Skeletonization
Prerequisites
Before starting this lesson, you should be familiar with:
Learning Objectives
After completing this lesson, learners should be able to:
Apply a skeletonization algorithm to a binary image to view its internal skeleton
Count the number of branches and branch lengths to obtain morphological information from the image
Motivation
For objects that contain protrusions, it can be helpful to look at the object’s internal skeleton. This reveals the inner branches that make up the object. Measuring the number of branches and their lengths can provide useful morphological information of irregularly shaped objects with protrusions, such as glial cells. Skeletonization algorithms work by applying sequential erosions to remove pixels from the boundary of the objects to the center, stopping when the remaining structure is only one pixel wide.
Concept map
Figure
Activities
- Open the binary image xy_8bit_glialcells.tif.
- Perform skeletonization.
- Analyze branch properties in the two different cells.
Show activity for:
ImageJ GUI
- Skeletonize a binary imgae and obtain branch information.
- Open xy_8bit_glialcells.tif
- Perform skeletonization: [ Process › Binary › Skeletonize]
- Obtain branch information by analyzing the skeleton: [Analyze › Skeleton › Analyze Skeleton (2D/3D)]
- ‘Prune ends’
- ‘Calculate largest shortest path’, ‘Show detailed info’, ‘Display labeled skeletons’.
- In the results window, you can find information about the different skeletons and their branches, such as the number of branches and junctions, the longest shortest path, and the average branch length.
ImageJ Macro
ImageJ Jython
Exercises
Perform skeletonization and skeleton analysis on this image: xy_8bit_glialcells2.tif.
Try to answer the following questions:
-
Which cell has the largest number of branches?
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Which cell has the longest “longest shortest path”?
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Which cell has the highest average branch length?
ImageJ GUI
- Open xy_8bit_glialcells2.tif
- Perform skeletonization: [ Process › Binary › Skeletonize]
- Obtain branch information by analyzing the skeleton: [Analyze › Skeleton › Analyze Skeleton (2D/3D)]
- ‘Prune ends’
- ‘Calculate largest shortest path’, ‘Show detailed info’, ‘Display labeled skeletons’.
- In the results window you can find that cell 3 has the largest number of branches, cell 1 has the longest “longest shortest path” and cell 2 has the highest average branch length.
ImageJ Macro
// Open binary image and perform skeletonization // Open image open("https://github.com/NEUBIAS/training-resources/raw/master/image_data/xy_8bit_glialcells2.tif"); // Duplicate the image run("Duplicate...", " "); // Perform skeletonization run("Skeletonize"); // Obtain branch properties run("Analyze Skeleton (2D/3D)", "prune=none calculate show display"); run("Tile") // check the data in the results window to answer the questions.
ImageJ Jython
# Open image and perform skeletonization # import classes from ij import IJ # open image imp = IJ.openImage("https://github.com/NEUBIAS/training-resources/raw/master/image_data/xy_8bit_glialcells2.tif") imp.show() # perform skeletonization skeleton = imp.duplicate() IJ.run(skeleton, "Skeletonize", "") skeleton.show() # analyze the skeleton IJ.run(skeleton, "Analyze Skeleton (2D/3D)", "prune=none calculate show display") IJ.run("Tile") # check the data in the results window to answer the questions.
Assessment
True or False
- Slab pixels never overlap with boundary pixels in the original binary image.
- Branches in the skeleton can be more than 1 pixel thick.
- The longest shortest path is the longest branch in the skeleton.
Solution
- Slab pixels never overlap with boundary pixels in the original binary image. True
- Branches in the skeleton can be more than 1 pixel thick. False. They can be longer than 1 pixel, but the branch thickness is always 1 pixel.
- The longest shortest path is the longest branch in the skeleton. False
Explanations
Follow-up material
Recommended follow-up modules:
Learn more: