UnivariateSpline in SciPy - Mini Project: Build & Apply

Smooth Data with UnivariateSpline

📖 Scenario: You have collected daily temperature data for a week. The data is a bit noisy, and you want to create a smooth curve to better understand the trend.

🎯 Goal: Use UnivariateSpline from scipy.interpolate to create a smooth curve from the temperature data and visualize the result.

📋 What You'll Learn

Create a dictionary called temperature_data with days as keys and temperatures as values

Create a variable called smoothing_factor to control the smoothness

Use UnivariateSpline with the data and smoothing factor to create a spline object

Generate smooth temperature values for the days using the spline

Print the list of smoothed temperature values

💡 Why This Matters

🌍 Real World

Smoothing noisy data is common in weather forecasting, stock price analysis, and sensor data cleaning.

💼 Career

Data scientists often use smoothing techniques like splines to prepare data for modeling and to visualize trends clearly.

1

Create the temperature data dictionary

Create a dictionary called temperature_data with these exact entries: 1: 22.1, 2: 21.8, 3: 23.0, 4: 22.5, 5: 24.0, 6: 23.5, 7: 24.2.

SciPy

# Create the temperature_data dictionary with days and temperatures
# Your code here

Hint

Use curly braces {} to create a dictionary with keys as days (1 to 7) and values as temperatures.

2

Set the smoothing factor

Create a variable called smoothing_factor and set it to 1 to control the smoothness of the spline.

SciPy

temperature_data = {1: 22.1, 2: 21.8, 3: 23.0, 4: 22.5, 5: 24.0, 6: 23.5, 7: 24.2}
# Create smoothing_factor and set it to 1
# Your code here

Hint

Just create a variable named smoothing_factor and assign the number 1 to it.

3

Create the UnivariateSpline and generate smooth values

Import UnivariateSpline from scipy.interpolate. Then create a spline object called spline using the days and temperatures from temperature_data and the smoothing_factor. Finally, create a list called smoothed_temps by applying spline to each day in temperature_data.

SciPy

temperature_data = {1: 22.1, 2: 21.8, 3: 23.0, 4: 22.5, 5: 24.0, 6: 23.5, 7: 24.2}
smoothing_factor = 1
# Import UnivariateSpline
# Create spline with days, temperatures, and smoothing_factor
# Create smoothed_temps list by applying spline to each day
# Your code here

Hint

Use list(temperature_data.keys()) and list(temperature_data.values()) to get days and temperatures. Then create the spline with s=smoothing_factor. Use a list comprehension to get smoothed values.

4

Print the smoothed temperature values

Write a print statement to display the smoothed_temps list.

SciPy

from scipy.interpolate import UnivariateSpline

days = list(temperature_data.keys())
temps = list(temperature_data.values())
spline = UnivariateSpline(days, temps, s=smoothing_factor)
smoothed_temps = [spline(day) for day in days]
# Print the smoothed_temps list
# Your code here

Hint

Use print(smoothed_temps) to show the smooth temperature values.

Practice

1. What is the primary purpose of UnivariateSpline in scipy?

easy

A. To fit a smooth curve through one-dimensional data points

B. To perform classification on datasets

C. To calculate the mean of data points

D. To generate random numbers

Solution

1. Step 1: Understand the function's goal

   UnivariateSpline is designed to fit a smooth curve through 1D data points, helping to model trends smoothly.

2. Step 2: Compare with other options

   Options B, C, and D describe unrelated tasks like classification, mean calculation, or random number generation, which are not the purpose of UnivariateSpline.

3. Final Answer:

   To fit a smooth curve through one-dimensional data points -> Option A

4. Quick Check:

   UnivariateSpline = smooth curve fitting [OK]

Hint: Remember: UnivariateSpline smooths 1D data points [OK]

Common Mistakes:

• Confusing smoothing with classification
• Thinking it calculates simple statistics
• Assuming it generates random data

2. Which of the following is the correct way to create a UnivariateSpline object with data arrays x and y?

easy

A. spline = UnivariateSpline(x, y, smoothness=1)

B. spline = UnivariateSpline(y, x, s=1)

C. spline = UnivariateSpline(x, y, s=1)

D. spline = UnivariateSpline(x, y, smooth=1)

Solution

1. Step 1: Check parameter order and names

   The correct order is x (independent variable), then y (dependent variable). The smoothing parameter is s, not smooth or smoothness.

