How to measure the euclidean distance of points continuously from point A to B (and point B to C)?

Hi,

Currently, I have the following code, which I use to measure the euclidean distances from a 3-dimensional point to a reference point (here it is (:,10)) for each cell of doubles in cell_of_double_baskets like so:

H = @(x) sqrt(((x(:,1)-x(:,10)).^2) ...                                         % euclidian distance formula
            + ((x(:,2)-x(:,10)).^2) ...
            + ((x(:,3)-x(:,10)).^2));

baskets_xyz_h_ref = cellfun( H, cell_of_double_baskets, 'uni',false );          % compute the distances

Now, I am trying to measure the distance from each point (x,y and z) to its previous point for every data point in the list in every cell of doubles.

Basically,

H = @(x) sqrt((x2-x1).^2) ...
            + (y2-y1).^2) ...
            + (z2-z1).^2));

but for the entire length of each cell in cell_of_double_baskets. In this case I would disregard column 10 of course.

How would that look? Do I need a for loop?

Thank you!

Answer 1

[Voss]

load('cell_of_double_baskets.mat')
cell_of_double_baskets{1}

ans = 1617×10

0.6930 1.6917 -2.0863 0.5340 0.7811 -2.1244 0.8558 0.8125 -1.9208 0 0.6930 1.6917 -2.0863 0.5343 0.7810 -2.1240 0.8552 0.8125 -1.9210 0 0.6929 1.6918 -2.0864 0.5344 0.7815 -2.1238 0.8553 0.8126 -1.9209 0 0.6927 1.6918 -2.0865 0.5349 0.7820 -2.1232 0.8546 0.8125 -1.9211 0 0.6926 1.6918 -2.0866 0.5354 0.7826 -2.1225 0.8540 0.8126 -1.9212 0 0.6925 1.6917 -2.0867 0.5359 0.7827 -2.1216 0.8537 0.8127 -1.9212 0 0.6923 1.6917 -2.0870 0.5367 0.7827 -2.1200 0.8529 0.8130 -1.9214 0 0.6923 1.6917 -2.0870 0.5371 0.7827 -2.1193 0.8525 0.8130 -1.9215 0 0.6921 1.6916 -2.0872 0.5371 0.7827 -2.1192 0.8526 0.8130 -1.9214 0 0.6921 1.6916 -2.0874 0.5375 0.7826 -2.1185 0.8523 0.8129 -1.9214 0

n = numel(cell_of_double_baskets);
d = cell(1,n);
for kk = 1:n
    % take the difference of adjacent rows of cell_of_double_baskets{kk},
    % first 3 columns only, square the differences, sum them along 2nd 
    % dimension (x,y,z), take the square root, store the result as d{kk}
    d{kk} = sqrt(sum(diff(cell_of_double_baskets{kk}(:,1:3),1,1).^2,2));
end
d

d = 1×19 cell array
    {1616×1 double}    {845×1 double}    {3811×1 double}    {529×1 double}    {5462×1 double}    {600×1 double}    {997×1 double}    {1124×1 double}    {644×1 double}    {954×1 double}    {1746×1 double}    {964×1 double}    {545×1 double}    {1210×1 double}    {3022×1 double}    {940×1 double}    {914×1 double}    {2941×1 double}    {978×1 double}

d{1}

ans = 1616×1

1.0e+00 * 0 0.0001 0.0002 0.0002 0.0002 0.0004 0 0.0002 0.0002 0.0002

[lil brain]

@_ unfortunately you understood that slightly wrong. I am only interested in the first three columns of each cell [x y z]. Sorry if this wasnt made clear by me.

I am trying to get the distance/difference between the values in each consecutive row.

So basically

cell_of_double_baskets{1, 1} (2,x) - cell_of_double_baskets{1, 1} (1,x)

cell_of_double_baskets{1, 1} (3,x) - cell_of_double_baskets{1, 1} (2,x)

cell_of_double_baskets{1, 1} (4,x) - cell_of_double_baskets{1, 1} (3,x)

...

Does that make it clearer? (Also have a look at my last comment on the above answer @Image Analyst)

[Voss]

I edited my answer after seeing your comment on the other answer. Please check my answer again.

