Linear vs cubic interpolation in SciPy - Visual Side-by-Side Comparison

Concept Flow - Linear vs cubic interpolation

Start with known data points

↓

Choose interpolation method

↓

Linear interpolation

↓

Calculate interpolated values

↓

Compare smoothness and accuracy

↓

End

We start with known points, pick linear or cubic interpolation, compute values, then compare results.

Execution Sample

SciPy

import numpy as np
from scipy.interpolate import interp1d

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

f_linear = interp1d(x, y, kind='linear')
f_cubic = interp1d(x, y, kind='cubic')

x_new = np.linspace(0, 3, 7)
y_linear = f_linear(x_new)
y_cubic = f_cubic(x_new)

This code creates linear and cubic interpolation functions and computes interpolated values at new points.

Execution Table

Step	Action	Input x_new	Linear Output	Cubic Output
1	Evaluate at 0.0	0.0	0.0	0.0
2	Evaluate at 0.5	0.5	0.5	0.6875
3	Evaluate at 1.0	1.0	1.0	1.0
4	Evaluate at 1.5	1.5	0.5	0.3125
5	Evaluate at 2.0	2.0	0.0	0.0
6	Evaluate at 2.5	2.5	0.5	0.6875
7	Evaluate at 3.0	3.0	1.0	1.0
8	End of evaluation	-	-	-

💡 All x_new points evaluated; interpolation complete.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	Final
x_new	[0.0, 0.5, 1.0, 1.5, 2.0, 2.5, 3.0]	0.0	0.5	1.0	1.5	2.0	2.5	3.0	All points processed
y_linear	[]	[0.0]	[0.0, 0.5]	[0.0, 0.5, 1.0]	[0.0, 0.5, 1.0, 0.5]	[0.0, 0.5, 1.0, 0.5, 0.0]	[0.0, 0.5, 1.0, 0.5, 0.0, 0.5]	[0.0, 0.5, 1.0, 0.5, 0.0, 0.5, 1.0]	Final linear interpolated values
y_cubic	[]	[0.0]	[0.0, 0.6875]	[0.0, 0.6875, 1.0]	[0.0, 0.6875, 1.0, 0.3125]	[0.0, 0.6875, 1.0, 0.3125, 0.0]	[0.0, 0.6875, 1.0, 0.3125, 0.0, 0.6875]	[0.0, 0.6875, 1.0, 0.3125, 0.0, 0.6875, 1.0]	Final cubic interpolated values

Key Moments - 2 Insights

Why are the cubic interpolation values different from linear at the midpoints?

Because cubic interpolation fits a smooth curve considering neighbors, it produces values that can curve up or down between points, unlike linear which connects points with straight lines (see execution_table rows 2,4,6).

Why does linear interpolation output exactly the input y values at known points?

Linear interpolation always matches the original data points exactly, so at x=0,1,2,3 the outputs equal y values (see execution_table rows 1,3,5,7).

Visual Quiz - 3 Questions

Look at the execution_table, what is the cubic interpolation output at x=1.5?

A 0.3125

B 1.0

C 0.5

D 0.6875

💡 Hint

Check execution_table row 4 under 'Cubic Output'

At which x_new value does linear interpolation output 0.0?

A 1.5

B 2.0

C 0.0

D 2.5

💡 Hint

Look at execution_table rows for 'Linear Output' equal to 0.0

If we add more points between known x, how would cubic interpolation outputs change?

A They would become straight lines

B They would stay the same

C They would become smoother and more detailed

D They would become less accurate

💡 Hint

Cubic interpolation fits smooth curves, so more points give finer smoothness (concept_flow)

Concept Snapshot

Linear vs Cubic Interpolation:
- Linear connects points with straight lines.
- Cubic fits smooth curves through points.
- Use scipy interp1d with kind='linear' or 'cubic'.
- Cubic is smoother but needs more computation.
- Both match original points exactly.

Full Transcript

We start with known data points and choose either linear or cubic interpolation. Linear interpolation connects points with straight lines, producing piecewise linear outputs. Cubic interpolation fits a smooth curve considering neighboring points, producing smoother outputs. We evaluate both methods at new points between known data. Linear outputs are simple averages between points, while cubic outputs curve smoothly. The execution table shows step-by-step outputs for each method at each new point. Variables x_new, y_linear, and y_cubic track inputs and outputs. Key moments clarify why cubic differs from linear at midpoints and why linear matches original points exactly. The visual quiz tests understanding of outputs at specific points and effects of adding more points. The snapshot summarizes key differences and usage.

Practice

1. What is the main difference between linear and cubic interpolation in SciPy?

easy

A. Linear interpolation connects points with straight lines, cubic fits a smooth curve.

B. Linear interpolation fits a smooth curve, cubic connects points with straight lines.

C. Both methods produce the same results for any data.

D. Cubic interpolation is faster than linear interpolation.

Solution

1. Step 1: Understand linear interpolation

   Linear interpolation connects known data points with straight lines, estimating values between points by simple linear scaling.

