{"id":434,"date":"2011-12-28T07:00:45","date_gmt":"2011-12-28T12:00:45","guid":{"rendered":"https:\/\/blogs.mathworks.com\/steve\/?p=434"},"modified":"2019-10-29T17:05:26","modified_gmt":"2019-10-29T21:05:26","slug":"modifying-centroid-locations-in-an-image-an-application-of-linear-indexing","status":"publish","type":"post","link":"https:\/\/blogs.mathworks.com\/steve\/2011\/12\/28\/modifying-centroid-locations-in-an-image-an-application-of-linear-indexing\/","title":{"rendered":"Modifying centroid locations in an image – an application of linear indexing"},"content":{"rendered":"
\r\n

Blog reader Mike posed the following question recently<\/a>:\r\n <\/p>\r\n

If you have a bunch of point locations (for example, object centroids), how you make a binary image containing just those\r\n points?\r\n <\/p>\r\n

For example, consider this image:<\/p>

bw = imread('text.png'<\/span>);\r\nimshow(bw, 'InitialMagnification'<\/span>, 200)<\/pre> 
How can we make an image like this, where the dots are located at the centroids of the objects?<\/p>\r\n   
 <\/p>\r\n   
Solving this problem is a nice application of linear indexing, something I wrote about in this blog a long time ago<\/a>. Let's see how it can work for us here.\r\n   <\/p>\r\n   
First, let's find the centroids using regionprops<\/tt>:\r\n   <\/p>
s = regionprops(bw, 'Centroid'<\/span>);<\/pre>
s<\/tt> is a struct array. Since we just asked for one measurement, the centroid, each element of s is a struct containing just one\r\n      field, 'Centroid'<\/tt>.\r\n   <\/p>
s(1)<\/pre>
ans = \r\n    Centroid: [11 13.5000]\r\n<\/pre>
s(2)<\/pre>
ans = \r\n    Centroid: [7.6829 38.1707]\r\n<\/pre>
The length of s<\/tt> is the number of objects in the image.\r\n   <\/p>
num_objects = length(s)<\/pre>
num_objects =\r\n    88\r\n<\/pre>
Next, we gather all the individual centroid locations into x and y vectors. To accomplish this I use the comma-separated list syntax for struct arrays<\/a>.\r\n   <\/p>
centroids = cat(1, s.Centroid);\r\nx = centroids(:,1);\r\ny = centroids(:,2);<\/pre>
If the comma-separated list syntax makes your brain hurt, you can use a loop instead:<\/p>
  centroids = zeros(length(s), 2);\r\n  for k = 1:length(s)\r\n      centroids(k,:) = s(k).Centroid;\r\n  end<\/pre>
Now let's round the centroid locations to get row and column subscripts.<\/p>
r = round(y);\r\nc = round(x);<\/pre>
Here's where linear indexing comes into play. In order to assign to a bunch of scattered locations like this, you want to\r\n      use a single subscript. That's what we call linear indexing. You can use the function sub2ind<\/tt> to convert a set of subscripts to linear indices.\r\n   <\/p>
ind = sub2ind(size(bw), r, c);<\/pre>
And finally we can use the linear indices to assign a value to a bunch of image pixel locations all at once.<\/p>
bw2 = false(size(bw));\r\nbw2(ind) = true;\r\n\r\nimshow(bw2, 'InitialMagnification'<\/span>, 200)<\/pre> 
See my 08-Feb-2008 blog post<\/a> for more about linear indexing.\r\n   <\/p>