Answer 2

[Image Analyst]

I think you could use a loop

curveDistance = 0;
dx = diff(x);
dy = diff(y);
dz = diff(z);
for k = 1 : length(dx)
    curveDistance = curveDistance + sqrt(dx(k)^2 + dy(k)^2 + dz(k)^2);
end

[lil brain]

Hi @Image Analyst

thanks!

But if I run

dx = diff(x);

shouldnt I get half the number of rows?

dx = diff(cell_of_double_baskets{1,1}(:,1)) has the same amount of rows as cell_of_double_baskets{1,1}(:,1).

What am I doing wrong?

[Image Analyst]

Why? It's just the difference between adjacent elements so you would have one less. Look at this example for 4 elements

x = sort(randi(50, 1, 4))

x = 1×4

24 28 29 43

dx = diff(x)

dx = 1×3

4 1 14

[lil brain]

Yes you are right. I was confusing diff with something else.

And how would I write your code if I want to apply it for every of the 19 cells in cell_of_doubles_baskets?

I would need another for loop inside the for loop right?

Thanks!

[Image Analyst]

I think that's simplest - easier to understand than cellfun(). So something like

for k2 = 1 : numel(cell_of_double_baskets) % For every cell in the cell array
    thisCellContents = cell_of_double_baskets{k2}; % Get contents of cell
    % For example if thisCellContents is an N-by-3 array of x,y,z columns
    x = thisCellContents(:, 1);
    y = thisCellContents(:, 2);
    z = thisCellContents(:, 3);
    curveDistance(k2) = 0;
    dx = diff(x);
    dy = diff(y);
    dz = diff(z);
    for k = 1 : length(dx)
        curveDistance(k2) = curveDistance(k2) + sqrt(dx(k)^2 + dy(k)^2 + dz(k)^2);
    end
end

[lil brain]

Hi @Image Analyst,

thank you. I tried your code but I think there might be a misunderstanding because it only yields one measurement for each cell.

What I am trying to achieve is one list of measurements for each cell like so:

cell_1_distances = sqrt(dx.^2 + dy.^2 + dz.^2);

This way the output would be a cell of doubles where each cell contains a list of distances with different lengths for each cell.

How would your for loop need to be changed in that case?

Sorry for the many questions!

[Image Analyst]

Maybe a diagram would help. I'm assuming each cell has a list of x,y,z coordinates and you want to add up the length of all the line segments as you daisy-chain your way from one point to the next. So each cell would have one curve length. Apparently that is not what you have so can you make up a line diagram showing what distances are to be measured?

[lil brain]

I see, you thought I wanted to add them up.

However, I need each distance individually.

I was trying to get a list of distances between each data point (x1-x2, x3-x2, x4-x3, etc.) for each cell.

Basically,

sqrt(dx.^2 + dy.^2 + dz.^2);

for each row for the entire length of each cell.

The result would be a list of distances (or differences) that looks like so:

list_of_distances_per_cell =

1×19 cell array

Columns 1 through 5

{1616×1 double} {845×1 double} {3811×1 double} {529×1 double} {5462×1 double} ... {nx1 double}

I hope this is more clear than before.

Ps: in cell_of_double_baskets only the first three columns for each cell are of importance. The others can be disregarded.

[Image Analyst]

What about using a 2-D array

% Make 2d array with enough columns to hold any array, like the max of
% length(dx) you ever expect to encounter.
curveDistance = zeros(numel(cell_of_double_baskets), 100000)
lastColumn = 100000;
for k2 = 1 : numel(cell_of_double_baskets) % For every cell in the cell array
    thisCellContents = cell_of_double_baskets{k2}; % Get contents of cell
    % For example if thisCellContents is an N-by-3 array of x,y,z columns
    x = thisCellContents(:, 1);
    y = thisCellContents(:, 2);
    z = thisCellContents(:, 3);
    curveDistance(k2) = 0;
    dx = diff(x);
    dy = diff(y);
    dz = diff(z);
    curveDistance(k2, 1) = sqrt(dx(1)^2 + dy(1)^2 + dz(1)^2);
    for k = 2 : length(dx)
        curveDistance(k2, k) = curveDistance(k2, k-1) + sqrt(dx(k)^2 + dy(k)^2 + dz(k)^2);
    end
    lastColumn = max([length(dx), lastColumn]); % Keep track of the most number of columns we're going to use.
end
% Crop off unused columns.
curveDistance = curveDistance(:, 1:lastColumn);