2D interpolation (interp2d, griddata) in SciPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of 2D interpolation using interp2d

What is the output of the following code snippet that uses interp2d for interpolation?

SciPy

import numpy as np
from scipy.interpolate import interp2d

x = np.array([0, 1, 2])
y = np.array([0, 1, 2])
z = np.array([[0, 1, 4], [1, 2, 5], [4, 5, 8]])

f = interp2d(x, y, z, kind='linear')
result = f(1.5, 1.5)
print(result[0])

A 6.0

B 5.0

C 5.5

D Error: interp2d deprecated

💡 Hint

Recall that interp2d returns an array even for single points; check the interpolation formula for linear kind.

The interp2d function performs bilinear interpolation. At point (1.5, 1.5), the interpolated value is 5.0. The function returns an array, so we print the first element.

❓ data_output

intermediate

Number of points interpolated with griddata

Given scattered points and values, how many points will be interpolated by the following code using griddata?

SciPy

import numpy as np
from scipy.interpolate import griddata

points = np.array([[0, 0], [0, 1], [1, 0], [1, 1]])
values = np.array([0, 1, 1, 0])
xi = np.array([[0.5, 0.5], [0.2, 0.8], [1.5, 1.5]])

zi = griddata(points, values, xi, method='linear')
print(len(zi))

A 1

B 2

C 3

D Error: xi shape mismatch

💡 Hint

Count how many points are in xi and how griddata handles points outside convex hull.

The input xi has 3 points, so griddata returns 3 interpolated values. Points outside the convex hull get NaN but are still counted.

🔧 Debug

advanced

Identify the error in griddata usage

What error does the following code raise when using griddata?

SciPy

import numpy as np
from scipy.interpolate import griddata

points = np.array([[0, 0], [1, 1]])
values = np.array([1, 2])
xi = np.array([0.5, 0.5])

zi = griddata(points, values, xi, method='nearest')
print(zi)

A TypeError: xi must be 2D array with shape (n, 2)

B ValueError: points and values length mismatch

C IndexError: index out of range

D No error, output: [1. 1.]

💡 Hint

Check the shape of xi; griddata expects a 2D array with each row as a point.

The xi array is 1D with shape (2,), but griddata expects a 2D array with shape (n, 2). This causes a TypeError.

❓ visualization

advanced

Visualizing interpolation results with griddata

Which option correctly describes the plot generated by this code using griddata with cubic interpolation?

SciPy

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

points = np.random.rand(10, 2)
values = np.sin(points[:,0]*10) + np.cos(points[:,1]*10)

grid_x, grid_y = np.mgrid[0:1:100j, 0:1:100j]

grid_z = griddata(points, values, (grid_x, grid_y), method='cubic')

plt.imshow(grid_z.T, extent=(0,1,0,1), origin='lower')
plt.scatter(points[:,0], points[:,1], c='red')
plt.title('Cubic interpolation of scattered data')
plt.colorbar()
plt.show()

A A grid of flat colors with sharp edges and no smooth transitions

B A scatter plot of points only, no color gradient

C An error due to incompatible shapes in griddata

D A smooth color gradient surface with red dots marking original points

💡 Hint

Cubic interpolation creates smooth surfaces; red dots show original data points.

The plot shows a smooth color gradient from cubic interpolation with red dots marking the scattered points. This visualizes how interpolation fills the gaps.

🧠 Conceptual

expert

Choosing interpolation method for irregular data

You have irregularly spaced 2D data points with noisy measurements. Which interpolation method is best to estimate values on a regular grid while minimizing oscillations?

A Use <code>griddata</code> with 'linear' method to balance smoothness and noise

B Use <code>interp2d</code> with 'cubic' method for smooth results

C Use <code>griddata</code> with 'nearest' method for smoothness

D Use <code>interp2d</code> with 'nearest' method to avoid oscillations

💡 Hint

Consider how each method handles noise and irregular spacing.

For irregular noisy data, 'linear' method in griddata balances smoothness and noise without introducing oscillations that cubic might cause. Nearest can be too rough.

Practice

1. What is the main difference between interp2d and griddata in SciPy for 2D interpolation?

easy

A. interp2d returns values directly; griddata returns a function.

B. interp2d works on regular grids; griddata works on scattered points.

C. interp2d only works for 1D data; griddata works for 2D data.

D. interp2d is faster than griddata in all cases.

Solution

1. Step 1: Understand interp2d usage

   interp2d requires data on a regular grid and returns an interpolation function.

2. Step 2: Understand griddata usage

   griddata works with scattered (irregular) points and returns interpolated values directly.

