I was wondering if someone could point out why I keep failing Test 1 and 2 despite my plot appearing as it should be.
My goal is to plot a bacteria population using three different K values on the same plot however I am failing due to the Index Exceeding Matrix Dimensions.
I am unsure what this means and how it relates to the plot.
I appreciate any help.
Thank you.
`function [bacteria_plot] = BacteriaPop(K,BacteriaPop)
C = 100;
r = 1;
t = 0:20;
K = 1000;
Kk = 3000;
kK = 5000;
BacteriaPop = (C.*exp(r.*t))./(1+(C./K).*exp(r.*t));
BacteriaPop = (C.*exp(r.*t))./(1+(C./Kk).*exp(r.*t));
BacteriaPop = (C.*exp(r.*t))./(1+(C./kK).*exp(r.*t));
hold on
plot(t,BacteriaPop)
end
`
[![Code, Plot and Error][2]][2]
Thanks to some help I have changed my code to
function [bacteria_plot] = bacteria_growth(K,Kk,kK,Y)
C = 100;
r = 1;
t = 0:20;
K = 1000;
Kk = 3000;
kK = 5000;
y(:,1) = (C.*exp(r.*t))./(1+(C./K).*exp(r.*t));
y(:,2) = (C.*exp(r.*t))./(1+(C./Kk).*exp(r.*t));
y(:,3) = (C.*exp(r.*t))./(1+(C./kK).*exp(r.*t));
hold on
plot(t,y)
hold off
end
It is still not perfect but it passed all the tests.
Any constructive criticism to streamline this code will be welcomed.
Related
function [V,H] = Arnoldi(A,v,m)
[n,~] = size(A);
V = zeros(n,m+1);
H = zeros(n,n);
V(:,1) = v/norm(v);
for k = 2:m
V(:,k) = A*V(:,k-1);
for j = 1:(k-1);
H(j,k-1) = V(:,j).'*V(:,k);
V(:,k) = V(:,k)- H(j,k-1)*V(:,j);
end
H(k,k-1) = norm(V(:,k));
V(:,k) = V(:,k)/H(k,k-1);
end
end
This is my implementation of the Arnoldi algorithm. We already used wikipedia but did not find an answer there.
A is a square matrix, v is our starting vector and m is the number of iterations/ dimension of Krylov subspace. It does not give the wanted Hessian H and we can not figure out where we go wrong. Can anybody help us?
I am currently looking for the most efficient way to shift and rearrange large matrices. Essentially, I have data with some parabolic shift that needs to be corrected in order to shift the "signal" to a linear event.
I have currently tried the following solutions and tried timing them. Is there any other method that may prove to be more efficient?
DATA = ones(100000,501);
DATA(10000,251) = 100;
for i=1:250
DATA(10000+i^2-1000:10000+i^2+1000,251-i) = 100;
DATA(10000+i^2-1000:10000+i^2+1000,251+i) = 100;
end
k = abs(-250:1:250).^2;
d = size(DATA,1);
figure(99)
imagesc(DATA)
t_INDEX = timeit(#()fun_INDEX(DATA,k))
t_SNIPPET = timeit(#()fun_SNIPPET(DATA,k))
t_CIRCSHIFT = timeit(#()fun_CIRCSHIFT(DATA,k))
t_INDEX_clean = timeit(#()fun_INDEX_clean(DATA,k))
t_SPARSE = timeit(#()fun_SPARSE(DATA,k))
t_BSXFUN = timeit(#()fun_BSXFUN(DATA,k))
function fun_INDEX(DATA,k)
DATA_1 = zeros(size(DATA));
for i=1:size(DATA,2)
DATA_1(:,i) = DATA([k(i)+1:end 1:k(i)],i);
end
figure(1)
imagesc(DATA_1)
end
function fun_SNIPPET(DATA,k)
kmax = max(k);
DATA_2 = zeros(size(DATA,1)-kmax,size(DATA,2));
for i=1:size(DATA,2)
DATA_2(:,i) = DATA(k(i)+1:end-kmax+k(i),i);
end
figure(2)
imagesc(DATA_2)
end
function fun_CIRCSHIFT(DATA,k)
DATA_3 = zeros(size(DATA));
for i=1:size(DATA,2)
