I need some advice regarding a problem I encountered in MATLAB:
I have 4 variables, I'm not sure what is the best methodology to go about doing this. I initially thought about just computing the GreatCircle distance from each grid point to the specified location and return the corresponding row/column index that has the minimum distance. But doing it this way, I'm not sure how I can compute the interpolation.
I tried reshaping the data into a vector data of 4 columns and running meshgrid to possibly utilize interp2. But I ended up with this error:
Requested 109620x109620 (44.8GB) array exceeds maximum array size preference
What could be the most efficient way to do this?
You are working on large arrays. If interp2 cannot handle it, work on a subarray instead:
% Create data
format shortG
[Latitude,Longitude] = meshgrid(1:12,1:12);
Altitude = floor(1000+sortrows(rand(12,12))*1000);
Temperature = 10+20*rand(12,12);
Lat = 2.1;
Lon = 11.8;
% Find closest match point
[~,i_Lat] = min(abs(Latitude(1,:)-Lat));
[~,i_Lon] = min(abs(Longitude(:,1)-Lon));
% Select subarrays around this point.
% Minimum size of these matrices depend on the type of interpolation you perform
ia1 = max(1,i_Lat-5);
ia2 = min(size(Latitude,1),i_Lat+5);
io1 = max(1,i_Lon-5);
io2 = min(size(Latitude,2),i_Lon+5);
subLatitude = Latitude(io1:io2,ia1:ia2);
subLongitude = Longitude(io1:io2,ia1:ia2);
subAltitude = Altitude(io1:io2,ia1:ia2);
subTemperature = Temperature(io1:io2,ia1:ia2);
% Interpolate on these small arrays, and evaluate at target (Lat, Lon) point
A_out = interp2(subLatitude, subLongitude, subAltitude, Lat, Lon)
T_out = interp2(subLatitude, subLongitude, subTemperature, Lat, Lon)
Related
Im trying to write a code in Matlab to calculate an area of influence type question. This is an exert from my data (Weighting, x-coord, y-coord):
M =
15072.00 486.00 -292
13269.00 486.00 -292
12843.00 414.00 -267
10969.00 496.00 -287
9907.00 411.00 -274
9718.00 440.00 -265
9233.00 446.00 -253
9138.00 462.00 -275
8830.00 496.00 -257
8632.00 432.00 -253
R =
-13891.00 452.00 -398
-13471.00 461.00 -356
-12035.00 492.00 -329
-11309.00 413.00 -353
-11079.00 467.00 -375
-10659.00 493.00 -333
-10643.00 495.00 -338
-10121.00 455.00 -346
-9795.00 456.00 -367
-8927.00 485.00 -361
-8765.00 467.00 -351
I want to make a function to calculate the sum of the weightings at any given position based on a circle of influence of 30 for each coordinate.
I have thought of using a for loop to calculate each point independently and summing the result but seems unnecessarily complicated and inefficient.
I also thought of assigning an intensity of color to each circle and overlaying them but I dont know how to change color intensity based on value here is my attempt so far (I would like to have a visual of the result):
function [] = Influence()
M = xlsread('MR.xlsx','A4:C310');
R = xlsread('MR.xlsx','E4:G368');
%these are my values around 300 coordinates
%M are negative values and R positive, I want to see which are dominant in their regions
hold on
scatter(M(:,2),M(:,3),3000,'b','filled')
scatter(R(:,2),R(:,3),3000,'y','filled')
axis([350 650 -450 -200])
hold off
end
%had to use a scalar of 3000 for some reason as it isnt correlated to the graph size
I'd appreciate any ideas/solutions thank you
This is the same but with ca. 2000 data points
How about this:
r_influence = 30; % radius of influence
r = #(p,A) sqrt((p(1)-A(:,2)).^2 + (p(2)-A(:,3)).^2); % distance
wsum = #(p,A) sum(A(find(r(p,A)<=r_influence),1)); % sum where distance less than roi
% compute sum on a grid
xrange = linspace(350,550,201);
yrange = linspace(-200,-450,201);
[XY,YX] = meshgrid(xrange,yrange);
map_M = arrayfun(#(p1,p2) wsum([p1,p2],M),XY,YX);
map_R = arrayfun(#(p1,p2) wsum([p1,p2],R),XY,YX);
figure(1);
clf;
imagesc(xrange,yrange,map_M + map_R);
colorbar;
Gives a picture like this:
Is that what you are looking for?
I'm trying to estimate the (unknown) original datapoints that went into calculating a (known) moving average. However, I do know some of the original datapoints, and I'm not sure how to use that information.
