I need to calculate the distance matrix between these locations in ev charging stations placement

1 次查看（过去 30 天）
chaitanya
chaitanya 2023-11-30
回答： recent works 2023-11-30
I'm working on an EV charging station project and have coordinates of potential station locations. I need to calculate the distance matrix between these locations using MATLAB. How can I efficiently compute this matrix based on the coordinates provided

请先登录，再回答此问题。

采纳的回答

recent works
recent works 2023-11-30
% Assuming chargingStationCoordinates contains the coordinates of potential charging stations
% Calculate distance matrix between charging stations
numStations = size(chargingStationCoordinates, 1);
distanceMatrix = zeros(numStations);
for i = 1:numStations
for j = 1:numStations
distanceMatrix(i, j) = norm(chargingStationCoordinates(i, :) - chargingStationCoordinates(j, :));
end
end
disp('Distance Matrix:');
disp(distanceMatrix);

更多回答（1 个）

Konrad
Konrad 2023-11-30
Hi,
let's say you have 10 2d coordinates of charging stations
pos = rand(10,2);
Then one efficient way of computing the distance matrix would be:
m = pos * pos';
dist = sqrt(bsxfun(@plus, diag(m), diag(m)') - 2*m);
  1 个评论
Konrad
Konrad 2023-11-30
编辑：Konrad 2023-11-30
Ran in:
Note that since 2016 or so, Matlab does implicit singleton expansion, i.e. you don't need the bsxfun() but you can write:
%sqrt(diag(m) + diag(m)' - 2*m)
which seems to be considerably faster.
EDIT: After some more testing it's not much of a difference:
tic;
for k = 1:1000
pos = rand(200,2);
m = pos * pos';
dist = sqrt(diag(m)+diag(m)' - 2*m);
end
toc
Elapsed time is 0.130854 seconds.
tic;
for k = 1:1000
pos = rand(200,2);
m = pos * pos';
dist = sqrt(bsxfun(@plus, diag(m), diag(m)') - 2*m);
end
toc
Elapsed time is 0.148703 seconds.
A different approach using the Pythagorean theorem:
tic;
for k = 1:1000
pos = rand(200,2);
dist = sqrt(sum( (permute(pos,[1 3 2])-permute(pos,[3 1 2])).^2 ,3));
end
toc
Elapsed time is 0.166725 seconds.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Mathematics 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by