Interpolation for smoothing data in SciPy - Mini Project: Build & Apply

Interpolation for smoothing data

📖 Scenario: You have some noisy temperature data collected every hour. You want to smooth the data to see a clearer trend.

🎯 Goal: Use interpolation to smooth the noisy temperature data and display the smoothed values.

📋 What You'll Learn

Create a dictionary with hourly temperature data

Create a variable for new time points to interpolate

Use scipy's interp1d to create an interpolation function

Print the smoothed temperature values for the new time points

💡 Why This Matters

🌍 Real World

Smoothing noisy sensor data to better understand trends in temperature, stock prices, or other measurements.

💼 Career

Data scientists often need to clean and smooth data before analysis or modeling to improve accuracy and insights.

1

Create the noisy temperature data

Create a dictionary called temp_data with these exact entries: 0: 15.0, 1: 15.5, 2: 14.8, 3: 16.2, 4: 15.9, 5: 16.5 representing temperature readings at hours 0 to 5.

SciPy

# Create the temperature data dictionary
# Your code here

Hint

Use curly braces to create a dictionary with keys 0 to 5 and the given temperature values.

2

Create new time points for interpolation

Create a list called new_hours with these exact values: 0.5, 1.5, 2.5, 3.5, 4.5 representing the half-hour points where you want to estimate temperatures.

SciPy

temp_data = {0: 15.0, 1: 15.5, 2: 14.8, 3: 16.2, 4: 15.9, 5: 16.5}
# Create the list of new hours for interpolation
# Your code here

Hint

Use square brackets to create a list with the given float values.

3

Create interpolation function using scipy

Import interp1d from scipy.interpolate. Then create a variable called interp_func by calling interp1d with the keys and values of temp_data as x and y, respectively, and set kind='linear'.

SciPy

temp_data = {0: 15.0, 1: 15.5, 2: 14.8, 3: 16.2, 4: 15.9, 5: 16.5}
new_hours = [0.5, 1.5, 2.5, 3.5, 4.5]
# Import interp1d and create interp_func with linear kind
# Your code here

Hint

Use list(temp_data.keys()) and list(temp_data.values()) to get x and y values for interpolation.

4

Print the smoothed temperature values

Use print() to display the result of calling interp_func with new_hours as input.

SciPy

from scipy.interpolate import interp1d

temp_data = {0: 15.0, 1: 15.5, 2: 14.8, 3: 16.2, 4: 15.9, 5: 16.5}
new_hours = [0.5, 1.5, 2.5, 3.5, 4.5]

interp_func = interp1d(list(temp_data.keys()), list(temp_data.values()), kind='linear')
# Print the interpolated temperatures for new_hours
# Your code here

Hint

Call interp_func(new_hours) inside print() to see the smoothed values.

Practice

1. What is the main purpose of using interpolation in data smoothing with SciPy?

easy

A. To convert data into categorical variables

B. To remove all data points from a dataset

C. To sort data points in ascending order

D. To estimate values between known data points

Solution

1. Step 1: Understand interpolation concept

   Interpolation is used to estimate or predict values between existing data points to create a smooth curve.

2. Step 2: Identify the correct purpose

   Among the options, only estimating values between known points matches interpolation's goal.

3. Final Answer:

   To estimate values between known data points -> Option D

4. Quick Check:

   Interpolation = estimate between points [OK]

Hint: Interpolation fills gaps between points [OK]

Common Mistakes:

• Confusing interpolation with data sorting
• Thinking interpolation removes data points
• Mixing interpolation with data categorization

2. Which of the following is the correct way to create a linear interpolation function using SciPy's interp1d?

easy

A. interp1d(x, y, type='linear')

B. interp1d(x, y, kind='linear')

C. interp1d(x, y, method='linear')

D. interp1d(x, y, mode='linear')

Solution

1. Step 1: Recall interp1d syntax

   The correct parameter to specify interpolation type is kind, not method, type, or mode.

2. Step 2: Match correct parameter usage

   Using kind='linear' is the proper way to create a linear interpolation function.

