Segmentation evaluation (IoU, Dice) in Computer Vision

Introduction

We use segmentation evaluation to check how well a model finds the exact shapes in images. IoU and Dice scores tell us how close the model's guess is to the real shape.

When you want to measure how well a model separates objects from the background in photos.

When comparing different models to see which one finds shapes more accurately.

When improving a model that detects tumors or organs in medical images.

When checking if a self-driving car correctly identifies road lanes or obstacles.

When validating models that segment animals or plants in nature photos.

Syntax

Computer Vision

def iou_score(pred_mask, true_mask):
    intersection = (pred_mask & true_mask).sum()
    union = (pred_mask | true_mask).sum()
    return intersection / union if union != 0 else 1.0

def dice_score(pred_mask, true_mask):
    intersection = (pred_mask & true_mask).sum()
    total = pred_mask.sum() + true_mask.sum()
    return 2 * intersection / total if total != 0 else 1.0

pred_mask and true_mask are binary masks (arrays of True/False or 1/0) of the same size.

IoU is also called Jaccard Index; Dice is similar but gives more weight to overlap.

Examples

This example calculates IoU for small masks where some pixels overlap.

Computer Vision

pred_mask = [0, 1, 1, 0]
true_mask = [1, 1, 0, 0]
iou = iou_score(pred_mask, true_mask)

Dice score is 1.0 here because the prediction perfectly matches the truth.

Computer Vision

pred_mask = [1, 1, 1, 1]
true_mask = [1, 1, 1, 1]
dice = dice_score(pred_mask, true_mask)

Sample Model

This program calculates IoU and Dice scores for two small example masks. It shows how much the predicted shape overlaps with the true shape.

Computer Vision

import numpy as np

def iou_score(pred_mask, true_mask):
    intersection = np.logical_and(pred_mask, true_mask).sum()
    union = np.logical_or(pred_mask, true_mask).sum()
    return intersection / union if union != 0 else 1.0

def dice_score(pred_mask, true_mask):
    intersection = np.logical_and(pred_mask, true_mask).sum()
    total = pred_mask.sum() + true_mask.sum()
    return 2 * intersection / total if total != 0 else 1.0

# Example masks
pred_mask = np.array([[0, 1, 1],
                      [0, 1, 0],
                      [0, 0, 0]], dtype=bool)
true_mask = np.array([[1, 1, 0],
                      [0, 1, 0],
                      [0, 0, 0]], dtype=bool)

print(f"IoU score: {iou_score(pred_mask, true_mask):.2f}")
print(f"Dice score: {dice_score(pred_mask, true_mask):.2f}")

Important Notes

IoU and Dice scores range from 0 to 1, where 1 means perfect overlap.

Always use masks of the same size and shape for correct evaluation.

Dice score tends to be higher than IoU for the same masks because it doubles the intersection.

Summary

IoU and Dice are simple ways to measure how well a model finds shapes in images.

They compare predicted and true masks by looking at overlap and total area.

Higher scores mean better segmentation quality.

Practice

1. What does the Intersection over Union (IoU) metric measure in image segmentation?

easy

A. The overlap between predicted and true masks divided by their union

B. The difference between predicted and true masks

C. The total number of pixels in the predicted mask

D. The ratio of true positives to false negatives

Solution

1. Step 1: Understand IoU definition

   IoU measures how much the predicted mask and true mask overlap compared to their combined area.

2. Step 2: Apply IoU formula

   IoU = (Area of Overlap) / (Area of Union) between predicted and true masks.

3. Final Answer:

   The overlap between predicted and true masks divided by their union -> Option A

4. Quick Check:

   IoU = Overlap / Union [OK]

Hint: IoU = Overlap divided by combined area [OK]

Common Mistakes:

• Confusing IoU with difference or subtraction
• Using only predicted mask area
• Mixing IoU with accuracy or precision

2. Which of the following is the correct formula for Dice coefficient between predicted mask P and true mask T?

easy

A. Dice = |P| / |T|

B. Dice = (2 * |P ∩ T|) / (|P| + |T|)

C. Dice = |P ∪ T| / (|P| + |T|)

D. Dice = |P ∩ T| / |P ∪ T|

Solution

1. Step 1: Recall Dice coefficient formula

   Dice coefficient is defined as twice the overlap divided by the sum of sizes of predicted and true masks.

