I have a signal that looks like this :
I would like to find a way to locate the start and end of the portion of the middle.
What I did, is for the values above and below 0.5 to be a constant ==> 1, and if I find many times 1 in the row it means that it's my signal... but it's not a good way I guess! First my "threshold" would not be every time 0.5, and I am sure it exists some better way to do so.
If you guys have some documentations or ideas about that..
Thank you very much.
As mentioned by the others it is more of a DSP question and dsp.stackexchange.com will propably give you a better answer, but until then this might help:
data=csvread('acceleration.txt',1)
threshold_y=max(data)*0.5; %Thanks to GameOfThrows
thershold_x=101; %how many zeros can be between to ones to still count as continuous
addframe=50; %if you want a little bit of data before and after the active area
logic_index=data>threshold_y;
num_index=find(logic_index);
distance=diff(num_index);
gaps=[1 ; find(distance>thershold_x)]; %find the gaps bigger than your threshold
final_index=false(length(data),1);
for i=1:length(gaps)-1 %add ones between
final_index(num_index(gaps(i)+1)-addframe:num_index(gaps(i+1))+addframe)=true;
end
plot(x,data,x,final_index);
it is basically what you decribed in your question but with the addition of dealing with the zeros inbetween an area. thanks to #GameofThrows for the threshold idea.
Related
I am trying to avoid the for loops and I have been reading through all the old posts there are about it but I am not able to solve my problem. I am new in MATLAB, so apologies for my ignorance.
The thing is that I have a 300x2 cell and in each one I have a 128x128x256 matrix. Each one is an image with 128x128 pixels and 256 channels per pixel. In the first column of the 300x2 cell I have my parallel intensity values and in the second one my perpendicular intensity values.
What I want to do is to take every pixel of every image (for each component) and sum the intensity values channel by channel.
The code I have is the following:
Image_par_channels=zeros(128,128,256);
Image_per_channels=zeros(128,128,256);
Image_tot_channels=zeros(128,128,256);
for a=1:128
for b=1:128
for j=1:256
for i=1:numfiles
Image_par_channels(a,b,j)=Image_par_channels(a,b,j)+Image_cell_par_per{i,1}(a,b,j);
Image_per_channels(a,b,j)=Image_per_channels(a,b,j)+Image_cell_par_per{i,2}(a,b,j);
end
Image_tot_channels(a,b,j)=Image_par_channels(a,b,j)+2*G*Image_per_channels(a,b,j);
end
end
end
I think I could speed it up introducing (:,:,j) instead of specifying a and b. But still a for loop. I am trying to use cellfun without any success due to my lack of expertise. Could you please give me a hand?
I would really appreciate it.
Many thanks and have a nice day!
Y
I believe you could do something like
Image_par_channels=zeros(128,128,256);
Image_per_channels=zeros(128,128,256);
Image_tot_channels=zeros(128,128,256);
for i=1:numfiles
Image_par_channels = Image_par_channels + Image_cell_par_per{i,1};
Image_per_channels = Image_per_channels + Image_cell_par_per{i,2};
end
Image_tot_channels = Image_par_channels + 2*G*Image_per_channels;
I haven't work with matlab in a long time, but I seem to recall you can do something like this. g is a constant.
EDIT:
Removed the +=. Incremental assignment is not an operator available in matlab. You should also note that Image_tot_channels can be build directly in the loop, if you don't need the other two variables later.
I have a linear system to solve, written as Ax=b
A is a 175 by 175 symmetric square, with ones at it's diagonal (i.e., aii=1), and other entries ranges from 0 to 1(i.e., 0
A is very ill conditioned, and not positive definite, its rank is 162 and its condition number is 3.5869e+16
I spent several days to solve this in MATLAB, I've tried almost every method I can find, including \, pcg, bicg, bicgstab, bicgstabl, cgs, gmres, lsqr, minres, qmr, symmlq, tfqmr
These methods gave me some solutions. But I don't know how to trust them, or which solution to trust. Is there a criteria to determine?
I'll appreciate someone who can give me a solution that I can trust.
Thanks!
A and b are stored in .mat files and can be download from dropbox links:
https://www.dropbox.com/s/s6xlbq68juqs6xi/A.mat?dl=0
https://www.dropbox.com/s/pxl0hdup20hf2lr/b.mat?dl=0
use like this:
load('A.mat');
load('b.mat');
x = A\b;
Not sure if this will help, but give it a go:
Tikhonov regularization
Basically, when the following is hard to compute due to ill-conditiones:
You minimize the following instead
Being \Gamma generally the identity matrix.
In the end, you get the following equation for x:
To add to that, you will generally want to add an "hyperparameter", to control how much you regularize the problem. So \Gamma instead of being just the identity matrix, it will be a number (i.e. 0.001) multiplied by the identity matrix of size(A).
The last equation should be straightforward to try in Matlab. Give it a go.
NOTE: this is not the answer. Actually probably there is no unique answer to solving an ill posed problem. This is just a way to go.
Pretty simple, and I thought I knew what I was doing, but apparently not. Anyways.
