Linear vs cubic interpolation in SciPy - Practice Questions

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of linear interpolation with scipy

What is the output of this code snippet that uses linear interpolation to estimate values?

SciPy

import numpy as np
from scipy.interpolate import interp1d

x = np.array([0, 1, 2, 3])
y = np.array([0, 1, 4, 9])
linear_interp = interp1d(x, y, kind='linear')
result = linear_interp(1.5)
print(result)

A 2.5

B 2.75

C Error: interp1d requires integer input

D 3.0

💡 Hint

Linear interpolation finds the value on the straight line between two known points.

The value at 1.5 is halfway between y=1 at x=1 and y=4 at x=2, so linear interpolation gives (1 + 4)/2 = 2.5.

❓ Predict Output

intermediate

Output of cubic interpolation with scipy

What value does this cubic interpolation code produce at x=1.5?

SciPy

import numpy as np
from scipy.interpolate import interp1d

x = np.array([0, 1, 2, 3])
y = np.array([0, 1, 4, 9])
cubic_interp = interp1d(x, y, kind='cubic')
result = cubic_interp(1.5)
print(round(result, 2))

A 2.5

B 4.5

C 3.38

D Error: cubic interpolation needs more points

💡 Hint

Cubic interpolation fits a smooth curve through the points, not just straight lines.

Cubic interpolation estimates a smoother curve, giving approximately 3.375 at 1.5, rounded to 3.38.

❓ data_output

advanced

Number of points generated by linear vs cubic interpolation

Given 5 data points, how many points will the following code generate when interpolating 10 new points between the minimum and maximum x values?

SciPy

import numpy as np
from scipy.interpolate import interp1d

x = np.linspace(0, 4, 5)
y = x**2
linear_interp = interp1d(x, y, kind='linear')
cubic_interp = interp1d(x, y, kind='cubic')
new_x = np.linspace(0, 4, 10)
linear_y = linear_interp(new_x)
cubic_y = cubic_interp(new_x)
print(len(linear_y), len(cubic_y))

A 10 10

B 5 5

C 10 5

D 5 10

💡 Hint

The interpolation returns values for each input in new_x.

Both linear and cubic interpolation return one value for each new x input, so both arrays have length 10.

❓ visualization

advanced

Visual difference between linear and cubic interpolation

Which plot correctly shows the difference between linear and cubic interpolation for the points (0,0), (1,1), (2,4), (3,9)?

SciPy

import numpy as np
import matplotlib.pyplot as plt
from scipy.interpolate import interp1d

x = np.array([0, 1, 2, 3])
y = np.array([0, 1, 4, 9])
new_x = np.linspace(0, 3, 100)
linear_interp = interp1d(x, y, kind='linear')
cubic_interp = interp1d(x, y, kind='cubic')
plt.plot(x, y, 'o', label='Data points')
plt.plot(new_x, linear_interp(new_x), label='Linear')
plt.plot(new_x, cubic_interp(new_x), label='Cubic')
plt.legend()
plt.show()

A Linear interpolation produces a curve with sharp peaks; cubic interpolation is flat.

B Linear interpolation is a smooth curve; cubic is piecewise straight lines.

C Both linear and cubic interpolation produce the same straight line.

D Linear interpolation is piecewise straight lines; cubic is a smooth curve passing through points.

💡 Hint

Think about how linear interpolation connects points versus cubic interpolation.

Linear interpolation connects points with straight lines, while cubic interpolation fits a smooth curve through all points.

🧠 Conceptual

expert

Why cubic interpolation can fail with few points

Why might cubic interpolation raise an error or produce unreliable results when given only two data points?

A Cubic interpolation only works with evenly spaced points.

B Cubic interpolation requires at least four points to fit a cubic polynomial.

C Two points are enough for cubic interpolation but it always returns linear results.

D Cubic interpolation cannot handle numeric data.

💡 Hint

Think about the minimum number of points needed to define a cubic curve.

A cubic polynomial has four coefficients, so at least four points are needed to define it uniquely.