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?
Related
I wrote a program implementing Gaussian Elimination with Complete Pivoting:
function x = gecp(A,b)
x = b;
n = length(A);
p = 1:n;
l = b;
for k = 1:n
[i,j] = find(A(k:n,k:n)==max(abs(A(k:n,k:n)),[],'all'));
i = i+k-1;
j = j+k-1;
[A(k,:),A(i,:)] = deal(A(i,:),A(k,:));
[A(:,j),A(:,k)] = deal(A(:,k),A(:,j));
[b(i),b(k)] = deal(b(k),b(i));
[p(k),p(j)] = deal(p(j),p(k));
temp = (k+1):n;
l(temp) = A(temp,k)/A(k,k);
b(temp) = b(temp)-l(temp).*b(k);
A(temp,temp) = A(temp,temp)-l(temp).*A(k,temp);
end
x(n) = b(n)/A(n,n);
for k = (n-1):-1:1
s = 0;
for h = (k+1):n
s = s+A(k,h)*x(h);
end
x(k) = (b(k)-s)/A(k,k);
end
x(p) = x;
And it is called like this:
N = 5; A = randn(N); b = randn(N,1); x = gecp(A,b)
Unfortunately all lines containing deal function (used for swapping rows of columns of matrices), give me following (or similar) error: "Unable to perform assignment because the size of the left side is 1-by-5 and the size of the right side is 0-by-5."
Unfortunately I have no idead why would the width of these vectors be changed to 0 as I wrote excatly the same thing on both sides.
To compute the mean of every bins along a dimension of a nd array in matlab, for example, average every 10 elements along dim 4 of a 4d array
x = reshape(1:30*30*20*300,30,30,20,300);
n = 10;
m = size(x,4)/10;
y = nan(30,30,20,m);
for ii = 1 : m
y(:,:,:,ii) = mean(x(:,:,:,(1:n)+(ii-1)*n),4);
end
It looks a bit silly. I think there must be better ways to average the bins?
Besides, is it possible to make the script applicable to general cases, namely, arbitray ndims of array and along an arbitray dim to average?
For the second part of your question you can use this:
x = reshape(1:30*30*20*300,30,30,20,300);
dim = 4;
n = 10;
m = size(x,dim)/10;
y = nan(30,30,20,m);
idx1 = repmat({':'},1,ndims(x));
idx2 = repmat({':'},1,ndims(x));
for ii = 1 : m
idx1{dim} = ii;
idx2{dim} = (1:n)+(ii-1)*n;
y(idx1{:}) = mean(x(idx2{:}),dim);
end
For the first part of the question here is an alternative using cumsum and diff, but it may not be better then the loop solution:
function y = slicedmean(x,slice_size,dim)
s = cumsum(x,dim);
idx1 = repmat({':'},1,ndims(x));
idx2 = repmat({':'},1,ndims(x));
idx1{dim} = slice_size;
idx2{dim} = slice_size:slice_size:size(x,dim);
y = cat(dim,s(idx1{:}),diff(s(idx2{:}),[],dim))/slice_size;
end
Here is a generic solution, using the accumarray function. I haven't tested how fast it is. There might be some room for improvement though.
Basically, accumarray groups the value in x following a matrix of customized index for your question
x = reshape(1:30*30*20*300,30,30,20,300);
s = size(x);
% parameters for averaging
dimAv = 4;
n = 10;
% get linear index
ix = (1:numel(x))';
% transform them to a matrix of index per dimension
% this is a customized version of ind2sub
pcum = [1 cumprod(s(1:end-1))];
sub = zeros(numel(ix),numel(s));
for i = numel(s):-1:1,
ixtmp = rem(ix-1, pcum(i)) + 1;
sub(:,i) = (ix - ixtmp)/pcum(i) + 1;
ix = ixtmp;
end
% correct index for the given dimension
sub(:,dimAv) = floor((sub(:,dimAv)-1)/n)+1;
% run the accumarray to compute the average
sout = s;
sout(dimAv) = ceil(sout(dimAv)/n);
y = accumarray(sub,x(:), sout, #mean);
If you need a faster and memory efficient operation, you'll have to write your own mex function. It shouldn't be so difficult, I think !
