Spline interpolation in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of cubic spline interpolation at given points

Given the code below using SciPy's CubicSpline, what is the output array printed?

SciPy

import numpy as np
from scipy.interpolate import CubicSpline

x = np.array([0, 1, 2, 3, 4])
y = np.array([0, 1, 0, 1, 0])
cs = CubicSpline(x, y)
print(cs([1.5, 2.5, 3.5]))

A [0.75 0.75 0.25]

B [1.0 0.0 1.0]

C [0.5 0.5 0.5]

D [0.6875 0.6875 0.3125]

💡 Hint

Think about how cubic splines smoothly connect points and evaluate at intermediate x values.

The CubicSpline fits a smooth curve through the points. Evaluating at 1.5, 2.5, and 3.5 gives values close to 0.6875, 0.6875, and 0.3125 respectively.

❓ data_output

intermediate

Number of points in spline evaluation output

Using the following code, how many points are in the output array from the spline evaluation?

SciPy

import numpy as np
from scipy.interpolate import CubicSpline

x = np.linspace(0, 10, 6)
y = np.sin(x)
cs = CubicSpline(x, y)
result = cs(np.linspace(0, 10, 50))
print(len(result))

A 6

B 60

C 50

D 10

💡 Hint

The output length matches the number of points you evaluate the spline at.

The spline is evaluated at 50 points, so the output array length is 50.

🔧 Debug

advanced

Identify the error in spline interpolation code

What error does the following code raise when run?

SciPy

import numpy as np
from scipy.interpolate import CubicSpline

x = np.array([0, 1, 2, 3])
y = np.array([1, 2])
cs = CubicSpline(x, y)

A ValueError: x and y must have the same length

B TypeError: unsupported operand type(s) for +: 'int' and 'str'

C IndexError: index out of range

D No error, code runs successfully

💡 Hint

Check if the input arrays x and y have matching lengths.

CubicSpline requires x and y to have the same length. Here y has length 2 but x has length 4, causing a ValueError.

🧠 Conceptual

advanced

Effect of boundary conditions in spline interpolation

Which boundary condition in CubicSpline ensures that the second derivative at the endpoints is zero (natural spline)?

A bc_type='not-a-knot'

B bc_type='natural'

C bc_type='clamped'

D bc_type='periodic'

💡 Hint

Natural splines have zero curvature at the ends.

The 'natural' boundary condition sets the second derivatives at the endpoints to zero, producing a natural spline.

🚀 Application

expert

Choosing spline type for periodic data

You have temperature data measured hourly over 24 hours and want to interpolate smoothly assuming the temperature at hour 0 equals hour 24. Which CubicSpline boundary condition should you use?

A bc_type='periodic'

B bc_type='natural'

C bc_type='clamped'

D bc_type='not-a-knot'

💡 Hint

Periodic data means the start and end points match in value and derivatives.

The 'periodic' boundary condition enforces that function values and derivatives match at the endpoints, ideal for cyclic data like daily temperature.

Practice

1. What is the main purpose of spline interpolation in data science?

easy

A. To calculate the mean of data points

B. To sort data points in ascending order

C. To create a smooth curve that passes through given data points

D. To remove outliers from data

Solution

1. Step 1: Understand spline interpolation

   Spline interpolation fits a smooth curve through known data points to estimate values between them.

2. Step 2: Compare options with definition

   Only To create a smooth curve that passes through given data points describes creating a smooth curve through data points, which matches spline interpolation.

3. Final Answer:

   To create a smooth curve that passes through given data points -> Option C

4. Quick Check:

   Spline interpolation = smooth curve through points [OK]

Hint: Spline interpolation = smooth curve fitting through points [OK]

Common Mistakes:

• Confusing interpolation with sorting
• Thinking it removes outliers
• Mixing interpolation with averaging

2. Which of the following is the correct way to import the spline interpolation function from scipy?

easy

A. import scipy.spline as make_interp_spline

B. from scipy.interpolate import make_interp_spline

C. from scipy import spline_interpolation

D. import make_interp_spline from scipy.interpolate

Solution

1. Step 1: Recall correct import syntax

   The function make_interp_spline is in scipy.interpolate and imported using 'from scipy.interpolate import make_interp_spline'.

