In order to find best fit (thru polyfit), i am getting negative p value but matlab is not accepting it (Subscript indices must either be real positive integers or logicals.). Is there any way that I can use it ? I can't think of alternative method. I'll always get negative values.
EDIT:
I am trying to flattening baseline of a curve, for that. I am running for loop to have fit from 1 to 3 order. And then I am using smallest normr s value to to find the best fit and then subtract it from the whole curve to get baseline straight. I tried with few curves it works well but not with all of the data because of above describes issue.
part of the code I am working on:
for i=1:3
[p,s]=polyfit(x,y,i);
a=s.normr;
b(i,1)=p(1);
normr(i,1)=a;
ind=find(b==min(b));
mn=b(ind,1);
Yflat=y-mn(1)*(x-mean(x));
ca{2,2}=Yflat;
clear a b normr p s rte ind ind2 Yflat
end
When I translate an image into negative coordinates,
I usually record an offset e.g.
offset = [ -5, -8.5 ]
and save the intensity values in matrix begin with (1, 1) as usual,
But when comes to calculation, let the coordinates array add up with the offset
e.g. [ actualX, actualY ] = [ x, y ] + offset ;
It may need extra efforts, but it works.
Good Luck!
The code below (your code from the comments + initialization of x, y) executes. What is the problem?
x = 1:50;
y = randn(size(x));
for i=1:3
[p,s]=polyfit(x,y,i);
a=s.normr;
b(i,1)=p(1);
normr(i,1)=a;
ind=find(b==min(b));
mn=b(ind,1);
Yflat=y-mn(1)*(x-mean(x));
ca{2,2}=Yflat;
end
Related
I want to compute using MATLAB all the integers x such that x^2-2x<2000, but I am having problems in the displaying part because when I run my code MATLAB seems not to finish. My code is as follows:
x=100;
while x^2+2x-2000>=10^-6
x=x-20;
if x^2+2x-2000<10^-6
disp(x)
end
end
I think the wrong part is when I type disp(x) in the while loop, but I don't know how to fix it.
NOTE: I am using the 10^-6 to avoid rounding errors
Can someone help me to fix this code please?
Computationally, finding all integers that satisfy this condition will need some help from a quick insight into this problem. Otherwise, you'll have to test all the integers, which is impossible since there is infinite number of integers. Analytically, finding all the integers that satisfy the condition x^2-2x < 2000 means finding the integers that lies within the intersection of the curve x^2 - 2x and y = 2000.
Let's first take a look at the problem by plotting it:
x = -500:.1:500;
y = x.^2 - 2*x;
plot(x, y, 'color', 'g')
hold on
line([-200 200], [2000 2000], 'color', 'r')
You can easily see that you can limit your search to at least between -100 and 100. You can store the value in an array
results = [] % Declare empty array first and append value later.
for x = -100:100
if x^2 - 2*x < 2000
disp(x)
results = [results,x]; % Using a bracket to append value to the array.
end
end
And a faster way to get results using logical indexing.
x = -100:100
results = x(x.^2 - 2*x<2000)
In the above code, x.^2 - 2*x < 2000 generates a logical array that has the same size as x. Each element stores the logical value that comes from the evaluating each element in x with the expression. You then use this as a mask to pick out only elements in x that satisfy this condition.
If you add in some parentheses and use the correct syntax for 2*x, it works.
x=100;
while (x^2+2*x-2000)>= (10^-6)
x=(x-20);
if (x^2+2*x-2000) <10^-6
disp(x)
end
end
Building upon #User3667217's solution, you can also vectorise this:
x = -100:100;
y = x.^2-2*x;
tmp = y<2000;
results = y(tmp);
this will give you a speed up over the for-loop solution.
I have written a code where i have to control, if the position (x,y) (saved in the Matrix Mat) is inside of a circular object which is centered at (posx,posy). If so the point gets a value val otherwise its zero.
My Code looks like this but as a matter of fact it is advertised to NOT use loops in matlab. Since i use not 1 but 2 loops, i was wondering if there is a more effective way for solving my problem.
Mat = zeros(300); %creates my coordinate system with zeros
...
for i =lowlimitx:highlimitx %variable boundary of my object
for j=lowlimity:highlimity
helpsqrdstnc = abs(posx-i)^2 + abs(posy-j)^2; %square distance from center
if helpsqrdstnc < radius^2
Mat(i,j)= val(helpsqrdstnc);
end
end
end
the usual way to optimize matlab code is to vectorize the operations. This is because built in functions and operators is in general much faster. For your case this would leave you with this code:
Mat = zeros(300); %creates my coordinate system with zeros
...
xSq = abs(posx-(lowlimitx:highlimitx)).^2;
ySq = abs(posy-(lowlimity:highlimity)).^2;
helpsqrdstnc = bsxfun(#plus,xSq,ySq.'); %bsxfun to do [xSq(1)+ySq(1),xSq(2)+ySq(1),...; xSq(1)+ySq(2),xSq(2)+ySq(2)...; ...]
