Hello I'm new to Matlab.
I've written this script :
k2=2*pi();
z1 = 1;
z2 = 2;
z3 = 4;
for l = linspace(0,1,11)
A = [ -1 1 1 0 ; 1 z1/z2 -z1/z2 0 ; 0 exp(-i*k2*l) exp(i*k2*l) -1 ; 0 exp(- i*k2*l) -exp(i*k2*l) -z2/z3];
B = [ 1 ; 1 ; 0 ; 0];
D = inv(A);
C = mtimes(D,B) ;
display(C)
r = C(1,1); % this is supposed to set r = the 1,1 element in the matrix C
t = C(1,4); % see above
end
My idea for taking the values of r and t from C didnt appear to work. How can I do this properly?
Also I want to plot a graph of |r|,|t|, arg(r) and arg(t) for each value of l, my for loop overwrites the values of r and t? how can I either plot one point per loop or make r and t assign the new values so that they become lists of data.
Thanks a lot!
Matlab sets the first dimension of a matrix as row number (i.e. y position).
So you want t=C(4, 1), as you should see that the size of C is 4x1. As a note Matlab is quite good at suppressing singleton dimensions so you could do also do C(1) and C(4).
For your second point you want to set a particular element of r and t in each loop. This is the same as when you access at particular element of C when setting the values. For your case you can use the index l to determine the element. Remembering that in matlab arrays start at element 1 (not 0 as in many other languages). So you want something like r(l+1)=C(1); (or change l to start at 1).
In the more general case if you are not looping over an integer for some reason you may need to create a separate counter variable which you increase in the loop. Also it is good practice to preallocate such arrays when the size is known beforehand, often by r=zeros(11, 1) or similar (note: zeros(11) is an 11x11 matrix). This isn't significant in this case but can drastically increase execution time for large multi-dimensional arrays so is a good practice.
Related
I got a large vector vec of size n x 1 (with n a huge number). All values in vec are 0 or 1. The vector consists out of blocks of size m and in each block, only one of these two values shall be included.
So as to ensure this condition, I loop over each start element within an block and copy this value into all the entries of the respective block:
for i = 1:m:n
if vec(i) == 1
vec(i:i+m-1) = 1;
else
vec(i:i+m-1) = 0;
end
end
How can I achieve this without loosing a loop so as to better use MATLABs vectorization parallelization? Thanks in advance
You can use repelem:
vec = repelem(vec(1:m:n), m);
I have a Matlab time series data set, which consist of a signal that can only be 1 or 0. How can I get rid of all the values except for the changing ones?
For example:
1
1
1
0
1
0
0
0
should ideally result in
1
0
1
0
while keeping the correct time values as well of course.
Thing is, that I need to find the frequency of the signal. The time should be measured from 0->1 to the next time 0->1 occurs. The smallest time / highest frequency is what I need in the end.
Thanks!
You can use the getsamples method to get a time series which contains a subset of the original samples. Remains to identify the indices where the time series has changed, for this purpose you can use diff and logical indexing:
ts = timeseries([1 1 1 0 1 0 0 0],1:8)
ts.getsamples([true;squeeze(diff(ts.Data)) ~= 0])
A simple and clever call to to diff should be sufficient:
>> A = [1; 1; 1; 0; 1; 0; 0; 0];
>> B = A(diff([-Inf; A]) ~= 0)
B =
1
0
1
0
The code is quite simple. diff finds pairs of differences in an array. Concretely, given an array A, the output is of the following structure:
B = [A(2) - A(1), A(3) - A(2), ..., A(N) - A(N-1)];
N is the total length of the signal. This results in a N-1 length signal. As such, a trick that you can use is to append the array A with -Inf (or some high non-zero value) so that when you find the difference between the first element of this appended array and the actual first element of the true array, you will get some non-zero change. That is registered with diff([-Inf; A]). The next thing you'll want is to check is to see where the differences are non-zero. Whenever there is a non-zero difference, that is a position that you want to keep because there has been a change that occurred. This produces a logical array and so the last step is to use this to index into your array A and thus get the result.
