interp1d for 1D interpolation in SciPy - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to import the interpolation function from scipy.

SciPy

from scipy.interpolate import [1]

Drag options to blanks, or click blank then click option'

Ainterp1d

Binterpolate1d

Cinterp2d

Dinterpolate

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'interpolate1d' which does not exist.

Confusing with 'interp2d' which is for 2D interpolation.

The function interp1d is the correct import for 1D interpolation in scipy.

2fill in blank

medium

Complete the code to create an interpolation function for given x and y arrays.

SciPy

f = [1](x, y)

Drag options to blanks, or click blank then click option'

Ainterpolate1d

Binterp1d

Cinterp2d

Dinterpolate

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'interp2d' which is for 2D interpolation.

Using 'interpolate' which is a module, not a function.

The interp1d function creates a callable interpolation function from x and y data.

3fill in blank

hard

Fix the error in the code to evaluate the interpolation function at new points.

SciPy

y_new = f([1])

Drag options to blanks, or click blank then click option'

Ay_new

Bx

Cx_new

Df

Attempts:

3 left

💡 Hint

Common Mistakes

Passing y-values instead of x-values to the interpolation function.

Passing the function itself as argument.

To get interpolated values, call the function f with new x-values, here x_new.

4fill in blank

hard

Fill both blanks to create an interpolation function with linear method and evaluate it at new points.

SciPy

f = interp1d(x, y, kind=[1])
y_new = f([2])

Drag options to blanks, or click blank then click option'

A'linear'

Bx_new

C'nearest'

Dy_new

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'nearest' instead of 'linear' for the kind parameter.

Passing y_new instead of x_new to the function.

Use kind='linear' to specify linear interpolation and call the function with x_new to get interpolated values.

5fill in blank

hard

Fill all three blanks to create a cubic interpolation function, evaluate it at new points, and convert the result to a list.

SciPy

f = interp1d(x, y, kind=[1])
y_new = f([2])
y_list = [3](y_new)

Drag options to blanks, or click blank then click option'

A'cubic'

Bx_new

Clist

Darray

Attempts:

3 left

💡 Hint

Common Mistakes

Using 'linear' or 'nearest' instead of 'cubic' for kind.

Not converting the result to a list.

Passing y_new instead of x_new to the function.

Use kind='cubic' for cubic interpolation, evaluate at x_new, and convert the result to a list with list().

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