Distance computation (distance.cdist) in SciPy - Cheat Sheet & Quick Revision

Recall & Review

beginner

What does the function scipy.spatial.distance.cdist compute?

cdist computes the distance between each pair of points from two collections of points. It returns a matrix where each element is the distance between a point in the first set and a point in the second set.

beginner

What are the inputs required by distance.cdist?

Two arrays of points (2D arrays), where each row is a point and each column is a dimension. Optionally, a metric string to specify the type of distance (like 'euclidean', 'cityblock').

beginner

Name two common distance metrics you can use with cdist.

Euclidean distance (straight line) and Manhattan distance (sum of absolute differences).

intermediate

How does cdist differ from computing distances in a loop?

cdist is optimized and vectorized, so it is much faster and simpler than manually looping over points to compute distances.

intermediate

What shape does the output of cdist have if the first input has shape (m, n) and the second input has shape (p, n)?

The output shape is (m, p), where each element [i, j] is the distance between the i-th point in the first set and the j-th point in the second set.

What does scipy.spatial.distance.cdist return?

A A single distance value between two points

B A matrix of distances between points in two sets

C A list of points sorted by distance

D A boolean matrix indicating if points are equal

cdist returns a matrix where each element is the distance between a point in the first set and a point in the second set.

Which of these is NOT a valid metric for cdist?

A euclidean

B cityblock

C random

D cosine

'random' is not a valid distance metric for cdist. Common metrics include 'euclidean', 'cityblock', and 'cosine'.

If you have 5 points in set A and 3 points in set B, what will be the shape of the output from cdist(A, B)?

A (1, 15)

B (3, 5)

C (8, 1)

D (5, 3)

The output shape is (number of points in A, number of points in B), so (5, 3).

Why is cdist preferred over manual loops for distance calculation?

A It is faster and simpler to use

B It only works for 1D points

C It returns distances as strings

D It uses less memory

cdist is optimized and vectorized, making it faster and easier than manual loops.

Which Python library provides the cdist function?

A scipy

B pandas

C numpy

D matplotlib

cdist is part of scipy.spatial.distance.

Explain how you would use scipy.spatial.distance.cdist to find distances between two sets of points.

Think about inputs, metric, and output shape.

Describe the difference between Euclidean and Manhattan distance metrics used in cdist.

Imagine walking in a city grid vs flying straight.

Practice

1.

What does the scipy.spatial.distance.cdist function do?

easy

A. Generates random points within a specified range.

B. Calculates the sum of all points in a single array.

C. Computes distances between two sets of points and returns a matrix of distances.

D. Sorts points in ascending order based on their coordinates.

Solution

1. Step 1: Understand the function purpose

   cdist calculates distances between two groups of points, not sums or sorting.

2. Step 2: Identify the output type

   The function returns a matrix showing distances from each point in the first group to each point in the second.

3. Final Answer:

   Computes distances between two sets of points and returns a matrix of distances. -> Option C

4. Quick Check:

   cdist = distance matrix [OK]

Hint: Remember: cdist = cross distances between two point sets [OK]

Common Mistakes:

• Thinking cdist sums points instead of distances
• Confusing cdist with sorting or random generation
• Assuming cdist returns a single distance value

2.

Which of the following is the correct way to import and use cdist from scipy.spatial.distance?

?

easy

A. from scipy import cdist
result = cdist(A, B, metric='euclidean')

B. import scipy.distance.cdist
result = cdist(A, B, metric='euclidean')

C. import cdist from scipy.spatial.distance
result = cdist(A, B, metric='euclidean')

D. from scipy.spatial.distance import cdist
result = cdist(A, B, metric='euclidean')

Solution

1. Step 1: Check correct import syntax

   The correct import is from scipy.spatial.distance import cdist. Other forms are invalid or cause errors.

2. Step 2: Confirm usage syntax

   Calling cdist(A, B, metric='euclidean') is correct to compute Euclidean distances.

3. Final Answer:

   from scipy.spatial.distance import cdist result = cdist(A, B, metric='euclidean') -> Option D

4. Quick Check:

   Correct import and call = from scipy.spatial.distance import cdist result = cdist(A, B, metric='euclidean') [OK]

Hint: Use 'from scipy.spatial.distance import cdist' to import [OK]

Common Mistakes:

• Using incorrect import statements causing ImportError
• Trying to import cdist directly from scipy
• Using wrong syntax like 'import cdist from ...'