This only extracts out the signal you need however. If you'd like to extract the time in between unique elements, supposing you had some time vector t that was as long as your signal stored in A. You would first record the logical vector in a separate variable, then index into both your time array and the signal array to extract out what you need (original idea from user dfri):
ind = diff([-Inf; A]) ~= 0;
times = t(ind);
B = A(ind);
You can make use of diff and logical to save the results as a logical array, used as a subsequent index filter in your data (say t for time and y for boolean values ))
%// example
t = 0:0.01:0.07;
y = [1,1,1,0,1,0,0,0];
%// find indices to keep
keep = [true logical(diff(y))];
%// truncated data
tTrunc = t(keep)
yTrunc = y(keep)
with the results for the example as follows
tTrunc =
0 0.0300 0.0400 0.0500
yTrunc =
1 0 1 0
So I am very new to Matlab and I have been tasked with implementing LU factorization. I have to do it recursively because we are not allowed to use for loops in our code, and recursion will give us optimal marks. I have written this code. The code works for the first step, and does what it is supposed to, but for the next two steps, the matrix is not modified at all. I have this code:
function[L, U] = myLU(A,B, pos)
%A = Mtrix that becomes U
%B = Matrix that becomes L
tmp_L = B;
[x,y] = size(A);
if pos > x
L = B;
U = A;
return
else
pos %<-- just to see if it iterates through everything
[tmp_U,tmp_L] = elimMat(A,pos);
myLU(tmp_U,tmp_L, pos+1);
end
L = tmp_L;
U = tmp_U;
end
I where elimMat(A, pos) returns the elimination matrix for column pos. as well as another matrix, which will end up being the matrix of multipliers. What i tried doing is then finding the LU factorization of this matrix A. since elimMat returns L and U(this works, if i do it manually it works), i had to make a function that allows me to do it automatically without using a for loop. I thought i would do it recursively. What i ended up doing is adding another variable B to the function so that i can store intermediate values of my matrix obtained in each step and put it all together later.
So here is my question. Am i implementing the recursion wrong? and if i am how can i fix it? The other thing i wanted to ask is how can i implement this so i do not need variable B as an additional imput, and only use the existing variables, or variables previously defined, to find the solution? I would really like only two inputs in my function: The matrix name and the starting index.
here is elimMat if if helps:
function [M,L] = elimMat(A,k)
%find the size of the matrix
[x,y] = size(A);
tmp_mat = zeros(x,y);
%M = The current matrix we are working on for Elimination -> going to
%become U.
%L = The L part of the matrix we are working on. Contains all the
%multipliers. This is going to be our L matrix.
for i = 1:x
mult = A(i,k)/A(k,k);
if i > k
tmp_mat(i,k) = mult;
P = A(k,:)*mult;
A(i,:) = A(i,:)-P;
elseif i == k
tmp_mat(k,k) = 1;
end
end
M = A;
L = tmp_mat;
end
thanks for any feedback you can provide.
Here is the output: WHAT I GET VS what it should be:
[U = VS [U =
1 2 2 1 2 2
0 -4 -6 0 -4 -6
0 -2 -4] 0 0 2
L = VS [L=
1 0 0 1 0 0
4 0 0 4 1 0
4 0 0] 4 0.5 1
As you can see only the first column is changed
You forgot to catch the output of your recursive call:
[tmp_L, tmp_U] = myLU(tmp_U,tmp_L, pos+1);
Matlab passes variables by value, so a function cannot change its input variable itself (ok, it can, but it's tricky and unsafe).
Your original version didn't return the updated matrices, so the outermost function call encountered the myLU() call, let the recursion unfold and finish, and then went on to use tmp_L and tmp_U as returned from the very first call to elimMAT(A,1).
Note that you might want to standardize your functions such that they return U and L in the same order to avoid confusion.
I have a 512X512 size of image and I have made 4x4 overlapping blocks for the entire image.How can i count the number of overlapping blocks and save it in an Array in matlab.
I have done like below for 4x4 overlapping blocks. Now how to count the no of blocks and store it using an Array.
[e f] = size(outImg);
l=0;
for i=2:e-2
for j=2:f-2
H =double(outImg((i-1:i+2),(j-1:j+2)));
eval(['out_' num2str(l) '=H']);
l=l+1
end;
end;
From what I understand the question, you want to know how many blocks of 4x4 can fit in the image, and then store them.
Calculating the number of blocks is trivial, in the code that you give as example, l is the number of element counted. Of course, that its value is deterministic (determined by f and e). No need to loop over them to get the value of the count.
count = (f-3)*(e-3);
If you want to save the values in an array (assuming that you mean here a matrix and not a cell array) you need to decide how to represent it, you can store it as a 4D e-3 x f-3 x 4 x 4 matrix (as #Steffen suggested), or as a 3D 4 x 4 x count matrix, I think that the later is more intuitive. In any case you should assign the memory for the matrix in advance and not on the fly:
[e f] = size(outImg);
count = (f-3)*(e-3);
outMat = zeros(4,4,count); % assign the memory for the matrix
l = 0;
for i=2:e-2
for j=2:f-2
l = l + 1;
outMat(:,:,l) = double(outImg((i-1:i+2),(j-1:j+2)));
end;
end;
The number of blocks is stored as both count and l, but calculating count in advance allows to assign the needed memory in advance, the i block is stored as outMat(:,:,i).
An implementation using the 4D matrix would be:
[e f] = size(outImg);
count = (f-3)*(e-3);
outMat = zeros((f-3),(e-3),4,4); % assign the memory for the matrix
for i=2:e-2
for j=2:f-2
outMat(i,j,:,:) = double(outImg((i-1:i+2),(j-1:j+2)));
end;
end;
In this case, l isn't needed and each block (indexed i,j) is located at outMat(i,j,:,:)
Regarding cell array vs. a matrix, since a matrix requires a continuous place in the memory, you may want to consider using a cell array instead of a matrix. A 512x512x4 matrix of doubles requires (assuming 8 Byte representation) 8MB (512*512*8*4 = 8*1024*1024). If the dimensions were bigger, or if you are strapped for (continuous) memory a cell array may be a better solution. You can read more about the difference at Difference between cell and matrix in matlab?.