DATA_3(:,i) = circshift(DATA(:,i),-k(i),1);
end
figure(3)
imagesc(DATA_3)
end
function fun_INDEX_clean(DATA,k)
[m, n] = size(DATA);
k = size(DATA,1)-k;
DATA_4 = zeros(m, n);
for i = (1 : n)
DATA_4(:, i) = [DATA((m - k(i) + 1 : m), i); DATA((1 : m - k(i) ), i)];
end
figure(4)
imagesc(DATA_4)
end
function fun_SPARSE(DATA,k)
[m,n] = size(DATA);
k = -k;
S = full(sparse(mod(k,m)+1,1:n,1,m,n));
DATA_5 = ifft(fft(DATA).*fft(S),'symmetric');
figure(5)
imagesc(DATA_5)
end
function fun_BSXFUN(DATA,k)
DATA = DATA';
k = -k;
[m,n] = size(DATA);
idx0 = mod(bsxfun(#plus,n-k(:),1:n)-1,n);
DATA_6 = DATA(bsxfun(#plus,(idx0*m),(1:m)'));
figure(6)
imagesc(DATA_6)
end
Is there any way to decrease computation time for this kind of problem?
Thanks in advance for any tips!
One option would be to use MATLAB's GPU functions, if your workstation has a GPU. Depending on if the entire data fits on the GPU at once, it will start to outperform CPU circshift at 1000 X 1000 matrix size.
The implementation only requires you to copy your data to the GPU with a single statement, and then operate circshift on the newly created you array.
A small discussion on its performance can be found here: https://www.mathworks.com/matlabcentral/answers/274619-circshift-slower-on-gpu . Especially, the last post describes a much faster GPU implementation if you actually don't need to circularly shift, but can get away with zero passing on one side, which might be relevant.
My script is supposed to run Runge-Kutta and then interpolate around the tops using polyfit to calculate the max values of the tops. I seem to get the x-values of the max points correct but the y-values are off for some reason. Have sat with it for 3 days now. The problem should be In the last for-loop when I calculate py?
Function:
function funk = FU(t,u)
L0 = 1;
C = 1*10^-6;
funk = [u(2); 2.*u(1).*u(2).^2./(1+u(1).^2) - u(1).*(1+u(1).^2)./(L0.*C)];
Program:
%Runge kutta
clear all
close all
clc
clf
%Given values
U0 = [240 1200 2400];
L0 = 1;
C = 1*10^-6;
T = 0.003;
h = 0.000001;
W = [];
% Runge-Kutta 4
for i = 1:3
u0 = [0;U0(i)];
u = u0;
U = u;
tt = 0:h:T;
for t=tt(1:end-1)
k1 = FU(t,u);
k2 = FU(t+0.5*h,u+0.5*h*k1);
k3 = FU((t+0.5*h),(u+0.5*h*k2));
k4 = FU((t+h),(u+k3*h));
u = u + (1/6)*(k1+2*k2+2*k3+k4)*h;
U = [U u];
end
W = [W;U];
end
I1 = W(1,:); I2 = W(3,:); I3 = W(5,:);
dI1 = W(2,:); dI2 = W(4,:); dI3 = W(6,:);
I = [I1; I2; I3];
dI = [dI1; dI2; dI3];
%Plot of the currents
figure (1)
plot(tt,I1,'r',tt,I2,'b',tt,I3,'g')
hold on
legend('U0 = 240','U0 = 1200','U0 = 2400')
BB = [];
d = 2;
px = [];
py = [];
format short
for l = 1:3
[H,Index(l)]=max(I(l,:));
Area=[(Index(l)-2:Index(l)+2)*h];
p = polyfit(Area,I(Index(l)-2:Index(l)+2),4);
rotp(1,:) = roots([4*p(1),3*p(2),2*p(3),p(4)]);
B = rotp(1,2);
BB = [BB B];
Imax(l,:)=p(1).*B.^4+p(2).*B.^3+p(3).*B.^2+p(4).*B+p(5);
Tsv(i)=4*rotp(1,l);
%px1 = linspace(h*(Index(l)-d-1),h*(Index(l)+d-2));
px1 = BB;
py1 = polyval(p,px1(1,l));
px = [px px1];
py = [py py1];
end
% Plots the max points
figure(1)
plot(px1(1),py(1),'b*-',px1(2),py(2),'b*-',px1(3),py(3),'b*-')