3.

Given the code below, what is the output?

import numpy as np
from scipy.spatial.distance import cdist

A = np.array([[0, 0], [1, 1]])
B = np.array([[1, 0], [2, 2]])
result = cdist(A, B, metric='euclidean')
print(result)

medium

A. [[1. 2.82842712]
[1. 1.41421356]]

B. [[1.41421356 2.23606798]
[1. 1.41421356]]

C. [[1. 1.41421356]
[1.41421356 2.23606798]]

D. [[0. 1.41421356]
[1. 2.23606798]]

Solution

1. Step 1: Calculate distances from points in A to points in B

   Distances: - From [0,0] to [1,0] = sqrt((1-0)^2 + (0-0)^2) = 1.0 - From [0,0] to [2,2] = sqrt((2-0)^2 + (2-0)^2) ≈ 2.8284 - From [1,1] to [1,0] = sqrt((1-1)^2 + (0-1)^2) = 1.0 - From [1,1] to [2,2] = sqrt((2-1)^2 + (2-1)^2) ≈ 1.4142

2. Step 2: Form the distance matrix

   Matrix rows correspond to points in A, columns to points in B: [[1.0, 2.8284], [1.0, 1.4142]]

3. Final Answer:

   [[1. 2.82842712] [1. 1.41421356]] -> Option A

4. Quick Check:

   Euclidean distances matrix = [[1. 2.82842712] [1. 1.41421356]] [OK]

Hint: Calculate each pair distance carefully, row by row [OK]

Common Mistakes:

• Mixing row and column order in the output matrix
• Using wrong distance formula (e.g., Manhattan instead of Euclidean)
• Rounding errors causing confusion

4.

What is wrong with the following code snippet?

import numpy as np
from scipy.spatial.distance import cdist

A = np.array([[0, 0], [1, 1]])
B = np.array([1, 0])
result = cdist(A, B, metric='euclidean')
print(result)

medium

A. The metric 'euclidean' is not supported by cdist.

B. B should be a 2D array, but it is 1D, causing a shape error.

C. cdist requires both inputs to be lists, not numpy arrays.

D. The import statement for cdist is incorrect.

Solution

1. Step 1: Check input array shapes

   cdist expects both inputs to be 2D arrays representing sets of points. Here, B is 1D, which is invalid.

2. Step 2: Understand error cause

   Passing 1D array B causes a shape mismatch error because cdist tries to interpret rows as points.

3. Final Answer:

   B should be a 2D array, but it is 1D, causing a shape error. -> Option B

4. Quick Check:

   Input arrays must be 2D for cdist [OK]

Hint: Always use 2D arrays for both inputs in cdist [OK]

Common Mistakes:

• Passing 1D arrays instead of 2D arrays
• Assuming cdist accepts lists only
• Using unsupported metric names

5.

You have two sets of points representing locations of stores and customers:

stores = np.array([[0, 0], [5, 5], [10, 10]])
customers = np.array([[1, 1], [6, 6], [8, 8], [12, 12]])

Using cdist, how can you find the closest store for each customer?

hard

A. Compute the distance matrix with cdist(customers, stores) and find the index of the minimum distance in each row.

B. Compute the distance matrix with cdist(stores, customers) and find the index of the maximum distance in each column.

C. Use cdist(stores, customers) and sum all distances to find closest stores.

D. Calculate Euclidean distance manually for each pair without using cdist.

Solution

1. Step 1: Compute distance matrix from customers to stores

   Use cdist(customers, stores) to get distances where each row is a customer and each column is a store.

2. Step 2: Find closest store for each customer

   For each customer (row), find the index of the minimum distance value. This index corresponds to the closest store.

3. Final Answer:

   Compute the distance matrix with cdist(customers, stores) and find the index of the minimum distance in each row. -> Option A

4. Quick Check:

   Closest store = min distance per customer row [OK]

Hint: cdist(customers, stores), then argmin per row finds closest store [OK]

Common Mistakes:

• Swapping order of inputs causing wrong distance matrix shape
• Using max instead of min to find closest
• Summing distances instead of finding minimum