I want to do a simple loop in matlab. For example, v=[1,2,3], I want to get v(1)=v*2+[1,2,5];V(2)=2*v(1)+[1,2,5], and so on. Then v(1)=[3,6,11]
I have tried:
x=[1,2,3];
y=x;
for j=1:5
y(j+1)=2*y(j)+[1,2,5];
end
but is wrong.
How can I solve it?
How about:
N = 100;
B = rand(N, N);
A = B / norm(B); % substitute norm of your choice
Related
I'm trying to simulate some random variables Y such that P(Y=1)=P(y=-1)=0.5, and X_n = sum of Y_i (i from 1 to n). I want to use matlab to simulate X_n and plot it versus different n's, where n = 1,2,3,...100. Here is my matlab code:
N = 100;
for M = 1:N
y_i = randi([-1 1], M, 1);
X_n = sum(y_i);
end
plot(M, X_n)
But my plot looks like this, can someone help me fix it? Is there something wrong with my code? Thank you.
Seems like somebody provided you with the right answer already but let me explain and how i would go about it. The only thing you're doing wrong is about the indexing. Try this.
N = 100; % sets your maximum
for M = 1:N % loops from 1 - N
y_i = randi([-1 1], M, 1); % your formula
X(M) = sum(y_i); % stores your data in vectors with increasing index from 1 - 100
end
index = 1:N % generates a vector 1-100 to serve as indexes
plot(index, X) % plots each point of X a corresponding index
I have an N by 2 matrix called r (N is very large). r is the position of points in 2D. I searched for the best-optimized way of calculating distance between point. I find that dist function is the best on in less time-consuming if one doesn't try to change it to a square matrix. I wonder if I write
D= pdist(r, 'euclidean');
When I need distance between particle i and j, what is the best way to find it using D vector? I do not really any way without using if.
I know that I can do it by
if (i < j)
D((i–1)*(m–i/2)+j–i)
end
But as N is very large, this is not efficient. Could anyone help me, please?
I'm using ii and jj as row and column indices into the hypohetical distance matrix M = squareform(D) of size and N. The result is ind, such that D(ind) equals M(ii,jj).
t = sort([ii, jj]); % temporary variable
ii = t(2); % maximum of ii and jj
jj = t(1); % minimum of ii and jj
t = N-1:-1:1;
ind = sum(t(1:jj-1)) + ii - jj;
Suppose there are a matrix of three row vectors
A = [1,2;
1,3;
2,3]
I would like to create a new matrix B which draws two vectors from A with repetitions, and there are 3^2 possible combinations. Some simple implementation is as follows:
For i = 1:3
c = A(i,:);
for j=1:3
d = A(j,:);
B = [c;d];
end
end
But, in general, if I need to choose k vectors from n vectors, what is the more general way to write such loop? It's difficult to continue write loop using i, j, ... I guess. Thanks!
For sampling at random, matlab has randsample:
rowIdx = randsample( size(A,1), k, true );
B = A(rowIdx,:);
You can just use randi for this to pick k uniformly distributed numbers in the range 1:n (with replacement)
k = 2;
n = size(A,1);
rowIdx = randi(n,k)
B = A(rowIdx,:)
Thanks for all previous suggestions.
In the end I figure out what I want is called permutations with repetitions. The matlab function permun from file exchange solves my problem.
I am trying to use meshgrid in Matlab together with Chebfun to get rid of double for loops. I first define a quasi-matrix of N functions,
%Define functions of type Chebfun
N = 10; %number of functions
x = chebfun('x', [0 8]); %Domain
psi = [];
for i = 1:N
psi = [psi sin(i.*pi.*x./8)];
end
A sample calculation would be to compute the double sum $\sum_{i,j=1}^10 psi(:,i).*psi(:,j)$. I can achieve this using two for loops in Matlab,
h = 0;
for i = 1:N
for j = 1:N
h = h + psi(:,i).*psi(:,j);
end
end
I then tried to use meshgrid to vectorize in the following way:
[i j] = meshgrid(1:N,1:N);
h = psi(:,i).*psi(:,j);
I get the error "Column index must be a vector of integers". How can I overcome this issue so that I can get rid of my double for loops and make my code a bit more efficient?