I need to find the first element in a vector that is less than a specific value. Here is the code that I have been using:
t = 0:0.01:5;
u = ((2)*exp(-10.*t).*cos((4.*sqrt(6)).*t) + ((5)./sqrt(6)).*exp(-10.*t).*sin((4.*sqrt(6)).*t));
for a = 1:size(u)
if u(a) < (0.05)
disp(a)
break
end
end
The value that I'm trying to find is the first element less than 0.05, however, when I run my code, I don't get anything.
What could I be doing wrong?
Thanks!
#user2994291 has correctly pointed out where your loop based solution is going wrong (+1).
However I would also add that what you are trying to do can simply be accomplished by:
find(u < 0.05, 1, 'first')
Technically, the third input is not necessary - you could just use:
find(u < 0.05, 1)
However, I seem to recall reading at some point that find will work faster if you provide the third input.
The upper bound of your for loop is probably equal to 1.
In your case, u is a row vector (can't say 100% for sure in MATLAB as I only have access to GNU Octave right now), but calling size(u) will probably give back [1 501] as an answer. Your for-loop will select 1 as upper bound.
Try replacing size(u) with size(u,2) or, even better, with length(u). I get a = 24 as an answer.
Edit:
from your questions I assume you are a MATLAB beginner, therefore I strongly advise you to look into the built-in debugger (you can add breakpoints by clicking on the left vertical bar next to the desired line of code), this would have helped you identify the error with ease and will save you a lot of time in the future.
I am having trouble plotting a patch on a figure with transparency and timestamp index varying in datenum range of 735k, which is more or less the datenum you will acquire for the actual dates.
The following example illustrate my issue:
k=[735172.912437674 735172.941375799 0 7830];
figure;
axis(k);
p=patch([k(1) k(1) k(2) k(2)], [k(3) k(4) k(4) k(3)],[0 0 0 0],'r', 'FaceAlpha', 0.1);
Are you able to reproduce this bug? Any suggestions? I've seen this other post and searched about it, but no clue. I wouldn't like to remove the minimum from my time stamps, I would have to adapt all the code just because of this x(
I was learning about this other question and they led me to study a bit the new Handle Graphics that may finally appear this or next year (?) — looks like trying to discovering trends at financial business, guess when this will go out, if it will haha. It may be used by adding -hgVersion 2 option at your matlab launch. And by doing so, the resulting figure looks like this \o/:
Great… but beware that there are some differences on the usage from the HG1 to the HG2 that may lead your code not to work (specially if you are using listeners) but if you want to explore this even more undocumented code add an if in your codes where it breaks:
if feature('UseHG2')
% HG2 approach
else
% HG1 approach
end
One good thing it's that at least we can get the meta.class from the handles, so we can explore them easier x) and the usage is quite more intuitive with the matlab oop development way.
I have been happily using MATLAB to solve some project Euler problems. Yesterday, I wrote some code to solve one of these problems (14). When I write code containing long loops I always test the code by running it with short loops. If it runs fine and it does what it's supposed to do I assume this will also be the case when the length of the loop is longer.
This assumption turned out to be wrong. While executing the code below, MATLAB ran out of memory somewhere around the 75000th iteration.
c=1;
e=1000000;
for s=c:e
n=s;
t=1;
while n>1
a(s,t)=n;
if mod(n,2) == 0
n=n/2;
else
n=3*n+1;
end
a(s,t+1)=n;
t=t+1;
end
end
What can I do to prevent this from happening? Do I need to clear variables or free up memory somewhere in the process? Will saving the resulting matrix a to the hard drive help?
Here is the solution, staying as close as possible to your code (which is very close, the main difference is that you only need a 1D matrix):
c=1;
e=1000000;
a=zeros(e,1);
for s=c:e
n=s;
t=1;
while n>1
if mod(n,2) == 0
n=n/2;
else
n=3*n+1;
end
t=t+1;
end
a(s)=t;
end
[f g]=max(a);
This takes a few seconds (note the preallocation), and the result g unlocks the Euler 14 door.
Simply put, there's not enough memory to hold the matrix a.
Why are you making a two-dimensional matrix here anyway? You're storing information that you can compute just as fast as looking it up.
There's a much better thing to memoize here.
EDIT: Looking again, you're not even using the stuff you put in that matrix! Why are you bothering to create it?
The code appears to be storing every sequence in a different row of a matrix. The number of columns of that matrix will be equal to the length of the longest sequence currently found. This means that a sequence of two numbers will be padded with a bunch of right hand zeros.
I am sure you can see how this is incredibly inefficient. That may be the point of the exercise, or it will be for you in this implementation.
Better is to keep a variable like "Seed of longest solution found" which would store the seed for the longest solution. I would also keep a "length of longest solution found" keep the length. As you try every new seed, if it wins the title of longest, then update those variables.
This will keep only what you need in memory.
Short Answer:Use a 2d sparse matrix instead.
Long Answer: http://www.mathworks.com/access/helpdesk/help/techdoc/ref/sparse.html