Use conditional statements to separate elements from a single array into 2 new arrays
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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 个评论
Geoff Hayes
2019-8-15
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?
Walter Roberson
2019-8-15
If HS_elim has more than one row then the && will fail because && requires scalars on both sides of the &&
Vance Blake
2019-8-16
编辑：Vance Blake
2019-8-16
采纳的回答
Bjorn Gustavsson
2019-8-16
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 个评论
Guillaume
2019-8-16
Note that:
elim_dist2 = nan(numel(x));
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
is:
elim_dist2 = hypot(x - x.', y - y.'); %assuming >=R2016b
elim_dist2(logical(eye(elim_dist2))) = NaN;
Also, since 2015a, you no longer need nanmin (which requires the stats toolbox), the basic min can omit nan:
i2keep = min(elim_dist2) > HS_HS_threshold, 'omitnan'); %and the find is also not needed. Just use the logical vector directly
Bjorn Gustavsson
2019-8-16
@Vance: Guillaumes solution is neater and more easily maintained - in addition to most likely being more computationally efficient. It relies on the recent extension of '+', and '-' to automatically do outer sum and outer subtraction (working similarly to outer multiplication of matrices). I thought you should see some more low-level tricks of matlab-programming first.
@Guillaume: Stepwise towards golf.
Guillaume
2019-8-16
Well, on earlier than R2016b:
elim_dist2 = hypot(bxsfun(@minus, x, x.'), bsxfun(@minus, y, y'));
elim_dist2(logical(eye(elim_dist2))) = NaN;
and since we're golfing:
i2keep = min(elim_dist2) > HS_HS_threshold, 'omitnan');
keep_x = x(i2keep);
keep_y = y(i2keep);
x_close_neighbours = x(~i2keep);
y_close_neighbours = y(~i2keep);
Vance Blake
2019-8-28
Guillaume
2019-8-28
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.
Vance Blake
2019-8-28
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