3. Final Answer:

   interp1d(x, y, kind='linear') -> Option B

4. Quick Check:

   Parameter for interpolation type = kind [OK]

Hint: Use 'kind' to set interpolation type [OK]

Common Mistakes:

• Using 'method' instead of 'kind'
• Using 'type' or 'mode' parameters incorrectly
• Forgetting to specify interpolation kind

3. Given the code below, what will be the output of y_new?

import numpy as np
from scipy.interpolate import interp1d
x = np.array([0, 1, 2, 3])
y = np.array([0, 1, 4, 9])
f = interp1d(x, y, kind='quadratic')
y_new = f(1.5)

medium

A. 2.25

B. 3.5

C. 2.0

D. Error due to insufficient points

Solution

1. Step 1: Understand quadratic interpolation

   Quadratic interpolation fits a parabola through points. At x=1.5, it estimates y between points (1,1) and (2,4).

2. Step 2: Calculate expected value

   Since y = x^2, at 1.5, y = 1.5^2 = 2.25, matching quadratic interpolation output.

3. Final Answer:

   2.25 -> Option A

4. Quick Check:

   Quadratic interp at 1.5 = 2.25 [OK]

Hint: Quadratic interp matches square values here [OK]

Common Mistakes:

• Choosing linear interpolation value
• Assuming error due to points count
• Miscomputing interpolation value

4. Identify the error in the following code snippet for smoothing data using cubic interpolation:

import numpy as np
from scipy.interpolate import interp1d
x = np.array([0, 1, 2])
y = np.array([1, 3, 7])
f = interp1d(x, y, kind='cubic')
y_new = f(1.5)

medium

A. Not enough data points for cubic interpolation

B. Incorrect import statement for interp1d

C. Wrong parameter name for interpolation kind

D. x and y arrays have different lengths

Solution

1. Step 1: Check data points for cubic interpolation

   Cubic interpolation requires at least 4 data points; here only 3 points are provided.

2. Step 2: Identify the error cause

   Because of insufficient points, interp1d will raise an error for cubic kind.

3. Final Answer:

   Not enough data points for cubic interpolation -> Option A

4. Quick Check:

   Cubic interp needs ≥4 points [OK]

Hint: Cubic interpolation needs at least 4 points [OK]

Common Mistakes:

• Assuming import error
• Confusing parameter names
• Ignoring array length mismatch

5. You have noisy temperature data recorded hourly for 10 hours: x = np.arange(10) and y contains noisy values. You want a smooth curve to estimate temperatures every 0.1 hour using cubic interpolation. Which code snippet correctly achieves this smoothing?

hard

A. f = interp1d(x, y, kind='cubic'); x_new = np.arange(0, 9.1, 0.1); y_smooth = f(x_new)

B. f = interp1d(x, y, kind='linear'); x_new = np.linspace(0, 9, 91); y_smooth = f(x_new)

C. f = interp1d(x, y, kind='cubic'); x_new = np.linspace(0, 9, 91); y_smooth = f(x_new)

D. f = interp1d(x, y, kind='quadratic'); x_new = np.arange(0, 10, 0.1); y_smooth = f(x_new)

Solution

1. Step 1: Choose correct interpolation kind and new x values

   Cubic interpolation is requested, so kind='cubic' is correct. For smooth sampling every 0.1 hour over 0 to 9, use np.linspace(0, 9, 91) for evenly spaced points.

2. Step 2: Verify code correctness

   f = interp1d(x, y, kind='cubic'); x_new = np.linspace(0, 9, 91); y_smooth = f(x_new) uses cubic kind and linspace for x_new, then applies f(x_new) to get smooth y values, matching requirements.

3. Final Answer:

   f = interp1d(x, y, kind='cubic'); x_new = np.linspace(0, 9, 91); y_smooth = f(x_new) -> Option C

4. Quick Check:

   Cubic interp + linspace for smooth curve [OK]

Hint: Use cubic kind and linspace for smooth interpolation [OK]

Common Mistakes:

• Using linear or quadratic instead of cubic
• Using arange with wrong stop value
• Not matching x_new range with original data