hold on
disp(Imax)
Your polyfit line should read:
p = polyfit(Area,I(l, Index(l)-2:Index(l)+2),4);
More interestingly, take note of the warnings you get about poor conditioning of that polynomial (I presume you're seeing these). Why? Partly because of numerical precision (your numbers are very small, scaled around 10^-6) and partly because you're asking for a 4th-order fit to five points (which is singular). To do this "better", use more input points (more than 5), or a lower-order polynomial fit (quadratic is usually plenty), and (probably) rescale before you use the polyfit tool.
Having said that, in practice this problem is often solved using three points and a quadratic fit, because it's computationally cheap and gives very nearly the same answers as more complex approaches, but you didn't get that from me (with noiseless data like this, it doesn't much matter anyway).
I am writing a graphical representation of numerical stability of differential operators and I am having trouble removing a nested for loop. The code loops through all entries in the X,Y, plane and calculates the stability value for each point. This is done by finding the roots of a polynomial of a size dependent on an input variable (length of input vector results in a polynomial 3d matrix of size(m,n,(lenght of input vector)). The main nested for loop is as follows.
for m = 1:length(z2)
for n = 1:length(z1)
pointpoly(1,:) = p(m,n,:);
r = roots(pointpoly);
if isempty(r),r=1e10;end
z(m,n) = max(abs(r));
end
end
The full code of an example numerical method (Trapezoidal Rule) is as follows. Any and all help is appreciated.
alpha = [-1 1];
beta = [.5 .5];
Wind = 2;
Wsize = 500;
if numel(Wind) == 1
Wind(4) = Wind(1);
Wind(3) = -Wind(1);
Wind(2) = Wind(4);
Wind(1) = Wind(3);
end
if numel(Wsize) == 1
Wsize(2) = Wsize;
end
z1 = linspace(Wind(1),Wind(2),Wsize(1));
z2 = linspace(Wind(3),Wind(4),Wsize(2));
[Z1,Z2] = meshgrid(z1,z2);
z = Z1+1i*Z2;
p = zeros(Wsize(2),Wsize(1),length(alpha));
for n = length(alpha):-1:1
p(:,:,(length(alpha)-n+1)) = alpha(n)-z*beta(n);
end
for m = 1:length(z2)
for n = 1:length(z1)
pointpoly(1,:) = p(m,n,:);
r = roots(pointpoly);
if isempty(r),r=1e10;end
z(m,n) = max(abs(r));
end
end
figure()
surf(Z1,Z2,z,'EdgeColor','None');
caxis([0 2])
cmap = jet(255);
cmap((127:129),:) = 0;
colormap(cmap)
view(2);
title(['Alpha Values (',num2str(alpha),') Beta Values (',num2str(beta),')'])
EDIT::
I was able to remove one of the for loops using the reshape command. So;
for m = 1:length(z2)
for n = 1:length(z1)
pointpoly(1,:) = p(m,n,:);
r = roots(pointpoly);
if isempty(r),r=1e10;end
z(m,n) = max(abs(r));
end
end
has now become
gg = reshape(p,[numel(p)/length(alpha) length(alpha)]);
r = zeros(numel(p)/length(alpha),1);
for n = 1:numel(p)/length(alpha)
temp = roots(gg(n,:));
if isempty(temp),temp = 0;end
r(n,1) = max(abs(temp));
end
z = reshape(r,[Wsize(2),Wsize(1)]);
This might be one for loop, but I am still going through the same number of elements. Is there a way to use the roots command on all of my rows at the same time?
