In matlab usually we add a header using fprintf command.
This is a problem when the size of the table depends on the input and when it exceeds a certain range (more than total number of column able to be presented in the command window).
When this occurs the header which we specified previously using the fprintf command will not be compatible with the current output data.
I would like to know is there a way like adding a character string into the 1st row of the matrix during some kind of iteration process. I had tried hardly but still can't find a proper way to solve this issue.
Or it is actually cannot be done in matlab for this purpose.
Eg
clear;clc
A = [2 8 3 1;0 2 -1 4;7 -2 1 2;-1 0 5 2]
B = [-2;4;3;5]
Es = 1e-5
n = length(B);
x = zeros(n,1);
Ea = ones(n,1);
iter = 0;
while max(Ea) >= Es
if iter <= 30
iter = iter + 1;
x_old = x;
for i = 1:n
j = 1:n;
j(i) = [];
x_cal = x;
x_cal(i) = [];
x(i) = (B(i) - sum(A(i,j) * x_cal)) / A(i,i);
end
else
break
end
x_ans(:,iter) = x;
Ea(:,iter) =abs(( x - x_old) ./ x);
end
result = [1:iter; x_ans; Ea]'
for the coding above..how could I add the heading like iteration for 1st column, x1...x2...x3..xn for nth column and error x1..error x2..error xn for another n column. I would like to make this heading can be automated generated based on the input matrix
If the size of the table depends on the input, use a cell array, using c = cell(...).
In each iteration simply call c{i,j} instead of c[i,j].
Related
I want to compute the above summation, for a given 'x'. The summation is to be carried out over a block of lengths specified by an array , for example block_length = [5 4 3]. The summation is carried as follows: from -5 to 5 across one dimension, -4 to 4 in the second dimension and -3 to 3 in the last dimension.
The pseudo code will be something like this:
sum = 0;
for i = -5:5
for j = -4:4
for k = -3:3
vec = [i j k];
tv = vec * vec';
sum = sum + 1/(1+tv)*cos(2*pi*x*vec'));
end
end
end
The problem is that I want to find the sum when the number of dimensions are not known ahead of time, using some kind of variable nested loops hopefully. Matlab uses combvec, but it returns all possible combinations of vectors, which is not required as we only compute the sum. When there are many dimensions, combvec returning all combinations is not feasible memory wise.
Appreciate any ideas towards solutions.
PS: I want to do this at high number of dimensions, for example 650, as in machine learning.
Based on https://www.mathworks.com/matlabcentral/answers/345551-function-with-varying-number-of-for-loops I came up with the following code (I haven't tested it for very large number of indices!):
function sum = fun(x, block_length)
sum = 0;
n = numel(block_length); % Number of loops
vec = -ones(1, n) .* block_length; % Index vector
ready = false;
while ~ready
tv = vec * vec';
sum = sum + 1/(1+tv)*cos(2*pi*x*vec');
% Update the index vector:
ready = true; % Assume that the WHILE loop is ready
for k = 1:n
vec(k) = vec(k) + 1;
if vec(k) <= block_length(k)
ready = false;
break; % v(k) increased successfully, leave "for k" loop
end
vec(k) = -1 * block_length(k); % v(k) reached the limit, reset it
end
end
end
where x and block_length should be both 1-x-n vectors.
The idea is that, instead of using explicitly nested loops, we use a vector of indices.
How good/efficient is this when tackling the suggested use case where block_length can have 650 elements? Not much! Here's a "quick" test using merely 16 dimensions and a [-1, 1] range for the indices:
N = 16; tic; in = 0.1 * ones(1, N); sum = fun(in, ones(size(in))), toc;
which yields an elapsed time of 12.7 seconds on my laptop.
Consider a Matlab matrix B which lists all possible unordered pairs (without repetitions) from [1 2 ... n]. For example, if n=4,
B=[1 2;
1 3;
1 4;
2 3;
2 4;
3 4]
Note that B has size n(n-1)/2 x 2
I want to take a random draw of m rows from B and store them in a matrix C. Continuing the example above, I could do that as
m=2;
C=B(randi([1 size(B,1)],m,1),:);
However, in my actual case, n=371293. Hence, I cannot create B and, then, run the code above to obtain C. This is because storing B would require a huge amount of memory.
