Remove duplicate detections with non-maximum suppression
Keep one box per object by suppressing weaker boxes that overlap a stronger one.
- Explain why detectors produce duplicate boxes
- Run non-maximum suppression step by step
- Choose the overlap threshold
Detectors check many candidate positions, so one object usually sets off a cluster of overlapping boxes, each with a confidence score. Non-maximum suppression (NMS) cleans this up: keep the most confident box, delete the boxes that overlap it too much (they are probably the same object), and repeat with the next most confident box still standing.
Try it
Step through NMS
Two cats sitting close together, five boxes. Step through the algorithm and read why each box is kept or suppressed.
- How many objects does it find at threshold 0.5?
- Slide the threshold up to 0.9: what goes wrong?
- Slide it down to 0.1: what goes wrong now?
- Box 1 score 0.92next
- Box 4 score 0.88waiting
- Box 2 score 0.81waiting
- Box 3 score 0.67waiting
- Box 5 score 0.55waiting
Detectors often fire several times on the same object. Press “Next step” to process the boxes from highest score to lowest.
The overlap threshold is a trade-off. Too high, and duplicates survive (one cat counted three times). Too low, and genuinely different objects that overlap - two people in a crowd - get merged into one. 0.5 is a common default.
Key takeaways
Detectors fire several times per object; NMS keeps one box each.
Greedy loop: keep the highest score, drop boxes overlapping it above the threshold, repeat.
Too high a threshold keeps duplicates; too low merges neighbors.
Lesson quiz
6 questions · pass with 5 correct · up to 50 XP
Passing this quiz completes the lesson and keeps your streak going. Questions you miss come back in review sessions later.
Practice: apply computer vision with Python
Use small pixel arrays to explore vision concepts, run your code against sample images, and connect each result to the larger computer vision idea.
Implement NMS
The first line is N, then N boxes, one per line, as score x1 y1 x2 y2. The last line is the overlap threshold.
Run non-maximum suppression: go through the boxes from highest to lowest score; keep a box unless its IoU with an already kept box is above the threshold. Print the kept boxes’ numbers (1-based, in input order) in the order they were kept, separated by spaces.
- Two objects, one duplicate
- Best box listed last
- Generous threshold keeps both
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