This is a follow-up question to How to append an element to an array in MATLAB? That question addressed how to append an element to an array. Two approaches are discussed there:
A = [A elem] % for a row array
A = [A; elem] % for a column array
and
A(end+1) = elem;
The second approach has the obvious advantage of being compatible with both row and column arrays.
However, this question is: which of the two approaches is fastest? My intuition tells me that the second one is, but I'd like some evidence for or against that. Any idea?
The second approach (A(end+1) = elem) is faster
According to the benchmarks below (run with the timeit benchmarking function from File Exchange), the second approach (A(end+1) = elem) is faster and should therefore be preferred.
Interestingly, though, the performance gap between the two approaches is much narrower in older versions of MATLAB than it is in more recent versions.
R2008a
R2013a
Benchmark code
function benchmark
n = logspace(2, 5, 40);
% n = logspace(2, 4, 40);
tf = zeros(size(n));
tg = tf;
for k = 1 : numel(n)
x = rand(round(n(k)), 1);
f = #() append(x);
tf(k) = timeit(f);
g = #() addtoend(x);
tg(k) = timeit(g);
end
figure
hold on
plot(n, tf, 'bo')
plot(n, tg, 'ro')
hold off
xlabel('input size')
ylabel('time (s)')
leg = legend('y = [y, x(k)]', 'y(end + 1) = x(k)');
set(leg, 'Location', 'NorthWest');
end
% Approach 1: y = [y, x(k)];
function y = append(x)
y = [];
for k = 1 : numel(x);
y = [y, x(k)];
end
end
% Approach 2: y(end + 1) = x(k);
function y = addtoend(x)
y = [];
for k = 1 : numel(x);
y(end + 1) = x(k);
end
end
How about this?
function somescript
RStime = timeit(#RowSlow)
CStime = timeit(#ColSlow)
RFtime = timeit(#RowFast)
CFtime = timeit(#ColFast)
function RowSlow
rng(1)
A = zeros(1,2);
for i = 1:1e5
A = [A rand(1,1)];
end
end
function ColSlow
rng(1)
A = zeros(2,1);
for i = 1:1e5
A = [A; rand(1,1)];
end
end
function RowFast
rng(1)
A = zeros(1,2);
for i = 1:1e5
A(end+1) = rand(1,1);
end
end
function ColFast
rng(1)
A = zeros(2,1);
for i = 1:1e5
A(end+1) = rand(1,1);
end
end
end
For my machine, this yields the following timings:
RStime =
30.4064
CStime =
29.1075
RFtime =
0.3318
CFtime =
0.3351
The orientation of the vector does not seem to matter that much, but the second approach is about a factor 100 faster on my machine.
In addition to the fast growing method pointing out above (i.e., A(k+1)), you can also get a speed increase from increasing the array size by some multiple, so that allocations become less as the size increases.
On my laptop using R2014b, a conditional doubling of size results in about a factor of 6 speed increase:
>> SO
GATime =
0.0288
DWNTime =
0.0048
In a real application, the size of A would needed to be limited to the needed size or the unfilled results filtered out in some way.
The Code for the SO function is below. I note that I switched to cos(k) since, for some unknown reason, there is a large difference in performance between rand() and rand(1,1) on my machine. But I don't think this affects the outcome too much.
function [] = SO()
GATime = timeit(#GrowAlways)
DWNTime = timeit(#DoubleWhenNeeded)
end
function [] = DoubleWhenNeeded()
A = 0;
sizeA = 1;
for k = 1:1E5
if ((k+1) > sizeA)
A(2*sizeA) = 0;
sizeA = 2*sizeA;
end
A(k+1) = cos(k);
end
end
function [] = GrowAlways()
A = 0;
for k = 1:1E5
A(k+1) = cos(k);
end
end