Could you advise on how I could proceed to create C, without having to first store B? Comments on a different question suggest to
Draw at random m integers between 1 and n(n-1)/2.
I=randi([1 n*(n-1)/2],m,1);
Use ind2sub to obtain C.
Here, I'm struggling to implement the second step.
Thanks to the comments below, I wrote this
n=4;
m=10;
coord=NaN(m,2);
R= randi([1 n^2],m,1);
for i=1:m
[cr, cc]=ind2sub([n,n],R(i));
if cr>cc
coord(i,1)=cc;
coord(i,2)=cr;
elseif cr<cc
coord(i,1)=cr;
coord(i,2)=cc;
end
end
coord(any(isnan(coord),2),:) = []; %delete NaN rows from coord
I guess there are more efficient ways to implement the same thing.
You can use the function named myind2ind in this post to take random rows of all possible unordered pairs without generating all of them.
function [R , C] = myind2ind(ii, N)
jj = N * (N - 1) / 2 + 1 - ii;
r = (1 + sqrt(8 * jj)) / 2;
R = N -floor(r);
idx_first = (floor(r + 1) .* floor(r)) / 2;
C = idx_first-jj + R + 1;
end
I=randi([1 n*(n-1)/2],m,1);
[C1 C2] = myind2ind (I, n);
If you look at the odds, for i=1:n-1, the number of combinations where the first value is equal to i is (n-i) and the total number of cominations is n*(n-1)/2. You can use this law to generate the first column of C. The values of the second column of C can then be generated randomly as integers uniformly distributed in the range [i+1, n]. Here is a code that performs the desired tasks:
clc; clear all; close all;
% Parameters
n = 371293; m = 10;
% Generation of C
R = rand(m,1);
C = zeros(m,2);
s = 0;
t = n*(n-1)/2;
for i=1:n-1
if (i<n-1)
ind_i = R>=s/t & R<(s+n-i)/t;
else % To avoid rounding errors for n>>1, we impose (s+n-i)=t at the last iteration (R<(s+n-i)/t=1 always true)
ind_i = R>=s/t;
end
C(ind_i,1) = i;
C(ind_i,2) = randi([i+1,n],sum(ind_i),1);
s = s+n-i;
end
% Display
C
Output:
C =
84333 266452
46609 223000
176395 328914
84865 94391
104444 227034
221905 302546
227497 335959
188486 344305
164789 266497
153603 354932
Good luck!
I am trying to implement a simplex algorithm following the rules I was given at my optimization course. The problem is
min c'*x s.t.
Ax = b
x >= 0
All vectors are assumes to be columns, ' denotes the transpose. The algorithm should also return the solution to dual LP. The rules to follow are:
Here, A_J denotes columns from A with indices in J and x_J, x_K denotes elements of vector x with indices in J or K respectively. Vector a_s is column s of matrix A.
Now I do not understand how this algorithm takes care of condition x >= 0, but I decided to give it a try and follow it step by step. I used Matlab for this and got the following code.
X = zeros(n, 1);
Y = zeros(m, 1);
% i. Choose starting basis J and K = {1,2,...,n} \ J
J = [4 5 6] % for our problem
K = setdiff(1:n, J)
% this while is for goto
while 1
% ii. Solve system A_J*\bar{x}_J = b.
xbar = A(:,J) \ b
% iii. Calculate value of criterion function with respect to current x_J.
fval = c(J)' * xbar
% iv. Calculate dual solution y from A_J^T*y = c_J.
y = A(:,J)' \ c(J)
% v. Calculate \bar{c}^T = c_K^T - u^T A_K. If \bar{c}^T >= 0, we have
% found the optimal solution. If not, select the smallest s \in K, such
% that c_s < 0. Variable x_s enters basis.
cbar = c(K)' - c(J)' * inv(A(:,J)) * A(:,K)
cbar = cbar'
tmp = findnegative(cbar)
if tmp == -1 % we have found the optimal solution since cbar >= 0
X(J) = xbar;
Y = y;
FVAL = fval;
return
end
s = findnegative(c, K) %x_s enters basis
% vi. Solve system A_J*\bar{a} = a_s. If \bar{a} <= 0, then the problem is
% unbounded.
abar = A(:,J) \ A(:,s)
if findpositive(abar) == -1 % we failed to find positive number
disp('The problem is unbounded.')
return;
end
% vii. Calculate v = \bar{x}_J / \bar{a} and find the smallest rho \in J,
% such that v_rho > 0. Variable x_rho exits basis.
v = xbar ./ abar
rho = J(findpositive(v))
% viii. Update J and K and goto ii.