2. Step 2: Understand cubic interpolation

   Cubic interpolation fits a smooth curve through the points using cubic polynomials, providing smoother transitions between points.

3. Final Answer:

   Linear interpolation connects points with straight lines, cubic fits a smooth curve. -> Option A

4. Quick Check:

   Linear = straight lines, Cubic = smooth curve [OK]

Hint: Linear = straight lines, cubic = smooth curve [OK]

Common Mistakes:

• Confusing which method produces smooth curves
• Thinking both methods give identical results
• Assuming cubic is always faster

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

easy

A. interp1d(x, y, kind='cubic')

B. interp1d(x, y, method='cubic')

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

D. interp1d(x, y, type='cubic')

Solution

1. Step 1: Recall the parameter for interpolation kind

   SciPy's interp1d uses the parameter kind to specify interpolation type, such as 'linear' or 'cubic'.

2. Step 2: Identify correct syntax for cubic interpolation

   The correct syntax is interp1d(x, y, kind='cubic'). Other parameters like method or type are invalid here.

3. Final Answer:

   interp1d(x, y, kind='cubic') -> Option A

4. Quick Check:

   Use kind='cubic' in interp1d [OK]

Hint: Use kind='cubic' to specify cubic interpolation [OK]

Common Mistakes:

• Using 'method' or 'type' instead of 'kind'
• Confusing 'linear' with 'cubic'
• Missing quotes around 'cubic'

3. Given the code below, what is 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, 0, 1])

f_linear = interp1d(x, y, kind='linear')
y_new = f_linear(1.5)

medium

A. 0.75

B. 1.0

C. 0.5

D. 0.25

Solution

1. Step 1: Understand linear interpolation between points

   At x=1, y=1 and at x=2, y=0. Linear interpolation at x=1.5 is halfway between these points.

2. Step 2: Calculate interpolated value

   Value = y1 + (x_new - x1) * (y2 - y1) / (x2 - x1) = 1 + (1.5 - 1)*(0 - 1)/(2 - 1) = 1 - 0.5 = 0.5

3. Step 3: Re-check calculation

   Recalculate carefully: 1 + 0.5 * (-1) = 1 - 0.5 = 0.5. So output is 0.5, but 0.75 is 0.75, 0.5 is 0.5.

4. Final Answer:

   0.5 -> Option C

5. Quick Check:

   Linear interp at 1.5 = 0.5 [OK]

Hint: Linear interp averages between points linearly [OK]

Common Mistakes:

• Mixing up x and y values
• Using cubic formula instead of linear
• Incorrect arithmetic in interpolation

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

import numpy as np
from scipy.interpolate import interp1d

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

f = interp1d(x, y, kind='cubic')
print(f(1.5))

medium

A. The interp1d function cannot be used with numpy arrays.

B. The y array has incorrect values for cubic interpolation.

C. The kind parameter should be 'linear' instead of 'cubic'.

D. The x array must have at least 4 points for cubic interpolation.

Solution

1. Step 1: Check minimum points for cubic interpolation

   Cubic interpolation requires at least 4 data points to fit a cubic polynomial between points.

2. Step 2: Analyze given data points

   Here, x has only 3 points, which is insufficient for cubic interpolation, causing an error.

3. Final Answer:

   The x array must have at least 4 points for cubic interpolation. -> Option D

4. Quick Check:

   Cubic needs ≥4 points [OK]

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

Common Mistakes:

• Assuming cubic works with 3 points
• Blaming y values instead of data length
• Thinking interp1d can't use numpy arrays

5. You have noisy sensor data sampled at irregular intervals. You want to estimate smooth values between points for visualization. Which interpolation method and approach is best?

hard

A. Use linear interpolation with interp1d for simplicity and speed.

B. Use cubic interpolation with interp1d for smooth curves but ensure enough points.

C. Use nearest neighbor interpolation to avoid smoothing noise.

D. Use linear interpolation but only on regularly spaced data.

Solution

1. Step 1: Understand data and goal

   Data is noisy and irregularly spaced; goal is smooth estimation between points for visualization.

2. Step 2: Evaluate interpolation methods

   Linear interpolation is simple but produces straight lines, which may look jagged. Cubic interpolation fits smooth curves, better for visualization.

3. Step 3: Consider data requirements

   Cubic interpolation requires enough points and can handle irregular spacing with interp1d. Nearest neighbor is not smooth. Linear on irregular data is possible but less smooth.

4. Final Answer:

   Use cubic interpolation with interp1d for smooth curves but ensure enough points. -> Option B

5. Quick Check:

   For smooth visualization, cubic interpolation is best [OK]

Hint: Choose cubic for smooth curves if enough data points exist [OK]

Common Mistakes:

• Choosing linear for smoothness
• Ignoring data spacing irregularity
• Using nearest neighbor for smoothing