The implementation would be very similar.
[e f] = size(outImg);
count = (f-3)*(e-3);
outArray = cell(1,count);
l = 0;
for i=2:e-2
for j=2:f-2
l = l + 1;
outArray{1,l} = double(outImg((i-1:i+2),(j-1:j+2)));
end;
end;
The answer is very simple. Each loop iteration will access 1 overlapping block in your image. All you have to do is count how many times the loop iterates, which is ((e-2) - 2 + 1) x ((f - 2) - 2 + 1) = (e - 3) x (f - 3). There's no need to keep a loop iteration variable.
Minor note. Under no circumstances should you use eval unless absolutely necessary. The MATLAB gods will smite any offenders swiftly. That code to assign a new variable to each increment of l is absolutely unnecessary. If you were to have 10000 overlapping blocks, you would have 10000 variables. You can just simply take a look at l at the end and this would tell you how many overlapping blocks you have.
Remove that line of code. WE HATES IT. IT BURNS US.
See this post by Loren Shure for more details on why using eval is bad.
For the 1-D case, the correct formula to estimate the total number of overlapping blocks is:
(#Of_blocks) = (#Image_width - #Block_width) / (#Block_width - #Block_overlap) + 1
If the result is not integer you just take the floor of #Of_blocks, this means that the whole interval can't be covered using that setup.
I have a 3x3 matrix, A. I also compute a value, g, as the maximum eigen value of A. I am trying to change the element A(3,3) = 0 for all values from zero to one in 0.10 increments and then update g for each of the values. I'd like all of the other matrix elements to remain the same.
I thought a for loop would be the way to do this, but I do not know how to update only one element in a matrix without storing this update as one increasingly larger matrix. If I call the element at A(3,3) = p (thereby creating a new matrix Atry) I am able (below) to get all of the values from 0 to 1 that I desired. I do not know how to update Atry to get all of the values of g that I desire. The state of the code now will give me the same value of g for all iterations, as expected, as I do not know how to to update Atry with the different values of p to then compute the values for g.
Any suggestions on how to do this or suggestions for jargon or phrases for me to web search would be appreciated.
A = [1 1 1; 2 2 2; 3 3 0];
g = max(eig(A));
% This below is what I attempted to achieve my solution
clear all
p(1) = 0;
Atry = [1 1 1; 2 2 2; 3 3 p];
g(1) = max(eig(Atry));
for i=1:100;
p(i+1) = p(i)+ 0.01;
% this makes a one giant matrix, not many
%Atry(:,i+1) = Atry(:,i);
g(i+1) = max(eig(Atry));
end
This will also accomplish what you want to do:
A = #(x) [1 1 1; 2 2 2; 3 3 x];
p = 0:0.01:1;
g = arrayfun(#(x) eigs(A(x),1), p);
Breakdown:
Define A as an anonymous function. This means that the command A(x) will return your matrix A with the (3,3) element equal to x.
Define all steps you want to take in vector p
Then "loop" through all elements in p by using arrayfun instead of an actual loop.
The function looped over by arrayfun is not max(eig(A)) but eigs(A,1), i.e., the 1 largest eigenvalue. The result will be the same, but the algorithm used by eigs is more suited for your type of problem -- instead of computing all eigenvalues and then only using the maximum one, you only compute the maximum one. Needless to say, this is much faster.
First, you say 0.1 increments in the text of your question, but your code suggests you are actually interested in 0.01 increments? I'm going to operate under the assumption you mean 0.01 increments.
Now, with that out of the way, let me state what I believe you are after given my interpretation of your question. You want to iterate over the matrix A, where for each iteration you increase A(3, 3) by 0.01. Given that you want all values from 0 to 1, this implies 101 iterations. For each iteration, you want to calculate the maximum eigenvalue of A, and store all these eigenvalues in some vector (which I will call gVec). If this is correct, then I believe you just want the following:
% Specify the "Current" A
CurA = [1 1 1; 2 2 2; 3 3 0];
% Pre-allocate the values we want to iterate over for element (3, 3)
A33Vec = (0:0.01:1)';
% Pre-allocate a vector to store the maximum eigenvalues
gVec = NaN * ones(length(A33Vec), 1);
% Loop over A33Vec
for i = 1:1:length(A33Vec)
% Obtain the version of A that we want for the current i
CurA(3, 3) = A33Vec(i);
% Obtain the maximum eigen value of the current A, and store in gVec
gVec(i, 1) = max(eig(CurA));
end
EDIT: Probably best to paste this code into your matlab editor. The stack-overflow automatic text highlighting hasn't done it any favors :-)
EDIT: Go with Rody's solution (+1) - it is much better!