Mat(helpsqrdstnc < radius^2)= val(helpsqrdstnc(helpsqrdstnc < radius^2));
where helpsqrdstnc must be the same size as Mat. There may also be neseccary to do a reshape here, but you will notice that by yourself if you get a column vector.
This does of course assume that radius, posx and posy is constant, but reading the question this seems to be the case. However, I do not know exactly how val looks, so it I have not managed to test the code. I also think that val(helpsqrdstnc) is tedious, since this refer to the distance, which does not neseccarily need to be an integer.
I need to produce a signal x=-2*cos(100*pi*n)+2*cos(140*pi*n)+cos(200*pi*n)
So I put it like this :
N=1024;
for n=1:N
x=-2*cos(100*pi*n)+2*cos(140*pi*n)+cos(200*pi*n);
end
But What I get is that the result keeps giving out 1
I tried to test each values according to each n, and I get the same results for any n
For example -2*cos(100*pi*n) with n=1 has to be -1.393310473. Instead of that, Matlab gave the result -2 for it and it always gave -2 for any n
I don't know how to fix it, so I hope someone could help me out! Thank you!
Not sure where you get the idea that -2*cos(100*pi) should be anything other than -2. Maybe you are not aware that Matlab works in radians?
Look at your expression. Each term can be factored to contain 2*pi*(an integer). And you should know that cos(2*pi*(an integer)) = 1.
So the results are exactly as expected.
What you are seeing is basically what happens when you under-sample a waveform. You may know that the Nyquist criterion says that you need to have a sampling rate that is at least two times greater than the highest frequency component present; but in your case, you are sampling one point every 50, 70, 100 complete cycles. So you are "far beyond Nyquist". And that can only be solved by sampling more closely.
For example, you could do:
t = linspace(0, 1, 1024); % sample the waveform 1024 times between 0 and 1
f1 = 50;
f2 = 70;
f3 = 100;
signal = -2*cos(2*pi*f1*t) + 2*cos(2*pi*f2*t) + cos(2*pi*f3*t);
figure; plot(t, signal)
I think you are using degrees when you are doing your calculations, so do this:
n = 1:1024
x=-2*cosd(100*pi*n)+2*cosd(140*pi*n)+cosd(200*pi*n);
cosd uses degrees instead of radians. Radians is the default for cos so matlab has a separate function when degree input is used. For me this gave:
-2*cosd(100*pi*1) = -1.3933
The first term that I got using:
x=-2*cosd(100*pi*1)+2*cosd(140*pi*1)+cosd(200*pi*1)
x = -1.0693
Also notice that I defined n as n = 1:1024; this will give all integers from 1,2,...,1024,
there is no need to use a for loop since many of Matlab's built in functions are vectorized. Meaning you can just input a vector and it will calculate the function for every element in the vector.
I'm trying to remove outliers from a tick data series, following Brownlees & Gallo 2006 (if you may be interested).
The code works fine but given that I'm working on really long vectors (the biggest has 20m observations and after 20h it was not done computing) I was wondering how to speed it up.
What I did until now is:
I changed the time and date format to numeric double and I saw that it saves quite some time in processing and A LOT OF MEMORY.
I allocated memory for the vectors:
[n] = size(price);
x = price;
score = nan(n,'double'); %using tic and toc I saw that nan requires less time than zeros
trimmed_mean = nan(n,'double');
sd = nan(n,'double');
out_mat = nan(n,'double');
Here is the loop I'd love to remove. I read that vectorizing would speed up a lot, especially using long vectors.
for i = k+1:n
trimmed_mean(i) = trimmean(x(i-k:i-1 & i+1:i+k),10,'round'); %trimmed mean computed on the 'k' closest observations to 'i' (i is excluded)
score(i) = x(i) - trimmed_mean(i);
sd(i) = std(x(i-k:i-1 & i+1:i+k)); %same as the mean
tmp = abs(score(i)) > (alpha .* sd(i) + gamma);
out_mat(i) = tmp*1;
end
Here is what I was trying to do
trimmed_mean=trimmean(regroup_matrix,10,'round',2);
score=bsxfun(#minus,x,trimmed_mean);
sd=std(regroup_matrix,2);
temp = abs(score) > (alpha .* sd + gamma);
out_mat = temp*1;
But given that I'm totally new to Matlab, I don't know how to properly construct the matrix of neighbouring observations. I just think it should be shaped like: regroup_matrix= nan (n,2*k).
EDIT: To be specific, what I am trying to do (and I am not able to) is:
Given a column vector "x" (n,1) for each observation "i" in "x" I want to take the "k" neighbouring observations to "i" (from i-k to i-1 and from i+1 to i+k) and put these observations as rows of a matrix (n, 2*k).