2. Step 2: Match formula to options

   Dice = (2 * |P ∩ T|) / (|P| + |T|) matches Dice = (2 * intersection) / (sum of areas).

3. Final Answer:

   Dice = (2 * |P ∩ T|) / (|P| + |T|) -> Option B

4. Quick Check:

   Dice = 2*Overlap / (Sum areas) [OK]

Hint: Dice doubles overlap, divides by sum of mask sizes [OK]

Common Mistakes:

• Confusing Dice with IoU formula
• Using union instead of sum of areas
• Forgetting to multiply overlap by 2

3. Given predicted mask pixels: {1,2,3,4} and true mask pixels: {3,4,5,6}, what is the IoU score?

medium

A. 0.33

B. 0.25

C. 0.5

D. 0.4

Solution

1. Step 1: Calculate intersection and union

   Intersection = {3,4} size = 2; Union = {1,2,3,4,5,6} size = 6.

2. Step 2: Compute IoU

   IoU = Intersection / Union = 2 / 6 = 0.3333 (approx). Check options carefully.

3. Step 3: Re-check options

   0.33 is 0.33, closest to calculation. But 0.4 is 0.4, which is incorrect.

4. Final Answer:

   0.33 -> Option A

5. Quick Check:

   IoU = 2/6 = 0.33 [OK]

Hint: IoU = intersection size / union size [OK]

Common Mistakes:

• Counting union incorrectly
• Mixing intersection with union
• Rounding IoU incorrectly

4. You have this Python code to compute Dice score but it returns 0. Identify the error:

def dice_score(pred, true):
    intersection = len(pred & true)
    return intersection / (len(pred) + len(true))

pred = {1,2,3}
true = {2,3,4}
print(dice_score(pred, true))

medium

A. Wrong variable names in function

B. Using set intersection instead of union

C. Incorrect data type for masks

D. Missing multiplication by 2 in numerator

Solution

1. Step 1: Review Dice formula

   Dice = 2 * intersection / (sum of sizes). The code misses multiplying intersection by 2.

2. Step 2: Check code calculation

   Code returns intersection / sum, which is half the correct Dice score.

3. Final Answer:

   Missing multiplication by 2 in numerator -> Option D

4. Quick Check:

   Dice needs 2*intersection [OK]

Hint: Dice numerator must be 2 times intersection [OK]

Common Mistakes:

• Forgetting to multiply intersection by 2
• Confusing intersection with union
• Using wrong data types for masks

5. A segmentation model predicts a mask with 80 pixels, the true mask has 100 pixels, and their overlap is 60 pixels. What is the Dice score and IoU score respectively?

hard

A. Dice = 0.8, IoU = 0.7

B. Dice = 0.75, IoU = 0.6

C. Dice = 0.67, IoU = 0.5

D. Dice = 0.6, IoU = 0.75

Solution

1. Step 1: Calculate Dice score

   Dice = 2 * overlap / (predicted + true) = 2*60 / (80+100) = 120/180 = 0.6667 (approx). Check options carefully.

2. Step 2: Calculate IoU score

   IoU = overlap / union = 60 / (80 + 100 - 60) = 60 / 120 = 0.5.

3. Step 3: Re-check options

   Dice = 0.67, IoU = 0.5 matches Dice ~0.67 and IoU 0.5 exactly.

4. Final Answer:

   Dice = 0.67, IoU = 0.5 -> Option C

5. Quick Check:

   Dice = 2*60/180=0.67, IoU=60/120=0.5 [OK]

Hint: Dice doubles overlap; IoU divides by union [OK]

Common Mistakes:

• Using sum instead of union for IoU denominator
• Forgetting to multiply overlap by 2 for Dice
• Mixing Dice and IoU formulas