exceeds matrix dimensions
Please some one help as at this point i see the error: Index exceeds matrix dimension
end
end
vel=(0.1).*pos;
%% Phase 4 - Function Evaluation
for i=1:n
out(i,1)=fun(pos(i,:));
end
pbestval=out;
pbest=pos;
[fminval, index]=min(out);
gbest=pbest(index,:);
%PSO algorithm start here
iter=1;
while iter<=Maxiter
w=wmax-(iter/Maxiter).*(wmax-wmin);
%% Phase 5 - Compute PBest and GBest
X=pos;
Out=fun(X);
Har=find(out<=pbestval);
pbest(Har,:)=X(Har,:);
pbestval=out(Har);
[fbestval, ind1]=min(pbestval);
if fbestval<=fminval
fminvalue=fbestval;
gbest=pbest(ind1,:);
end
%% Phase 6 - Update Velocity & Position (Handling Boundary Constrains)
for i=1:n
for j=1:m
vel(i,j)=w.*vel(i,j)+c1.*rand().*(pbest(i,j)-pos(i,j))...
+c2.*rand().*(gbeat(1,j)-pos(i,j));
pos(i,j)=vel(i,j)+pos(i,j);
if pos(i,j)<LB(j)
pos(i,j)=LB(j);
elseif pos(i,j)>UB(j)
pos(i,j)=UB(j);
end
end
end
iter=iter+1;
end
%% Phase 7 - Store Best Value
F_ans(run)=fun(gbest);
F_gbest(run,:)=gbest;
% Loop for maximum run end
end
[bestFUN, bestRUN]=min(F_ans);
Best_X=F_gbest(bestRUN,:);
plot(F_ans)
function output = fun(X)
fx=10*(x1-1)^2+20.*x1.*x2+(x3-3)^2;
Con=[];
% Con(1)=x1+x2+x3-5;
% Con(2)=x1^2+x2^2-x3;
%
% for i=1:length(Con)
% if Con(i)>0
% Pen(i)=1;
% else
% Pen(i)=0;
% end
% end
% penalty=1000;
%output=fx+penalty*sum(pen);
output=fx;
end
12 个评论
aaliyan javaid
2021-3-29
hello brother have you implemented pso suceesfully i facing some issues can you help me
thanks
Walter Roberson
2021-3-29
The code executed for me without error (other than commenting out the "<== Error Here" phrase)
Mirghni Mohammed
2022-12-28
The program you will prepare must have an interface,
· Inputs (inputs.xlsx) and outputs (targets.xlsx) will be read from the files given to you in the program. The program will automatically determine the number of inputs and outputs according to the data read.
· There will be only one hidden layer in the program,
· The user will be able to determine the number of neurons in the hidden layer, the training rate, the number of epochs, and the data rates to be used for training and testing,
· Training and test data will be randomly selected according to the determined rate,
· The neurons in the hidden layer will use the sigmoid activation function, and the neuron in the output layer will use the linear activation (k=1) function,
· When the program starts, the initial weight values of the artificial neural network will be determined as random real numbers between -5 and 5.
· Backpropagation algorithm will be used for training.
· After the user-specified amount of training data and the desired epoch number are completed, the weights of the trained network should be displayed in the program.
· After the training is completed, the success of the test data will be graphically displayed using the trained network and test data. The metric to be used to determine success will be the Mean squared error.
1 个评论
Walter Roberson
2022-12-28
I do not understand how this will solve the problem that the original poster had.
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