J = setdiff(J, rho)
J = union(J, s)
K = setdiff(K, s)
K = union(K, rho)
end
Functions findpositive(x) and findnegative(x, S) return the first index of positive or negative value in x. S is the set of indices, over which we look at. If S is omitted, whole vector is checked. Semicolons are omitted for debugging purposes.
The problem I tested this code on is
c = [-3 -1 -3 zeros(1,3)];
A = [2 1 1; 1 2 3; 2 2 1];
A = [A eye(3)];
b = [2; 5; 6];
The reason for zeros(1,3) and eye(3) is that the problem is inequalities and we need slack variables. I have set starting basis to [4 5 6] because the notes say that starting basis should be set to slack variables.
Now, what happens during execution is that on first run of while, variable with index 1 enters basis (in Matlab, indices go from 1 on) and 4 exits it and that is reasonable. On the second run, 2 enters the basis (since it is the smallest index such that c(idx) < 0 and 1 leaves it. But now on the next iteration, 1 enters basis again and I understand why it enters, because it is the smallest index, such that c(idx) < 0. But here the looping starts. I assume that should not have happened, but following the rules I cannot see how to prevent this.
I guess that there has to be something wrong with my interpretation of the notes but I just cannot see where I am wrong. I also remember that when we solved LP on the paper, we were updating our subjective function on each go, since when a variable entered basis, we removed it from the subjective function and expressed that variable in subj. function with the expression from one of the equalities, but I assume that is different algorithm.
Any remarks or help will be highly appreciated.
The problem has been solved. Turned out that the point 7 in the notes was wrong. Instead, point 7 should be
In Matlab, I have two single row (1x249) vectors in a 2x249 matrix and I have to create a matrix A by replicating them many times, each time shifting the vectors of 2 positions to the right. I would like to fill the entries on the left with zeros. Is there a smart way to do this? Currently, I am using a for loop and circshift, and I add at each iteration I add the new row to A, but probably this is highly inefficient.
Code (myMat is the matrix I want to shift):
A = [];
myMat = [1 0 -1 zeros(1,246); 0 2 0 -2 zeros(1,245)];
N = 20;
for i=1:N-1
aux = circshift(myMat,[0,2*(i-1)]);
aux(:,1:2*(i-1)) = 0;
A =[A; aux];
end
As you are probably aware, loops in Matlab are not so efficient.
I know that the Mathworks keep saying this is no longer so with JIT
compilation, but I haven't experienced the fast loops yet.
I put your method for constructiong the matrix A in a function:
function A = replvector1(myMat,shift_right,width,N)
pre_alloc = true; % make implementation faster using pre-allocation yes/no
% Pad myMat with zeros to make it wide enough
myMat(1,width)=0;
% initialize A
if pre_alloc
A = zeros(size(myMat,1)*(N-1),width);
else
A = [];
end
% Fill A
for i=1:N-1
aux = circshift(myMat,[0,shift_right*(i-1)]);
aux(:,1:min(width,shift_right*(i-1))) = 0;
A(size(myMat,1)*(i-1)+1:size(myMat,1)*i,:) =aux;
end
Your matrix-operation looks a lot like a kronecker product, but the
block-matrixces have overlapping column ranges so a direct kronecker product
will not work. Instead, I constructed the following function:
function A = replvector2(myMat,shift_right,width,N)
[i,j,a] = find(myMat);
i = kron(ones(N-1,1),i) + kron([0:N-2]',ones(size(i))) * size(myMat,1);
j = kron(ones(N-1,1),j) + kron([0:N-2]',ones(size(j))) * shift_right;
a = kron(ones(N-1,1),a);
ok = j<=width;
A = full(sparse(i(ok),j(ok),a(ok),(N-1)*size(myMat,1),width));
You can follow the algorithm by removing semicolons and looking at intermediate
results.