BTW, Chebfun is not part of native MATLAB and you have to download it in order to run your code: http://www.chebfun.org/. However, that shouldn't affect how I answer your question.
Basically, psi is a N column matrix and it is your desire to add up products of all combinations of pairs of columns in psi. You have the right idea with meshgrid, but what you should do instead is unroll the 2D matrix of coordinates for both i and j so that they're single vectors. You'd then use this and create two N^2 column matrices that is in such a way where each column corresponds to that exact column numbers specified from i and j sampled from psi. You'd then do an element-wise multiplication between these two matrices and sum across all of the columns for each row. BTW, I'm going to use ii and jj as variables from the output of meshgrid instead of i and j. Those variables are reserved for the complex number in MATLAB and I don't want to overshadow those unintentionally.
Something like this:
%// Your code
N = 10; %number of functions
x = chebfun('x', [0 8]); %Domain
psi = [];
for i = 1:N
psi = [psi sin(i.*pi.*x./8)];
end
%// New code
[ii,jj] = meshgrid(1:N, 1:N);
%// Create two matrices and sum
matrixA = psi(:, ii(:));
matrixB = psi(:, jj(:));
h = sum(matrixA.*matrixB, 2);
If you want to do away with the temporary variables, you can do it in one statement after calling meshgrid:
h = sum(psi(:, ii(:)).*psi(:, jj(:)), 2);
I don't have Chebfun installed, but we can verify that this calculates what we need with a simple example:
rng(123);
N = 10;
psi = randi(20, N, N);
Running this code with the above more efficient solution gives us:
>> h
h =
8100
17161
10816
12100
14641
9216
10000
8649
9025
11664
Also, running the above double for loop code also gives us:
>> h
h =
8100
17161
10816
12100
14641
9216
10000
8649
9025
11664
If you want to be absolutely sure, we can have both codes run with the outputs as separate variables, then check if they're equal:
%// Setup
rng(123);
N = 10;
psi = randi(20, N, N);
%// Old code
h = 0;
for i = 1:N
for j = 1:N
h = h + psi(:,i).*psi(:,j);
end
end
%// New code
[ii,jj] = meshgrid(1:N, 1:N);
hnew = sum(psi(:, ii(:)).*psi(:, jj(:)), 2);
%// Check for equality
eql = isequal(h, hnew);
eql checks if both variables are equal, and we do get them as such:
>> eql
eql =
1
I am using the matlab code from this book: http://books.google.com/books/about/Probability_Markov_chains_queues_and_sim.html?id=HdAQdzAjl60C
Here is the Code:
function [pi] = GE(Q)
A = Q';
n = size(A);
for i=1:n-1
for j=i+1:n
A(j,i) = -A(j,i)/A(i,i);
end
for j =i+1:n
for k=i+1:n
A(j,k) = A(j,k)+ A(j,i) * A(i,k);
end
end
end
x(n) = 1;
for i = n-1:-1:1
for j= i+1:n
x(i) = x(i) + A(i,j)*x(j);
end
x(i) = -x(i)/A(i,i);
end
pi = x/norm(x,1);
Is there a faster code that I am not aware of? I am calling this functions millions of times, and it takes too much time.
MATLAB has a whole set of built-in linear algebra routines - type help slash, help lu or help chol to get started with a few of the common ways to efficiently solve linear equations in MATLAB.
Under the hood these functions are generally calling optimised LAPACK/BLAS library routines, which are generally the fastest way to do linear algebra in any programming language. Compared with a "slow" language like MATLAB it would not be unexpected if they were orders of magnitude faster than an m-file implementation.
Hope this helps.
Unless you are specifically looking to implement your own, you should use Matlab's backslash operator (mldivide) or, if you want the factors, lu. Note that mldivide can do more than Gaussian elimination (e.g., it does linear least squares, when appropriate).
