# Converting Cartesian Coordinates to Binary image

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Mohanad Alkhodari 2020 年 6 月 14 日
コメント済み: Ameer Hamza 2020 年 6 月 21 日
Hello
I have two vectors of X and Y cartesian values in +/-, how to convert them to a binary image of 0's and 1's
Thank you.
I attached my plot.

#### 2 件のコメント

Aditya Verma 2020 年 6 月 14 日
Hi, Could you specify how you want to map X and Y coordinates to the binary image. If I understand it correctly you want to create a 2-D matrix with binary values (Binary Image).
Mohanad Alkhodari 2020 年 6 月 14 日
Yes, I have the values as X and Y cartesian coordinates.
I want to convert them to pixel values. Like a 2D binary image where a 1 is located on the each location (X,Y).

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### 採用された回答

Ameer Hamza 2020 年 6 月 14 日
Something like this
x = randn(1, 5000); % generate random points
y = randn(1, 5000);
img_dimension = [500 500]; % [width height]
img = zeros(img_dimension);
xt = floor(rescale(x, 1, img_dimension(2))); % convert x-y in range of image coordinates
yt = floor(rescale(y, 1, img_dimension(1)));
idx = sub2ind(img_dimension, yt, xt);
img(idx) = 1;
imshow(img)

#### 10 件のコメント

Ameer Hamza 2020 年 6 月 21 日
I don't clearly understand what the required output is. Can you explain it with an example? Also, you can use imresize() to change the size of an image.
Mohanad Alkhodari 2020 年 6 月 21 日
A circle of radius 0.8 covers more space than a circle of radius 0.2 when converting them from cartesian to binary.
With the above code, both circles are converted for a 500x500 for example and have the same size.
How can I convert to binary while preserving the scaling differences?
Ameer Hamza 2020 年 6 月 21 日
You can make the circle small and large by specifying the range of rescale() function
xt = floor(rescale(x, 100, img_dimension(2)-100)); % convert x-y in range of image coordinates
yt = floor(rescale(y, 100, img_dimension(1)-100));
create a circle between 100 to 400 pixels, i.e., leave 100 pixels on all sides unoccupied.

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### その他の回答 (1 件)

Thiago Henrique Gomes Lobato 2020 年 6 月 14 日
This is an general example you could use:
Cartesian = [ 10,1;
15,10];
imSize=20;
binaryImage= false(imSize);
Cartesian(:,2) = imSize-(Cartesian(:,2)-1); % Comment if you want zero at the top-left corner
for idx=1:size(Cartesian,1)
binaryImage(Cartesian(idx,2),Cartesian(idx,1)) = true; % Note that X-Y index are inverted in indexing
end
figure,imshow(binaryImage)

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