2. Step 2: Validate each option

   spline = UnivariateSpline(x, y, s=1) uses correct order and parameter s. spline = UnivariateSpline(y, x, s=1) swaps x and y. Options A and C use incorrect parameter names.

3. Final Answer:

   spline = UnivariateSpline(x, y, s=1) -> Option C

4. Quick Check:

   Correct order and s parameter = spline = UnivariateSpline(x, y, s=1) [OK]

Hint: Remember: x first, y second, smoothing parameter is 's' [OK]

Common Mistakes:

• Swapping x and y arrays
• Using wrong parameter names like 'smooth'
• Omitting the smoothing parameter

3. Given the code:

import numpy as np
from scipy.interpolate import UnivariateSpline

x = np.array([0, 1, 2, 3, 4])
y = np.array([0, 1, 4, 9, 16])
spline = UnivariateSpline(x, y, s=0)
print(round(spline(2.5), 2))

What is the printed output?

medium

A. 8.75

B. 7.5

C. 9.0

D. 6.25

Solution

1. Step 1: Understand the data and spline creation

   The data points are from y = x^2. Using s=0 forces the spline to interpolate exactly through points.

2. Step 2: Calculate spline value at 2.5

   Since the data exactly follows y = x² (degree 2), and UnivariateSpline uses cubic splines (degree 3 ≤ which reproduce polynomials of degree ≤3 exactly), spline(2.5) = 2.5² = 6.25 exactly. round(6.25, 2) prints 6.25.

3. Final Answer:

   6.25 -> Option D

4. Quick Check:

   Exact interpolation with s=0 gives 6.25 at 2.5 [OK]

Hint: With s=0, spline passes through points exactly [OK]

Common Mistakes:

• Thinking spline overshoots or deviates from y = x^2
• Confusing with linear interpolation
• Rounding errors

4. What is wrong with this code snippet?

from scipy.interpolate import UnivariateSpline
x = [0, 1, 2]
y = [1, 2]
spline = UnivariateSpline(x, y)

medium

A. x and y have different lengths causing an error

B. UnivariateSpline requires numpy arrays, not lists

C. Missing smoothing parameter s

D. UnivariateSpline cannot handle integer inputs

Solution

1. Step 1: Check input data lengths

   x has 3 elements, y has 2 elements. They must be the same length for spline fitting.

2. Step 2: Validate other options

   UnivariateSpline accepts lists and integers; smoothing parameter s is optional. The main error is mismatched input lengths.

3. Final Answer:

   x and y have different lengths causing an error -> Option A

4. Quick Check:

   Input length mismatch = x and y have different lengths causing an error [OK]

Hint: Check that x and y arrays have equal length [OK]

Common Mistakes:

• Assuming lists are invalid inputs
• Forgetting that s is optional
• Thinking integer inputs cause errors

5. You have noisy data points x and y representing a sensor reading over time. You want to smooth the data but keep the curve close to points. Which UnivariateSpline parameter should you adjust, and how?

hard

A. Set s to zero for exact interpolation

B. Decrease s to reduce smoothing

C. Increase s to allow more smoothing

D. Remove s parameter to use default smoothing

Solution

1. Step 1: Understand smoothing parameter s

   The parameter s controls smoothness. Larger s means more smoothing, allowing the curve to ignore noise.

2. Step 2: Choose appropriate smoothing for noisy data

   To smooth noisy data but keep curve close, decrease s to reduce smoothing and fit closer to points. Setting s=0 forces exact fit, which keeps noise.

3. Final Answer:

   Decrease s to reduce smoothing -> Option B

4. Quick Check:

   Less smoothing = decrease s [OK]

Hint: Lower s means less smoothing, closer fit [OK]

Common Mistakes:

• Setting s=0 causes overfitting to noise
• Increasing s increases smoothing and may ignore data
• Ignoring s leads to default smoothing that may not fit needs