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Use conditional statements to separate elements from a single array into 2 new arrays

Asked by Vance Blake on 15 Aug 2019
Latest activity Commented on by Vance Blake on 28 Aug 2019
Hello, I am trying to separate elements from a matrix A (10x2) using conditional statements. I compare the distances between the all the points in A and if they are smaller than the threshold (16) then the pair of points that created that produced that result are isolated in another matrix (HS_elim). I want to use that newly created elim matrix to determine the points that wont be eliminated and instead sent to another matrix (B). I am able to isolated the points that meet my if condition but I cannot isolate the points that fail the condition. the snippet of code I am using is below. Any help would be greatly appreciated.
HS_HS_threshold = 16;
for i = 1:n
for j = 1:n
elim_dist2(i,j) = sqrt((x(i)-x(j)).^2 + (y(i)-y(j)).^2);
if (elim_dist2(i,j) ~= 0 && elim_dist2(i,j) < HS_HS_threshold)
HS_elim_x = x(j); HS_elim_y = y(j);
HS_elim(j,:) = [HS_elim_x, HS_elim_y];
elseif HS_elim(:,1) ~= x(i) && HS_elim(:,2) ~= y(i)
keep_x = x(i); keep_y = y(i);
B(i,:) = [keep_x, keep_y];
end
end
end

  3 Comments

Vance - what is the intent behind this condition
HS_elim(:,1) ~= x(i) && HS_elim(:,2) ~= y(i)
You are comparing two columns with (presumably) scalars x and y. If you just want to isolate the points that fail the condition then why not replace the elseif with an else?
If HS_elim has more than one row then the && will fail because && requires scalars on both sides of the &&
@Walter Oh so is that why the second matrix keeps collecting all the points instead of just the points that aren't too close to each other??
@Geoff I tried else first but the outcome was still incorrect because of what Walter explained if Im understanding correctly.

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1 Answer

Answer by Bjorn Gustavsson on 16 Aug 2019
 Accepted Answer

Are you trying to separate points that have no neighbours closer than HS_HS_threshold from those that have?
If so this should do something like that:
elim_dist2 = nan(numel(x)); % this to have nans on the diagonal after distance calculation
HS_HS_threshold = 16;
for i = n:-1:1 % looping from largest index lets you avoid calculating the size of the elim_dist2 matrix without pying the price of dynamic growing of a matrix
for j = (i-1):-1:1 % distance symmetric (l(i1,i2) == l(i2,i1)) so calculate it once
elim_dist2(i,j) = sqrt((x(i)-x(j)).^2 + (y(i)-y(j)).^2);
elim_dist2(j,i) = elim_dist2(i,j);
end
end
% find the points that have its nearest neighbour further away than HS_HS_threshold:
i2keep = find(nanmin(elim_dist2)>HS_HS_threshold);
% put those into one pair of arrays
keep_x = x(i2keep);
keep_y = y(i2keep);
% and the others into another pair of arrays
x_close_neighbours = x;
y_close_neighbours = y;
x_close_neighbours(i2keep) = [];
y_close_neighbours(i2keep) = [];
HTH

  6 Comments

Hi i know this is a late follow up but how do I modify the above code to work with a variable number of points because I have attempted to do so but it still gives me a 10x10 matrix despite only have 4 points in one matrix and 2 in the other matrix ?
None of the pieces of code you were given assume anything about the size of the inputs x and y (other than they are vectors). For a vector of N points, all the example will give you a NxN matrix for elim_dist2.
If that's not the case for you, then you're doing something different. It's clearly different since your original code worked with two vectors of the same length and you're now telling us that you have two matrices of different size instead. Please clarify what you're doing by showing us the actual code.
Oh I acutally figured out my problem. I needed to set up the initial NaN matrix for the bigger of the 2 sets of vectors not matrices (my bad on the wrong terminology) and then fill out the resulting matrix with the distance values accordingly. Therefore avoidng the problem of comparing vectors of unequal size and gettting 0s. Basically i didnt realize just how good of a solution you guys gave me. thanks for the help both now and then.

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