Spline interpolation in SciPy - Step-by-Step Execution

Concept Flow - Spline interpolation

Input data points (x, y)

↓

Choose spline type (e.g., cubic)

↓

Fit spline to data

↓

Evaluate spline at new points

↓

Output smooth interpolated values

Spline interpolation fits a smooth curve through given points by choosing a spline type, fitting it, then evaluating at new points.

Execution Sample

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)
x_new = np.linspace(0, 4, 9)
y_new = cs(x_new)

This code fits a cubic spline to 5 points and evaluates it at 9 new points between 0 and 4.

Execution Table

Step	Action	Input/Variable	Result/Output
1	Define original x points	x = [0,1,2,3,4]	Array x created
2	Define original y points	y = [0,1,0,1,0]	Array y created
3	Create CubicSpline object	cs = CubicSpline(x, y)	Spline fitted to points
4	Create new x points for evaluation	x_new = np.linspace(0,4,9)	x_new = [0.0,0.5,1.0,1.5,2.0,2.5,3.0,3.5,4.0]
5	Evaluate spline at x_new	y_new = cs(x_new)	y_new = [0.0,0.6875,1.0,0.6875,0.0,0.3125,1.0,0.6875,0.0]

💡 All new points evaluated, spline interpolation complete

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5
x	undefined	[0,1,2,3,4]	[0,1,2,3,4]	[0,1,2,3,4]	[0,1,2,3,4]	[0,1,2,3,4]
y	undefined	undefined	[0,1,0,1,0]	[0,1,0,1,0]	[0,1,0,1,0]	[0,1,0,1,0]
cs	undefined	undefined	undefined	CubicSpline object	CubicSpline object	CubicSpline object
x_new	undefined	undefined	undefined	undefined	[0.0,0.5,1.0,1.5,2.0,2.5,3.0,3.5,4.0]	[0.0,0.5,1.0,1.5,2.0,2.5,3.0,3.5,4.0]
y_new	undefined	undefined	undefined	undefined	undefined	[0.0,0.6875,1.0,0.6875,0.0,0.3125,1.0,0.6875,0.0]

Key Moments - 3 Insights

Why does the spline output y_new have values between the original y points?

Because spline interpolation fits a smooth curve, it estimates values between points, not just the original y values. See execution_table step 5 where y_new has intermediate values.

What happens if we evaluate the spline outside the original x range?

By default, CubicSpline extrapolates outside the original x range, which can give unexpected values. This is not shown here but important to know.

Why do we use np.linspace for x_new instead of just original x?

np.linspace creates evenly spaced points to see the smooth curve between original points, as shown in execution_table step 4.

Visual Quiz - 3 Questions

Look at the execution_table step 5, what is the interpolated y value at x_new = 2.0?

A 0.6875

B 1.0

C 0.0

D 0.3125

💡 Hint

Check the y_new array in step 5 of execution_table; the value at index corresponding to x_new=2.0 is 0.0

At which step is the spline function created and fitted to the data?

A Step 3

B Step 2

C Step 4

D Step 5

💡 Hint

Look at the Action column in execution_table; step 3 shows creation of CubicSpline object

If we change x_new to have fewer points, how does y_new change?

A y_new will have more values

B y_new will have fewer values

C y_new values will be unchanged

D y_new will be empty

💡 Hint

Refer to variable_tracker for x_new and y_new sizes; y_new matches length of x_new

Concept Snapshot

Spline interpolation fits a smooth curve through data points.
Use scipy.interpolate.CubicSpline(x, y) to create spline.
Evaluate spline at new points with cs(x_new).
Output is smooth y values between original points.
Useful for smooth data estimation and plotting.

Full Transcript

Spline interpolation is a way to draw a smooth curve through given points. We start with original x and y points. Then we create a cubic spline object using scipy's CubicSpline. This fits a smooth curve through the points. Next, we create new x points where we want to estimate y values. Finally, we evaluate the spline at these new points to get smooth y values. This helps us see smooth trends between data points instead of just dots.

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