I am programming a simple perceptron in matlab but it is not converging and I can't figure out why.
The goal is to binary classify 2D points.
%P1 Generate a dataset of two-dimensional points, and choose a random line in
%the plane as your target function f, where one side of the line maps to +1 and
%the other side to -1. Let the inputs xn 2 R2 be random points in the plane,
%and evaluate the target function f on each xn to get the corresponding output
%yn = f(xn).
clear all;
clc
clear
n = 20;
inputSize = 2; %number of inputs
dataset = generateRandom2DPointsDataset(n)';
[f , m , b] = targetFunction();
signs = classify(dataset,m,b);
weights=ones(1,2)*0.1;
threshold = 0;
fprintf('weights before:%d,%d\n',weights);
mistakes = 1;
numIterations = 0;
figure;
plotpv(dataset',(signs+1)/2);%mapping signs from -1:1 to 0:1 in order to use plotpv
hold on;
line(f(1,:),f(2,:));
pause(1)
while true
mistakes = 0;
for i = 1:n
if dataset(i,:)*weights' > threshold
result = 1;
else
result = -1;
end
error = signs(i) - result;
if error ~= 0
mistakes = mistakes + 1;
for j = 1:inputSize
weights(j) = weights(j) + error*dataset(i,j);
end
end
numIterations = numIterations + 1
end
if mistakes == 0
break
end
end
fprintf('weights after:%d,%d\n',weights);
random points and signs are fine since plotpv is working well
The code is based on that http://es.mathworks.com/matlabcentral/fileexchange/32949-a-perceptron-learns-to-perform-a-binary-nand-function?focused=5200056&tab=function.
When I pause the infinite loop, this is the status of my vairables:
I am not able to see why it is not converging.
Additional code( it is fine, just to avoid answers asking for that )
function [f,m,b] = targetFunction()
f = rand(2,2);
f(1,1) = 0;
f(1,2) = 1;
m = (f(2,2) - f(2,1));
b = f(2,1);
end
function dataset = generateRandom2DPointsDataset(n)
dataset = rand(2,n);
end
function values = classify(dataset,m,b)
for i=1:size(dataset,1)
y = m*dataset(i,1) + b;
if dataset(i,2) >= y, values(i) = 1;
else values(i) = -1;
end
end
end
I'm writing the code in Matlab to find interest point using DoG in the image.
Here is the main.m:
imTest1 = rgb2gray(imread('1.jpg'));
imTest1 = double(imTest1);
sigma = 0.6;
k = 5;
thresh = 3;
[x1,y1,r1] = DoG(k,sigma,thresh,imTest1);
%get the interest points and show it on the image with its scale
figure(1);
imshow(imTest1,[]), hold on, scatter(y1,x1,r1,'r');
And the function DoG is:
function [x,y,r] = DoG(k,sigma,thresh,imTest)
x = []; y = []; r = [];
%suppose 5 levels of gaussian blur
for i = 1:k
g{i} = fspecial('gaussian',size(imTest),i*sigma);
end
%so 4 levels of DoG
for i = 1:k-1
d{i} = imfilter(imTest,g{i+1}-g{i});
end
%compare the current pixel in the image to the surrounding pixels (26 points),if it is the maxima/minima, this pixel will be a interest point
for i = 2:k-2
for m = 2:size(imTest,1)-1
for n = 2:size(imTest,2)-1
id = 1;
compare = zeros(1,27);
for ii = i-1:i+1
for mm = m-1:m+1
for nn = n-1:n+1
compare(id) = d{ii}(mm,nn);
id = id+1;
end
end
end
compare_max = max(compare);
compare_min = min(compare);
if (compare_max == d{i}(m,n) || compare_min == d{i}(m,n))
if (compare_min < -thresh || compare_max > thresh)
x = [x;m];
y = [y;n];
r = [r;abs(d{i}(m,n))];
end
end
end
end
end
end
So there's a gaussian function and the sigma i set is 0.6. After running the code, I find the position is not correct and the scales looks almost the same for all interest points. I think my code should work but actually the result is not. Anybody know what's the problem?
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