The algorithms used by mldivide and lu are from C and Fortran libraries, and your own implementation in Matlab will never be as fast. If, however, you are determined to use your own implementation and want it to be faster, one option is to look for ways to vectorize your implementation (maybe start here).
One other thing to note: the implementation from the question does not do any pivoting, so its numerical stability will generally be worse than an implementation that does pivoting, and it will even fail for some nonsingular matrices.
Different variants of Gaussian elimination exist, but they are all O(n3) algorithms. If any one approach is better than another depends on your particular situation and is something you would need to investigate more.
function x = naiv_gauss(A,b);
n = length(b); x = zeros(n,1);
for k=1:n-1 % forward elimination
for i=k+1:n
xmult = A(i,k)/A(k,k);
for j=k+1:n
A(i,j) = A(i,j)-xmult*A(k,j);
end
b(i) = b(i)-xmult*b(k);
end
end
% back substitution
x(n) = b(n)/A(n,n);
for i=n-1:-1:1
sum = b(i);
for j=i+1:n
sum = sum-A(i,j)*x(j);
end
x(i) = sum/A(i,i);
end
end
Let's assume Ax=d
Where A and d are known matrices.
We want to represent "A" as "LU" using "LU decomposition" function embedded in matlab thus:
LUx = d
This can be done in matlab following:
[L,U] = lu(A)
which in terms returns an upper triangular matrix in U and a permuted lower triangular matrix in L such that A = LU. Return value L is a product of lower triangular and permutation matrices. (https://www.mathworks.com/help/matlab/ref/lu.html)
Then if we assume Ly = d where y=Ux.
Since x is Unknown, thus y is unknown too, by knowing y we find x as follows:
y=L\d;
x=U\y
and the solution is stored in x.
This is the simplest way to solve system of linear equations providing that the matrices are not singular (i.e. the determinant of matrix A and d is not zero), otherwise, the quality of the solution would not be as good as expected and might yield wrong results.
if the matrices are singular thus cannot be inversed, another method should be used to solve the system of the linear equations.
For the naive approach (aka without row swapping) for an n by n matrix:
function A = naiveGauss(A)
% find's the size
n = size(A);
n = n(1);
B = zeros(n,1);
% We have 3 steps for a 4x4 matrix so we have
% n-1 steps for an nxn matrix
for k = 1 : n-1
for i = k+1 : n
% step 1: Create multiples that would make the top left 1
% printf("multi = %d / %d\n", A(i,k), A(k,k), A(i,k)/A(k,k) )
for j = k : n
A(i,j) = A(i,j) - (A(i,k)/A(k,k)) * A(k,j);
end
B(i) = B(i) - (A(i,k)/A(k,k)) * B(k);
end
end
function Sol = GaussianElimination(A,b)
[i,j] = size(A);
for j = 1:i-1
for i = j+1:i
Sol(i,j) = Sol(i,:) -( Sol(i,j)/(Sol(j,j)*Sol(j,:)));
end
end
disp(Sol);
end
I think you can use the matlab function rref:
[R,jb] = rref(A,tol)
It produces a matrix in reduced row echelon form.
In my case it wasn't the fastest solution.
The solution below was faster in my case by about 30 percent.
function C = gauss_elimination(A,B)
i = 1; % loop variable
X = [ A B ];
[ nX mX ] = size( X); % determining the size of matrix
while i <= nX % start of loop
if X(i,i) == 0 % checking if the diagonal elements are zero or not
disp('Diagonal element zero') % displaying the result if there exists zero
return
end
X = elimination(X,i,i); % proceeding forward if diagonal elements are non-zero
i = i +1;
end
C = X(:,mX);
function X = elimination(X,i,j)
% Pivoting (i,j) element of matrix X and eliminating other column
% elements to zero
[ nX mX ] = size( X);
a = X(i,j);
X(i,:) = X(i,:)/a;
for k = 1:nX % loop to find triangular form
if k == i
continue
end
X(k,:) = X(k,:) - X(i,:)*X(k,j);
end