I am using the method given in the answers here: https://stats.stackexchange.com/questions/67907/extract-data-points-from-moving-average, but in MATLAB (my code below). This method works quite well for large numbers of data points (>1000), but less well with fewer data points, as you'd expect.
window = 3;
datapoints = 150;
data = 3*rand(1,datapoints)+50;
moving_averages = [];
for i = window:size(data,2)
moving_averages(i) = mean(data(i+1-window:i));
end
length = size(moving_averages,2)+(window-1);
a = (tril(ones(length,length),window-1) - tril(ones(length,length),-1))/window;
a = a(1:length-(window-1),:);
ai = pinv(a);
daily = mtimes(ai,moving_averages');
x = 1:size(data,2);
figure(1)
hold on
plot(x,data,'Color','b');
plot(x(window:end),moving_averages(window:end),'Linewidth',2,'Color','r');
plot(x,daily(window:end),'Color','g');
hold off
axis([0 size(x,2) min(daily(window:end))-1 max(daily(window:end))+1])
legend('original data','moving average','back-calculated')
Now, say I know a smattering of the original data points. I'm having trouble figuring how might I use that information to more accurately calculate the rest. Thank you for any assistance.
You should be able to calculate the original data exactly if you at any time can exactly determine one window's worth of data, i.e. in this case n-1 samples in a window of length n. (In your case) if you know A,B and (A+B+C)/3, you can solve now and know C. Now when you have (B+C+D)/3 (your moving average) you can exactly solve for D. Rinse and repeat. This logic works going backwards too.
Here is an example with the same idea:
% the actual vector of values
a = cumsum(rand(150,1) - 0.5);
% compute moving average
win = 3; % sliding window length
idx = hankel(1:win, win:numel(a));
m = mean(a(idx));
% coefficient matrix: m(i) = sum(a(i:i+win-1))/win
A = repmat([ones(1,win) zeros(1,numel(a)-win)], numel(a)-win+1, 1);
for i=2:size(A,1)
A(i,:) = circshift(A(i-1,:), [0 1]);
end
A = A / win;
% solve linear system
%x = A \ m(:);
x = pinv(A) * m(:);
% plot and compare
subplot(211), plot(1:numel(a),a, 1:numel(m),m)
legend({'original','moving average'})
title(sprintf('length = %d, window = %d',numel(a),win))
subplot(212), plot(1:numel(a),a, 1:numel(a),x)
legend({'original','reconstructed'})
title(sprintf('error = %f',norm(x(:)-a(:))))
You can see the reconstruction error is very small, even using the data sizes in your example (150 samples with a 3-samples moving average).
I am trying to generate a 1D array in Matlab and calling another function to do it. However, I am getting an error when I try to find out the minimum element and its position within the array. I want to find out the minimum value in maxSS and then find its index position and use that to find the corresponding angle. Here's my code:
angle = linspace(0,pi/2,1000);
for i = 1:length(angle)
[maxSS] = GetMaxShearStress(strainMatrix, complianceMatrix, angle(i));
M = min(maxSS);
[M,I] = min(maxSS);
minimum_angle = angle();
end
disp(' The fibre angle that minimises the shear stress is 'num2str(minimum_angle) 'radians');
Does this work?
[M,I] = min(maxSS(:))
I present my simple working Matlab code and will ask questions:
tic
nrand1 = 10000;
nrand2 = 20000;
% Location matrix 1: [longitude, latitude, w1]
lmat1=[rand(nrand1,1)-75 rand(nrand1,1)+39 round(rand(nrand1,1)*1000)+1];
% Location matrix 2: [longitude, latitude, w2]
lmat2=[rand(nrand2,1)-75 rand(nrand2,1)+39 round(rand(nrand2,1)*100)+1];
% The number of rows for each matrix = In fact it's nrand1 X nrand2, obviously
nobs1 = size(lmat1,1);
nobs2 = size(lmat2,1);
% The number of pair-wise distances
% between L1 locations X L2 locations
ndist = nobs1*nobs2;
% Initialization: Distance vector and weight vector
hdist = zeros(ndist,1);
weight = zeros(ndist,1);
% Double for loop -- for calculating the pair-wise distances and weights
k=1;
for i=1:nobs1
for j=1:nobs2
% distances in kilometers.
lonH = sin(0.5*(lmat1(i,1)-lmat2(j,1))*pi/180.0)^2;
latH = sin(0.5*(lmat1(i,2)-lmat2(j,2))*pi/180.0)^2;
hdist(k) = 0.001*6372797.560856*2 ...