2. Step 2: Check each option

   'from scipy.interpolate import make_interp_spline' matches the correct syntax. The other options use invalid import statements.

3. Final Answer:

   from scipy.interpolate import make_interp_spline -> Option B

4. Quick Check:

   Correct import syntax = from scipy.interpolate import make_interp_spline [OK]

Hint: Use 'from scipy.interpolate import make_interp_spline' [OK]

Common Mistakes:

• Using incorrect import keywords
• Trying to import non-existent modules
• Wrong order of import statement

3. What will be the output of the following code snippet?

import numpy as np
from scipy.interpolate import make_interp_spline

x = np.array([0, 1, 2, 3])
y = np.array([0, 1, 4, 9])
spline = make_interp_spline(x, y)
x_new = np.array([1.5, 2.5])
y_new = spline(x_new)
print(np.round(y_new, 2))

medium

A. [2.25 6.25]

B. [1.5 2.5]

C. [3.0 5.0]

D. [1.75 3.75]

Solution

1. Step 1: Understand the input data and spline creation

   x and y define points (0,0), (1,1), (2,4), (3,9). The spline fits a smooth curve through these points.

2. Step 2: Calculate spline values at x_new points

   At x=1.5, y is close to 2.25 (between 1 and 4). At x=2.5, y is close to 6.25 (between 4 and 9). Rounded values are [2.25, 6.25].

3. Final Answer:

   [2.25 6.25] -> Option A

4. Quick Check:

   Spline interpolation at 1.5 and 2.5 = [2.25 6.25] [OK]

Hint: Spline smooths between points; expect values between neighbors [OK]

Common Mistakes:

• Assuming linear interpolation instead of spline
• Not rounding output
• Confusing input and output arrays

4. Identify the error in the following code snippet for spline interpolation:

import numpy as np
from scipy.interpolate import make_interp_spline

x = np.array([0, 1, 2])
y = np.array([0, 1, 4, 9])
spline = make_interp_spline(x, y)
print(spline(1.5))

medium

A. x and y arrays have different lengths

B. make_interp_spline is not imported correctly

C. x array contains non-numeric values

D. spline function cannot take float inputs

Solution

1. Step 1: Check array lengths

   x has length 3, y has length 4. For spline interpolation, x and y must have the same length.

2. Step 2: Verify other parts

   Import is correct, x contains numeric values, and spline accepts float inputs. So, only length mismatch is error.

3. Final Answer:

   x and y arrays have different lengths -> Option A

4. Quick Check:

   Array length mismatch = x and y arrays have different lengths [OK]

Hint: Check x and y lengths match before spline creation [OK]

Common Mistakes:

• Ignoring length mismatch error
• Assuming spline can't take floats
• Misreading import statements

5. You have noisy temperature data recorded hourly: x = [0, 1, 2, 3, 4] and y = [15, 14, 16, 15, 17]. You want a smooth curve to estimate temperature every 0.5 hour. Which approach using scipy spline interpolation is best?

hard

A. Use make_interp_spline but only evaluate at original points

B. Use linear interpolation instead of spline for smoothness

C. Use spline interpolation but with reversed x and y arrays

D. Use make_interp_spline with x and y, then evaluate at new points with 0.5 increments

Solution

1. Step 1: Understand the goal

   We want a smooth curve estimating temperature every 0.5 hour, so we need interpolation at finer points.

2. Step 2: Choose correct method

   make_interp_spline fits a smooth curve through given points. Evaluating it at 0.5 increments gives smooth estimates. Linear interpolation is less smooth. Evaluating only at original points gives no new info. Reversing arrays is incorrect.

3. Final Answer:

   Use make_interp_spline with x and y, then evaluate at new points with 0.5 increments -> Option D

4. Quick Check:

   Spline + finer evaluation points = smooth estimates [OK]

Hint: Fit spline on data, evaluate at desired finer points [OK]

Common Mistakes:

• Using linear interpolation for smoothness
• Not evaluating spline at new points
• Reversing data arrays unnecessarily