EDIT 2: I made a few changes to the code and I think I am getting closer to the solution. I posted another question specific to what I think is the problem now:
Matlab: Filling up matrix rows using moving intervals from a column vector without a for loop
What I am trying to do now is:
[n] = size(price,1);
x = price;
[j1]=find(x);
matrix_left=zeros(n, k,'double');
matrix_right=zeros(n, k,'double');
toc
matrix_left(j1(k+1:end),:)=x(j1-k:j1-1);
matrix_right(j1(1:end-k),:)=x(j1+1:j1+k);
matrix_group=[matrix_left matrix_right];
trimmed_mean=trimmean(matrix_group,10,'round',2);
score=bsxfun(#minus,x,trimmed_mean);
sd=std(matrix_group,2);
temp = abs(score) > (alpha .* sd + gamma);
outmat = temp*1;
I have problems with the matrix_left and matrix_right creation.
j1, that I am using for indexing is a column vector with the indices of price's observations. The output is simply
j1=[1:1:n]
price is a column vector of double with size(n,1)
For your reshape, you can do the following:
idxArray = bsxfun(#plus,(k:n)',[-k:-1,1:k]);
reshapedArray = x(idxArray);
Thanks to Jonas that showed me the way to go I came up with this:
idxArray_left=bsxfun(#plus,(k+1:n)',[-k:-1]); %matrix with index of left neighbours observations
idxArray_fill_left=bsxfun(#plus,(1:k)',[1:k]); %for observations from 1:k I take the right neighbouring observations, this way when computing mean and standard deviations there will be no problems.
matrix_left=[idxArray_fill_left; idxArray_left]; %Just join the two matrices and I have the complete matrix of left neighbours
idxArray_right=bsxfun(#plus,(1:n-k)',[1:k]); %same thing as left but opposite.
idxArray_fill_right=bsxfun(#plus,(n-k+1:n)',[-k:-1]);
matrix_right=[idxArray_right; idxArray_fill_right];
idx_matrix=[matrix_left matrix_right]; %complete index matrix, joining left and right indices
neigh_matrix=x(idx_matrix); %exactly as proposed by Jonas, I fill up a matrix of observations from 'x', following idx_matrix indexing
trimmed_mean=trimmean(neigh_matrix,10,'round',2);
score=bsxfun(#minus,x,trimmed_mean);
sd=std(neigh_matrix,2);
temp = abs(score) > (alpha .* sd + gamma);
outmat = temp*1;
Again, thanks a lot to Jonas. You really made my day!
Thanks also to everyone that had a look to the question and tried to help!
I am not very sure how the hist function in MATLAB works. I seem to have few problems with it.
Bascially, in the code below, i am trying to run the rotation invariant Uniform Local Binary Pattern(LBP) code. I have no problem with the LBP code but the problem is with hist function(indicated in the code below).
The problem is that the range i should get is from 0:9 but when i apply the histogram function i get values greater than 9 such as 35, 27 and even values such as 178114.Not very sure how to correct it.
I2 = imread('test.png');
RIUniformHist=[];
m=size(I2,1);
n=size(I2,2);
for i=1:10:m
for j=1:10:n
for k=i+1:i+8
for l=j+1:j+8
J0=I2(k,l);
I3(k-1,l-1)=I2(k-1,l-1)>J0;
I3(k-1,l)=I2(k-1,l)>J0;
I3(k-1,l+1)=I2(k-1,l+1)>J0;
I3(k,l+1)=I2(k,l+1)>J0;
I3(k+1,l+1)=I2(k+1,l+1)>J0;
I3(k+1,l)=I2(k+1,l)>J0;
I3(k+1,l-1)=I2(k+1,l-1)>J0;
I3(k,l-1)=I2(k,l-1)>J0;
LBP=I3(k-1,l-1)*2^7+I3(k-1,l)*2^6+I3(k-1,l+1)*2^5+I3(k,l+1)*2^4+I3(k+1,l+1)*2^3+I3(k+1,l)*2^2+I3(k+1,l-1)*2^1+I3(k,l-1)*2^0;
bits = bitand(LBP, 2.^(7:-1:0))>0;
if nnz(diff(bits([1:end, 1]))) <= 2
RIULBP(k,l)=abs(I3(k-1,l-1)-I3(k-1,l))+ abs(I3(k-1,l)-I3(k-1,l+1))+ abs(I3(k-1,l+1)-I3(k,l+1))+ abs(I3(k,l+1)-I3(k+1,l+1))+abs(I3(k+1,l+1)-I3(k+1,l))+abs(I3(k+1,l)-I3(k+1,l-1))+abs(I3(k+1,l-1)-I3(k,l-1));
else
RIULBP(k,l)=9;
end
end
end
RIULBP=uint8(RIULBP);
RIULBPv=reshape(RIULBP,1,size(RIULBP,1)*size(RIULBP,2));
RIUHist=hist(RIULBPv,0:9); % problem
RIUniformHist = [RIUniformHist RIUHist];
end
end
The vector returned by
RIUHist=hist(data, bins)
is the count of how many elements of data are nearest the point identified by the bins vector. So if you have a value of 178114, that juts means that there were 178114 elements of data that were nearest to the matching index in bins.
You can use
[RIUHist, binsOut] = hist(data)
to let Matlab choose the bins (I believe it uses 20 bins) or
[RIUHist, binsOut] = hist(data, binCount)
To let Matlab choose the bins, but force a certain number of bins (I often use 100 or 200).