2D interpolation (interp2d, griddata) in SciPy

Introduction

2D interpolation helps you estimate values between known points on a flat surface. It fills gaps in data smoothly.

You have temperature readings at scattered locations and want to estimate temperature at other spots.

You want to create a smooth surface from a few measured heights on a map.

You need to fill missing pixels in an image based on nearby pixel values.

You want to predict values on a grid from irregularly spaced sensor data.

Syntax

SciPy

from scipy.interpolate import interp2d, griddata

# interp2d:
f = interp2d(x, y, z, kind='linear')
zi = f(xi, yi)

# griddata:
zi = griddata(points, values, (xi, yi), method='linear')

interp2d works with regular grids and returns a function to estimate values.

griddata works with scattered points and directly returns interpolated values.

Examples

Using interp2d to estimate value at (1.5, 1.5) on a regular grid.

SciPy

from scipy.interpolate import interp2d
import numpy as np

x = np.array([0, 1, 2])
y = np.array([0, 1, 2])
z = np.array([[0, 1, 4], [1, 2, 5], [4, 5, 6]])

f = interp2d(x, y, z, kind='linear')
zi = f(1.5, 1.5)
print(zi)

Using griddata to interpolate at point (0.5, 0.5) from scattered data.

SciPy

from scipy.interpolate import griddata
import numpy as np

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

xi = np.array([0.5])
yi = np.array([0.5])

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

Sample Program

This program shows how to use both interp2d and griddata to estimate values between known points.

SciPy

from scipy.interpolate import interp2d, griddata
import numpy as np

# Data on a regular grid
x = np.array([0, 1, 2])
y = np.array([0, 1, 2])
z = np.array([[0, 1, 4], [1, 2, 5], [4, 5, 6]])

# Create interp2d function
f = interp2d(x, y, z, kind='linear')

# Interpolate at new points
zi_interp2d = f(1.5, 1.5)

# Data as scattered points
points = np.array([[0, 0], [0, 1], [1, 0], [1, 1]])
values = np.array([0, 1, 1, 2])

# Points to interpolate
xi = np.array([0.5])
yi = np.array([0.5])

# Use griddata
zi_griddata = griddata(points, values, (xi, yi), method='linear')

print(f"interp2d result at (1.5,1.5): {zi_interp2d[0]}")
print(f"griddata result at (0.5,0.5): {zi_griddata[0]}")

Important Notes

interp2d is best for data on a regular grid but is deprecated in newer SciPy versions; consider RectBivariateSpline or RegularGridInterpolator for advanced use.

griddata works well with scattered data but can be slower for large datasets.

Choose interpolation method ('linear', 'nearest', 'cubic') based on smoothness and speed needs.

Summary

2D interpolation estimates values between known points on a surface.

interp2d works with regular grids and returns a function.

griddata works with scattered points and returns values directly.

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