Normalization › Visual Example

Michael BrenndoerferMay 19, 20255 min read
Machine Learning from ScratchAll chapters
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This example demonstrates how normalization maps features with different ranges into a common [0, 1] scale:

Out[2]:
Visualization
Scatter plot comparing age and income features with vastly different scales before normalization.
Original features with vastly different ranges. Age ranges from 20-80 years, while income ranges from 30,000-150,000 dollars. Without normalization, algorithms would be dominated by the income feature due to its larger numeric range.
Scatter plot showing age and income features normalized to same 0-1 scale for fair algorithm treatment.
Normalized features on the same [0, 1] scale. Both features now have minimum 0 and maximum 1. The relative relationships within each feature are preserved, making them comparable.
Original data:
Age - Min: 25 Max: 75
Income - Min: 30000 Max: 130000

Normalized data:
Age - Min: 0.0 Max: 1.0
Income - Min: 0.0 Max: 1.0

Example: Step-by-Step Calculation

Let's work through a detailed example with two features:

  • x1: age in years → [25, 35, 45, 55, 65, 75]
  • x2: income in thousands → [30, 50, 70, 90, 110, 130]

Step 1: Calculate min and max values

  • min⁡(x1)=25\min(x_1) = 25, max⁡(x1)=75\max(x_1) = 75
  • min⁡(x2)=30\min(x_2) = 30, max⁡(x2)=130\max(x_2) = 130

Step 2: Apply normalization formula

For the first feature (j=1j=1):

  • x11′=25−2575−25=050=0.000x'_{11} = \frac{25 - 25}{75 - 25} = \frac{0}{50} = 0.000
  • x21′=35−2575−25=1050=0.200x'_{21} = \frac{35 - 25}{75 - 25} = \frac{10}{50} = 0.200
  • x31′=45−2575−25=2050=0.400x'_{31} = \frac{45 - 25}{75 - 25} = \frac{20}{50} = 0.400
  • x41′=55−2575−25=3050=0.600x'_{41} = \frac{55 - 25}{75 - 25} = \frac{30}{50} = 0.600
  • x51′=65−2575−25=4050=0.800x'_{51} = \frac{65 - 25}{75 - 25} = \frac{40}{50} = 0.800
  • x61′=75−2575−25=5050=1.000x'_{61} = \frac{75 - 25}{75 - 25} = \frac{50}{50} = 1.000

For the second feature (j=2j=2):

  • x12′=30−30130−30=0100=0.000x'_{12} = \frac{30 - 30}{130 - 30} = \frac{0}{100} = 0.000
  • x22′=50−30130−30=20100=0.200x'_{22} = \frac{50 - 30}{130 - 30} = \frac{20}{100} = 0.200
  • x32′=70−30130−30=40100=0.400x'_{32} = \frac{70 - 30}{130 - 30} = \frac{40}{100} = 0.400
  • x42′=90−30130−30=60100=0.600x'_{42} = \frac{90 - 30}{130 - 30} = \frac{60}{100} = 0.600
  • x52′=110−30130−30=80100=0.800x'_{52} = \frac{110 - 30}{130 - 30} = \frac{80}{100} = 0.800
  • x62′=130−30130−30=100100=1.000x'_{62} = \frac{130 - 30}{130 - 30} = \frac{100}{100} = 1.000

Result: Both features are now on the same [0, 1] scale:

  • x1: [25, 35, 45, 55, 65, 75] → [0.000, 0.200, 0.400, 0.600, 0.800, 1.000]
  • x2: [30, 50, 70, 90, 110, 130] → [0.000, 0.200, 0.400, 0.600, 0.800, 1.000]