3. Final Answer:

   interp2d works on regular grids; griddata works on scattered points. -> Option B

4. Quick Check:

   Grid type difference = interp2d works on regular grids; griddata works on scattered points. [OK]

Hint: Remember: interp2d = grid, griddata = scattered points [OK]

Common Mistakes:

• Confusing which function returns a callable
• Thinking both require regular grids
• Assuming griddata returns a function

2. Which of the following is the correct way to create an interpolation function using interp2d with arrays x, y, and z?

easy

A. f = interp2d(z, x, y, kind='cubic')

B. f = interp2d(x, y)

C. f = interp2d(x, y, z, kind='linear')

D. f = interp2d(x, y, z, method='nearest')

Solution

1. Step 1: Check interp2d parameters

   interp2d requires x, y, and z arrays, plus optional kind for interpolation type.

2. Step 2: Validate option syntax

   f = interp2d(x, y, z, kind='linear') uses correct parameter order and valid kind argument. f = interp2d(z, x, y, kind='cubic') swaps parameters incorrectly. f = interp2d(x, y) misses z. f = interp2d(x, y, z, method='nearest') uses invalid method keyword.

3. Final Answer:

   f = interp2d(x, y, z, kind='linear') -> Option C

4. Quick Check:

   Correct parameters and keyword = f = interp2d(x, y, z, kind='linear') [OK]

Hint: Remember interp2d needs x, y, z arrays and 'kind' keyword [OK]

Common Mistakes:

• Swapping x, y, z order
• Using 'method' instead of 'kind'
• Omitting z array

3. What is the output of this code snippet?

import numpy as np
from scipy.interpolate import griddata
points = np.array([[0, 0], [1, 0], [0, 1]])
values = np.array([1, 2, 3])
xi = np.array([[0.5, 0.5]])
result = griddata(points, values, xi, method='linear')
print(result)

medium

A. [3.0]

B. [1.5]

C. [2.5]

D. [2.0]

Solution

1. Step 1: Understand input points and values

   Points are at (0,0)=1, (1,0)=2, (0,1)=3. We want value at (0.5,0.5).

2. Step 2: Apply linear interpolation

   Linear interpolation averages values weighted by distance. At (0.5,0.5), value is average of 1, 2, 3 weighted equally, resulting in 2.0.

3. Final Answer:

   [2.0] -> Option D

4. Quick Check:

   Linear interpolation at center = 2.0 [OK]

Hint: For triangle center, average vertex values linearly [OK]

Common Mistakes:

• Confusing method keyword
• Miscomputing weighted average
• Expecting a scalar instead of array

4. Identify the error in this code using interp2d:

import numpy as np
from scipy.interpolate import interp2d
x = np.array([0, 1, 2])
y = np.array([0, 1, 2])
z = np.array([1, 2, 3])
f = interp2d(x, y, z, kind='cubic')
print(f(1.5, 1.5))

medium

A. The shape of z does not match the grid defined by x and y.

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

C. interp2d does not accept numpy arrays as input.

D. The function call f(1.5, 1.5) should be f([1.5], [1.5]).

Solution

1. Step 1: Check shapes of x, y, and z

   x and y have length 3 each, so z must be 3x3 grid (2D array), but z is 1D with length 3.

2. Step 2: Understand interp2d requirements

   interp2d expects z shape to be (len(y), len(x)) matching grid points. Mismatch causes error.

3. Final Answer:

   The shape of z does not match the grid defined by x and y. -> Option A

4. Quick Check:

   z shape must match x,y grid [OK]

Hint: z must be 2D array matching x and y grid sizes [OK]

Common Mistakes:

• Passing 1D z instead of 2D
• Ignoring shape mismatch errors
• Misusing function call syntax

5. You have scattered temperature data at points pts = np.array([[0,0],[1,0],[0,1],[1,1]]) with values temps = np.array([30, 32, 31, 29]). You want to estimate temperature on a regular 5x5 grid covering [0,1]x[0,1]. Which approach correctly achieves this?

hard

A. Use griddata with pts, temps, and the 5x5 grid points to get interpolated values.

B. Use interp2d directly with pts and temps arrays.

C. Use griddata but only with the corner points of the grid.

D. Use interp2d with a 1D array of temps and x, y coordinates.

Solution

1. Step 1: Identify data type

   Data points are scattered (not on a regular grid), so griddata is suitable.

2. Step 2: Create regular grid and interpolate

   Create 5x5 grid points with np.meshgrid and use griddata with pts and temps to get interpolated temperatures on grid.

3. Final Answer:

   Use griddata with pts, temps, and the 5x5 grid points to get interpolated values. -> Option A

4. Quick Check:

   Scattered data + regular grid = griddata [OK]

Hint: For scattered data to grid, use griddata [OK]

Common Mistakes:

• Trying interp2d on scattered points
• Using incomplete grid points
• Passing 1D temps to interp2d