The following main program runs your example, and can easily be modified to
run similar examples:
% inputs (you may vary them to see that it always works)
shift_right = 2;
width = 249;
myMat1 = [ 1 0 -1 0 ;
0 2 0 -2 ];
N = 20;
% Run your implementation
tic;
A = replvector1(myMat,shift_right,width,N);
disp(sprintf('\n original implementation took %e sec',toc))
% Run the new implementation
tic;
B = replvector2(myMat,shift_right,width,N);
disp(sprintf(' new implementation took %e sec',toc))
disp(sprintf('\n norm(B-A)=%e\n',norm(B-A)))
I've taken Nathan's code (see his answer to this question), and added another possible implementation (replvector3).
My idea here stems from you not really needing a circular shift. You need to right-shift and add zeros to the left. If you start with a pre-allocated array (this is really where the big wins in time are for you, the rest is peanuts), then you already have the zeros. Now you just need to copy over myMat to the right locations.
These are the times I see (MATLAB R2017a):
OP's, with pre-allocation: 1.1730e-04
Nathan's: 5.1992e-05
Mine: 3.5426e-05
^ shift by one on purpose, to make comparison of times easier
This is the full copy, copy-paste into an M-file and run:
function so
shift_right = 2;
width = 249;
myMat = [ 1 0 -1 0 ;
0 2 0 -2 ];
N = 20;
A = replvector1(myMat,shift_right,width,N);
B = replvector2(myMat,shift_right,width,N);
norm(B(:)-A(:))
C = replvector3(myMat,shift_right,width,N);
norm(C(:)-A(:))
timeit(#()replvector1(myMat,shift_right,width,N))
timeit(#()replvector2(myMat,shift_right,width,N))
timeit(#()replvector3(myMat,shift_right,width,N))
% Original version, modified to pre-allocate
function A = replvector1(myMat,shift_right,width,N)
% Assuming width > shift_right * (N-1) + size(myMat,2)
myMat(1,width) = 0;
M = size(myMat,1);
A = zeros(M*(N-1),width);
for i = 1:N-1
aux = circshift(myMat,[0,shift_right*(i-1)]);
aux(:,1:shift_right*(i-1)) = 0;
A(M*(i-1)+(1:M),:) = aux;
end
% Nathan's version
function A = replvector2(myMat,shift_right,width,N)
[i,j,a] = find(myMat);
i = kron(ones(N-1,1),i) + kron((0:N-2)',ones(size(i))) * size(myMat,1);
j = kron(ones(N-1,1),j) + kron((0:N-2)',ones(size(j))) * shift_right;
a = kron(ones(N-1,1),a);
ok = j<=width;
A = full(sparse(i(ok),j(ok),a(ok),(N-1)*size(myMat,1),width));
% My trivial version with loops
function A = replvector3(myMat,shift_right,width,N)
% Assuming width > shift_right * (N-1) + size(myMat,2)
[M,K] = size(myMat);
A = zeros(M*(N-1),width);
for i = 1:N-1
A(M*(i-1)+(1:M),shift_right*(i-1)+(1:K)) = myMat;
end
I know how to write the program with a given number of entries, but not when if the number of entries are n:. Basically I think I need a loop but I can't get it to work properly.
This is a code that works fine when having 3 entries. % is what I put in. I want to create a vector with specific position and values, this is a example how the vector would look like:
X = [1;0;0;0;0;0;0;0;0;0;0;0;0;0;99;0;0;0;0;99]
and the code for it:
n1 = input('Determine value for case:'); %n1 = 1
n2 = input('Determine value for case:'); %n2 = 15
n3 = input('Determine value for case:'); %n3 = 20
X = zeros(20,1);
X(n1) = input('Determine position '); %1
X(n2) = input('Determine position '); %99
X(n3) = input('Determine position '); %99
But for n entries, I need a loop I figured. (the vector may still be 20x1)
for n = 1:entries (%entries are 3, so 3 loops)
n = n+1
n = input('Determine value for case :');
X =zeros(20,1);
X(n) = input('Determine position:')
end
But I can't just get it to work
thanks in advance
Not sure exactly what you are trying to achieve, but I think that you can fix your loop as follows:
entries = 3;
X =zeros(20,1);
for ni = 1:entries
n = input('Determine value for case :');
X(n) = input('Determine position:');
end