interp1d for 1D interpolation in SciPy - Mini Project: Build & Apply

Using interp1d for 1D Interpolation

📖 Scenario: Imagine you have temperature readings taken at certain hours of the day, but you want to estimate the temperature at times between those hours.

🎯 Goal: You will create a simple program that uses interp1d from scipy to estimate temperatures at new times based on existing data.

📋 What You'll Learn

Create arrays for known time points and their corresponding temperatures

Set up an interpolation function using interp1d

Use the interpolation function to find temperatures at new time points

Print the interpolated temperature values

💡 Why This Matters

🌍 Real World

Interpolation helps estimate unknown values between measured data points, useful in weather forecasting, finance, and engineering.

💼 Career

Data scientists often use interpolation to fill missing data or create smooth curves from discrete data points.

1

Create arrays for time and temperature data

Create a NumPy array called time_hours with values [0, 3, 6, 9, 12] representing hours of the day. Create another NumPy array called temperatures with values [15, 18, 21, 24, 27] representing temperatures at those hours.

SciPy

import numpy as np
# Create the time_hours array with values [0, 3, 6, 9, 12]
# Create the temperatures array with values [15, 18, 21, 24, 27]

Hint

Use np.array to create arrays with the exact values given.

2

Set up the interpolation function

Import interp1d from scipy.interpolate. Create a variable called interp_func by calling interp1d with time_hours and temperatures as arguments.

SciPy

import numpy as np
from scipy.interpolate import interp1d

time_hours = np.array([0, 3, 6, 9, 12])
temperatures = np.array([15, 18, 21, 24, 27])
# Create interp_func using interp1d with time_hours and temperatures

Hint

Use from scipy.interpolate import interp1d and then call interp1d(time_hours, temperatures).

3

Interpolate temperatures at new times

Create a NumPy array called new_times with values [1, 4, 7, 10]. Use interp_func to find interpolated temperatures at new_times and store the result in a variable called new_temperatures.

SciPy

import numpy as np
from scipy.interpolate import interp1d

time_hours = np.array([0, 3, 6, 9, 12])
temperatures = np.array([15, 18, 21, 24, 27])

interp_func = interp1d(time_hours, temperatures)
# Create new_times array with [1, 4, 7, 10]
# Use interp_func to get new_temperatures at new_times

Hint

Create new_times with the given values and call interp_func(new_times).

4

Print the interpolated temperatures

Write a print statement to display the new_temperatures array.

SciPy

import numpy as np
from scipy.interpolate import interp1d

time_hours = np.array([0, 3, 6, 9, 12])
temperatures = np.array([15, 18, 21, 24, 27])

interp_func = interp1d(time_hours, temperatures)

new_times = np.array([1, 4, 7, 10])
new_temperatures = interp_func(new_times)
# Print the new_temperatures array

Hint

Use print(new_temperatures) to show the interpolated values.

Practice

1. What is the main purpose of interp1d in scipy?

easy

A. To sort data points in ascending order

B. To estimate values between known data points

C. To calculate the derivative of a function

D. To perform matrix multiplication

Solution

1. Step 1: Understand the function role

   interp1d is used to create a function that estimates values between given points.

2. Step 2: Compare options with purpose

   Only To estimate values between known data points describes estimating values between known points, which matches interp1d's purpose.

3. Final Answer:

   To estimate values between known data points -> Option B

4. Quick Check:

   interp1d = estimate between points [OK]

Hint: interp1d fills gaps between points [OK]

Common Mistakes:

• Confusing interpolation with sorting
• Thinking it calculates derivatives
• Mixing it up with matrix operations

2. Which of the following is the correct way to create a linear interpolation function using interp1d with x and y arrays?

easy

A. f = interp1d(x, y, method='linear')

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

C. f = interp1d(x, y, kind='linear')

D. f = interp1d(x, y, kind='smooth')

Solution

1. Step 1: Check parameter order and names

   interp1d takes x and y arrays in order, and the interpolation type is set by kind.

