I having trying to run linear regression and plot the results on the below data.
x1 = [40 45 38 50 48 55 53 55 58 40 55 48 45 55 60 60 60 65 50 58];
x2 = [25 20 30 30 28 30 34 36 32 34 38 28 30 36 34 38 42 38 34 38];
y = [1 2 1 3 2 3 3 4 4 3 5 3 3 2 4 5 5 5 4 3];
I have attempted it with this code
X = [ones(size(x1)) x1 x2 (x1.*x2)];
b = regress(y, X);
scatter3(x1,x2,y,'filled')
I am getting this error
Error using regress
Y must be a vector and must have the same number of rows as X.
Both y and X have one row each so I am unsure how to go about this regression.
Any help is appreciated.
Related
I have implemented matching pursuit algorithm but i m unable to get the required result.
Here is my code:
D=[1 6 11 16 21 26 31 36 41 46
2 7 12 17 22 27 32 37 42 47
3 8 13 18 23 28 33 38 43 48
4 9 14 19 24 29 34 39 44 49
5 10 15 20 25 30 35 40 45 50];
b=[6;7;8;9;10];
n=size(D);
A1=zeros(n);
R=b;
H=10;
if(H <= 0)
error('The number of iterations needs to be greater then 0')
end;
for k=1:1:H
[c,d] = max(abs(D'*R)); %//'
A1(:,d)=D(:,d);
D(:,d)=0;
y = A1\b;
R = b-A1*y;
end
Output
y=
0.8889
0
0
0
0
0
0
0
0
0.1111
I should get only non-zero value at (2,1) and other values should be zero but I'm getting 2 non-zero value. Can you please help me find out where the error is?
Thanks.
I checked with:
http://www.scholarpedia.org/article/Matching_pursuit
Your functions need to be normalized!
D = D./repmat(sum(D,1),5,1);
I get the following algorithm:
D=[1 6 11 16 21 26 31 36 41 46
2 7 12 17 22 27 32 37 42 47
3 8 13 18 23 28 33 38 43 48
4 9 14 19 24 29 34 39 44 49
5 10 15 20 25 30 35 40 45 50];
D = D./repmat(sum(D,1),5,1);
b=[6;7;8;9;10];
n=size(D);
A1=zeros(n);
R=b;
H=100;
if(H <= 0)
error('The number of iterations needs to be greater then 0')
end;
a = zeros(1,H);
G = zeros(size(D,1),H);
for k=1:1:H
ip = D'*R;
[~,d] = max(abs(ip)); %//'
G(:,k) = D(:,d);
a(k) = ip(d);
R = R-a(k)*G(:,k);
end
% recover signal:
Rrec = zeros(size(R));
for i=1:H
Rrec = Rrec + a(i)*G(:,i);
end
figure();
plot(b);
hold on;
plot(Rrec)
It approximates the signal quite well. But not with D(:,2) at first as expected. Maybe it is a starting point...
Here is the updated code. This is based on the algorithm provided at https://en.wikipedia.org/wiki/Matching_pursuit
clc;
clear all;
D=[1 6 11 16 21 26 31 36 41 46
2 7 12 17 22 27 32 37 42 47
3 8 13 18 23 28 33 38 43 48
4 9 14 19 24 29 34 39 44 49
5 10 15 20 25 30 35 40 45 50];
b=[6;7;8;9;10];
H=10;
for index=1:10
G(:,index)=D(:,index)./norm(D(:,index));
end
G1=G;
n=size(G);
R=b;
if(H <= 0)
error('The number of iterations needs to be greater then 0')
end;
if(H >size(D,2))
error('The number of iterations needs to be less than dictionary size')
end;
bIndex=1:size(G,2);
for k=H:-1:1
innerProduct=[];
for index=1:size(G,2)
innerProduct(index)=dot(R,G(:,index));
end
[c,d] = max(abs(innerProduct));
An(H-k+1)=innerProduct(d);
R = R-(An(H-k+1)*G(:,d));
G(:,d)=[];
strong(H-k+1)=bIndex(d);
bIndex(d)=[];
end
G_new=G1(:,strong);
%% reconstruction
bReconstructed=zeros(size(G_new,1),1);
for index=1:size(G_new,2)
bReconstructed(:,index) = (An(index)*G_new(:,index));
end
b_new=sum(bReconstructed,2)
Yes the atoms in the dictionary must be normalized so that the inner products of the current residual with different atoms can be compared fairly.
You may want to check my OMP implementation which also includes incremental Cholesky updates for the least square step in OMP at https://github.com/indigits/sparse-plex/blob/master/library/%2Bspx/%2Bpursuit/%2Bsingle/omp_chol.m
I have written detailed tutorial notes on OMP in my library documentation at https://sparse-plex.readthedocs.io/en/latest/book/pursuit/omp/index.html
My library sparse-plex contains a C implementation of OMP which is close to 4 times faster than fastest MATLAB implementations. See the discussion here https://sparse-plex.readthedocs.io/en/latest/book/pursuit/omp/fast_omp.html
20 4 4 74 20 20 74 85 85 85
A = 36 1 1 11 36 36 11 66 66 66
77 1 1 15 77 77 15 11 11 11
3 4 2 6 7 8 10 10 15 17
how from the matrix A, I can extract the submatrix whose fourth line (end line) contains only the values [3 6 10]?
for a single value, I do:
B=A(:,A(4,:)==10)
but I do not know how to do this for several values.
