Easiest way to create a line from 2 XY coordinates

What is the fastest way or the best way to transfer these values to an array to obtain an output that is the distance between Point(X1,Y1) and Point(X2,Y2) for every iteration? The original data is 300,000 iterations, so I want to end up with a vector of 300,000 distance values

Iter 	X1	        Y1		        X2	        Y2
0	125.1673584	13.18046188		194.4534607	176.9597931
1	126.4874725	13.56817341		196.5686646	176.3899078
2	125.1508484	13.18618774		196.9122314	175.5451508
3	125.8279877	12.93647861		195.5947876	176.2555695

Answer 1

[Scott MacKenzie]

MATLAB's hypot function is probably your best bet:

M = [0	125.1673584	13.18046188		194.4534607	176.9597931;
    1	126.4874725	13.56817341		196.5686646	176.3899078;
    2	125.1508484	13.18618774		196.9122314	175.5451508;
    3	125.8279877	12.93647861		195.5947876	176.2555695]

M = 4×5

0 125.1674 13.1805 194.4535 176.9598 1.0000 126.4875 13.5682 196.5687 176.3899 2.0000 125.1508 13.1862 196.9122 175.5452 3.0000 125.8280 12.9365 195.5948 176.2556

d = hypot(M(:,2)-M(:,4), M(:,3)-M(:,5))

d = 4×1

177.8320 177.2633 177.5109 177.5965

[Lina Koronfel]

would it work with a huge matrix? I have total of 300,000 iterations.

[Scott MacKenzie]

I suspect "yes". I just tried the following simulation, and it worked fine:

M = [(1:300000)' rand(300000,4)];
d = hypot(M(:,2)-M(:,4), M(:,3)-M(:,5));
whos;

  Name           Size               Bytes  Class     Attributes

  M         300000x5             12000000  double              
  d         300000x1              2400000  double              

[Walter Roberson]

If you have a large matrix you are trying to do simultaneously, then sometimes you want to use pdist2()

[Lina Koronfel]

Dear @Walter Roberson, can you please specify why using pdist2 is better than hypot for large matrices? All I need is one value for the comparison between pt1(X,Y) and pt2(X,Y) for every iteration. pdist2 seems to give me multiple values for each comparison, and as a consequence the resulting matrix is bigger than using the hypot function. I don't understand how it is better if I have a larger matrix, could you please enlighten me? Thank you

[Walter Roberson]

@Lina Koronfel

One of the common tasks is to want to take distances between sets of points, including one point to a number of different points. pdist2() can help with that task of taking distances to a number of other points simultaneously.

[Walter Roberson]

hypot is consistently slower than square root of sum.

The difference is that hypot has code in order to be able to give useful answers for cases where the squares would overflow to infinity or underflow to 0, and for those cases it can be very valuable. But if you are certain you will never encounter those cases then manual calculation is faster.

N = 20;
t1 = zeros(N,1); t2 = zeros(N,1);

M = [(1:300000)' rand(300000,4)];

for K = 1 : N; start = tic; d = hypot(M(:,2)-M(:,4), M(:,3)-M(:,5)); stop = toc(start); t1(K) = stop; end
for K = 1 : N; start = tic; d = sqrt((M(:,2)-M(:,4)).^2 + (M(:,3)-M(:,5)).^2); stop = toc(start); t2(K) = stop; end

plot([t1, t2])
legend({'hypot', 'sqrt sum'})

[Lina Koronfel]

Thank you for explaining. I got my answer :)