*asin(sqrt(latH+(cos(lmat1(i,2)*pi/180.0) ...
*cos(lmat2(j,2)*pi/180.0))*lonH));
weight(k) = lmat1(i,3)*lmat2(j,3);
k=k+1;
end
end
toc
The code calculates 10000 X 20000 distances and weights.
Elapsed time is 67.124844 seconds.
Is there a way to vectorize the double-loop processing, or to perform a parallel computing? If there is no room for performance improvement in Matlab, I may have to write the double loops in C and call it from Matlab. I don't know how to call C from matlab, so I will ask a separate question. Thanks!
Using bsxfun, you can eliminate the for loops and the need for calculating matrices for each combination (this should reduce memory usage). The following is about six times faster than your original code on my computer using R2014b:
nrand1 = 10000;
nrand2 = 20000;
% Location matrix 1: [longitude, latitude, w1]
lmat1=[rand(nrand1,1)-75 rand(nrand1,1)+39 round(rand(nrand1,1)*1000)+1];
% Location matrix 2: [longitude, latitude, w2]
lmat2=[rand(nrand2,1)-75 rand(nrand2,1)+39 round(rand(nrand2,1)*100)+1];
p180 = pi/180;
lonH = sin(0.5*bsxfun(#minus,lmat1(:,1).',lmat2(:,1))*p180).^2;
latH = sin(0.5*bsxfun(#minus,lmat1(:,2).',lmat2(:,2))*p180).^2;
hdist = 0.001*6372797.560856*2*asin(sqrt(latH+bsxfun(#times,cos(lmat1(:,2).'*p180),cos(lmat2(:,2)*p180)).*lonH));
hdist1 = hdist(:);
weight1 = bsxfun(#times,lmat1(:,3).',lmat2(:,3));
weight1 = weight1(:);
Note that by using the variable p180, the math is changed slightly so you won't get precisely the same values, but they will be very close.
The solution is that your inputs (lmat1 and lmat2) do not need to be matrices like you have them. Each one is really three vectors. Once you've broken out the vectors, you can create arrays that have every permutation of lmat1 and lmat2 together (which is what your double loop is doing). At that point, you can call your math as single, fully-vectorized operations...
%make your vectors
lmat1A = rand(nrand1,1)-75;
lmat1B = rand(nrand1,1)+39;
lmat1C = round(rand(nrand1,1)*1000)+1
lmat2A = rand(nrand2,1)-75;
lmat2B = rand(nrand2,1)+39;
lmat2C = round(rand(nrand2,1)*1000)+1
%make every combination
lmat1A = lmat1A(:)*ones(1,nrand2);
lmat1B = lmat1B(:)*ones(1,nrand2);
lmat1C = lmat1C(:)*ones(1,nrand2);
lmat2A = ones(nrand1,1)*(lmat2A(:)');
lmat2B = ones(nrand1,1)*(lmat2B(:)');
lmat2C = ones(nrand1,1)*(lmat2C(:)');
%do your math
lonH = sin(0.5*(lmat1A-lmat2A)*pi/180.0).^2;
latH = sin(0.5*(lmat1B-lmat2B)*pi/180.0).^2;
hdist = 0.001*6372797.560856*2 ...
.*asin(sqrt(latH+(cos(lmat1B*pi/180.0) ...
.*cos(lmat2B*pi/180.0)).*lonH)); %use element-wise multiplication
weight = lmat1C.*lmat2C;
%reshape your output into vectors (not arrays), which is what your original code does
lonH = lonH(:)
latH = latH(:)
hdist = hdist(:);
weight = weight(:);
I have a trajectory defined in Lat/Long coordinates in "n" points. I need to estimate its length over the surface of the earth:
Lat = [la1 la2 la3 la4 la5 la6];
Lon = [lo1 lo2 lo3 lo4 lo5 lo6];
How can I do this in Matlab? I've tried to use the command distance, but it seems it doesn't do this?
Thanks!
[arclen,az] = distance(lat1,lon1,lat2,lon2)
[arclen,az] = distance(lat1,lon1,lat2,lon2,ellipsoid)
Check the matlab documentation
Something similar to this shouldn't be very hard to write:
sum = 0;
for i=1:(size(Lat)-1)
sum = sum + distance(Lat(i),Lon(i),Lat(i+1),Lon(i+1));
//end (Sorry I haven't used matlab in over a year, and forgot the syntax)