2. Step 2: Validate options

   f = interp1d(x, y, kind='linear') uses correct order and kind='linear'. f = interp1d(y, x, kind='linear') swaps x and y, which is wrong. f = interp1d(x, y, method='linear') uses wrong parameter name method. f = interp1d(x, y, kind='smooth') uses invalid kind 'smooth'.

3. Final Answer:

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

4. Quick Check:

   Correct syntax uses kind='linear' [OK]

Hint: Use kind='linear' and x,y in order [OK]

Common Mistakes:

• Swapping x and y arrays
• Using wrong parameter name 'method'
• Using unsupported kind values

3. Given the code:

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='linear')
print(f(1.5))

What is the output?

medium

A. 2.5

B. 3.25

C. 4.5

D. Error

Solution

1. Step 1: Understand interpolation at 1.5

   Between x=1 (y=1) and x=2 (y=4), linear interpolation at 1.5 is halfway: (1 + 4)/2 = 2.5.

2. Step 2: Confirm calculation

   Linear interpolation formula: y = y1 + (x - x1)*(y2 - y1)/(x2 - x1) = 1 + (1.5-1)*(4-1)/(2-1) = 1 + 0.5*3 = 2.5.

3. Final Answer:

   2.5 -> Option A

4. Quick Check:

   Linear interp at 1.5 = 2.5 [OK]

Hint: Linear interp averages between points [OK]

Common Mistakes:

• Using wrong points for interpolation
• Calculating midpoint incorrectly
• Expecting quadratic interpolation result

4. What is wrong with this code snippet?

from scipy.interpolate import interp1d
x = [0, 1, 2]
y = [1, 4]
f = interp1d(x, y)
print(f(1.5))

medium

A. x and y arrays have different lengths

B. interp1d requires x to be sorted descending

C. Missing kind parameter causes error

D. Cannot interpolate at 1.5 because it's outside x range

Solution

1. Step 1: Check lengths of x and y

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

2. Step 2: Understand interp1d requirements

   Different lengths cause a ValueError. Sorting or kind parameter is not mandatory. 1.5 is within x range (0 to 2).

3. Final Answer:

   x and y arrays have different lengths -> Option A

4. Quick Check:

   Equal length arrays needed [OK]

Hint: Check x and y lengths match [OK]

Common Mistakes:

• Assuming kind is mandatory
• Thinking x must be descending
• Confusing interpolation range limits

5. You have temperature data recorded every hour: x = [0, 1, 2, 3, 4] hours and y = [15, 17, 20, 22, 21] degrees Celsius. You want to estimate temperature at half-hour intervals using cubic interpolation. Which code correctly creates the interpolation function and estimates temperature at 2.5 hours?

hard

A. f = interp1d(x, y, kind='linear')\nprint(f(2.5))

B. f = interp1d(x, y, kind='quadratic')\nprint(f(2.5))

C. f = interp1d(y, x, kind='cubic')\nprint(f(2.5))

D. f = interp1d(x, y, kind='cubic')\nprint(f(2.5))

Solution

1. Step 1: Identify correct parameters for cubic interpolation

   Use kind='cubic' with x and y arrays in correct order.

2. Step 2: Check options

   f = interp1d(x, y, kind='cubic')\nprint(f(2.5)) uses correct order and cubic kind. f = interp1d(x, y, kind='linear')\nprint(f(2.5)) uses linear, not cubic. f = interp1d(y, x, kind='cubic')\nprint(f(2.5)) swaps x and y, which is wrong. f = interp1d(x, y, kind='quadratic')\nprint(f(2.5)) uses quadratic kind instead of cubic.

3. Final Answer:

   f = interp1d(x, y, kind='cubic') print(f(2.5)) -> Option D

4. Quick Check:

   Cubic kind with correct x,y order [OK]

Hint: Use kind='cubic' and correct x,y order [OK]

Common Mistakes:

• Swapping x and y arrays
• Using quadratic instead of cubic
• Choosing linear instead of cubic for smoothness