Use ismember -
search_array = [3 6 10]
subA = A(:,ismember(A(end,:),search_array))
Or bsxfun -
subA = A(:,any(bsxfun(#eq,A(end,:),search_array(:)),1))
consider i have an odd matrix with dimension:
x = <349*385 double>
and then i do this following step:
order = [1 9 2 3 10 17 25 18 11 4 5 12 19 26 33 ...
41 34 27 20 13 6 7 14 21 28 35 42 49 57 50 ...
43 36 29 22 15 8 16 23 30 37 44 51 58 59 52 ...
45 38 31 24 32 39 46 53 60 61 54 47 40 48 55 ...
62 63 56 64];
x = double(x);
y = im2col(x, [8 8],'distinct');
xb = size(y, 2);
y = y(order, :);
y = y(:)';
I get y :
y = <64x2156 double>
xb = 2156;
y = <1x137984 double>
and I transform y into cell :
for i=1:length(x)
r{i} = y(floor((end/length(x))*(i-1)+1):ceil((end/length(x))*i));
end
finally i reverse r into matrix :
y1 = cell2mat(r);
I get y1 :
y1 = <1x138308 double>
My question is why y1 can't get the same result with y? I want y1 and y same. Maybe i wrong when dividing matrix y into cell r. Please help me.
I have a vector of electrical power consumption data which consists of transient, steady and power off states. I would like to identify steady-state starting point by the following condition:
The 5 consecutive elements of the data have difference values between each adjacent element <= threshold value (for this case, let say =10 W)
The first element that meets the condition shows the starting point of steady-state.
Example:
data = [0 0 0 40 70 65 59 50 38 30 32 33 30 33 37 19 ...
0 0 0 41 73 58 43 34 25 39 33 38 34 31 35 38 19 0]
abs(diff(data)) = [0 0 40 30 15 7 9 12 8 3 2 1 3 4 18 19 ...
0 0 41 32 15 9 14 6 5 4 3 4 3 19 19 0]
The sequences of abs(diff(data)) that meet the condition are 8 3 2 1 3 and 6 5 4 3 4. Therefore, the output should show the 10th data element (=30) and 27th data element (=33) as starting point of steady-state (There are 2 times of steady-state detected).
How would I write MATLAB code for this scenario?
(PS: data = 0 shows power off state)
Here's one approach using nlfilter (if the function is not available, you can implement a sliding window yourself):
data = [0 0 0 40 70 65 59 50 38 30 32 33 30 33 37 19 0 0 0 41 73 58 43 34 25 39 33 38 34 31 35 38 19 0];
difs = abs(diff(data));
% Use sliding window to find windows of consecutive elements below threshold
steady = nlfilter(difs, [1, 5], #(x)all(x <= 10));
% Find where steady state starts (1) and ends (-1)
start = diff(steady);
% Return indices of starting steady state
ind = find(start == 1);
assume I have a 10x10 matrix M
M=[64 36 50 87 22 45 37 23 68 88;
33 23 87 49 54 25 35 98 78 52;
12 54 76 43 24 87 54 98 45 34;
77 87 23 45 34 65 23 76 12 76;
12 34 55 44 76 98 93 23 54 67;
22 55 78 90 88 56 34 23 12 76;
99 23 67 89 34 23 12 87 45 23;
22 54 76 89 65 23 45 12 93 12;
44 56 23 88 67 14 15 67 34 12;
11 44 77 99 34 23 78 34 12 79];
I want to first find out the local maximum in the matrix
and then according to the maximum position do a sum over a 3x3 region over M
For the first step, the code I used is local_max=imregionalmax(M). to find out the local maximum position, but how can I go further to use this coordination to sum over a 3x3 matrix over M?
Thanks for the help.
You can calculate the sum for the whole matrix and then only keep the values that you're interested in. This should work:
local_max=imregionalmax(M)
sums = imfilter(M, ones(3));
local_max_sums = sums(local_max);
And if what you want is a matrix with non-zero entries where the local maxima are located:
local_max_sums = sums .* local_max;
You seem to be looking for the matrix subset functionality of Matlab.
Basically, for
M = [ 1 2 3 4 5 6;
4 5 6 7 8 9;
7 8 9 0 1 2;
0 1 2 3 4 5;
3 4 5 6 7 8;
6 7 8 9 0 1];
If you have a max at (3,3), you can use M(2:4, 2:4) to get
N = [ 5 6 7;
8 9 0;
1 2 3];
Summing that matrix is all that remains - as simple as
total = sum(N(:));
This is kind of brute force for Matlab, but I think it works.
bw = imregionalmax(M);
[x,y] = find(bw);
s = [];
for i = 1:length(x)
startX = x(i)-2;
if(startX < 1)
startX = 1;
end
endX = x(i)+2;
if endX > 10
endX = 10;
end
startY = y(i)-2;
if startY < 1
startY = 1;
end
endY = y(i)+2;
if endY > 10
endY = 10;
end
s(i) = sum2(M(startX:endX, startY:endY));
end