CODEWITHSUNDEEP
pythonarraysnumpymatrix
Mederic Fourmy
Mederic Fourmy

Reputation: 331

Compute distance matrix with numpy

I am trying to compute a "distance matrix" matrix to position using numpy. To put it more clearly, I have a matrix representing positions in a 2-D grid: 

array([[[0, 0],
    [1, 0],
    [2, 0],
    [3, 0]],

   [[0, 1],
    [1, 1],
    [2, 1],
    [3, 1]],

   [[0, 2],
    [1, 2],
    [2, 2],
    [3, 2]],

   [[0, 3],
    [1, 3],
    [2, 3],
    [3, 3]],

   [[0, 4],
    [1, 4],
    [2, 4],
    [3, 4]]])

I can loop over the position and compute the norm of the difference between the goal position and each position of the position matrix like this:

pos_goal = np.array([1,2])
dist_matrix = np.zeros(l_arr.shape[:2])
for i, line in enumerate(l_arr):
    for j, pos in enumerate(line):
        dist_matrix[i,j] = np.linalg.norm(pos - pos_goal) 

dist_matrix

Result:

array([[ 2.23606798,  2.        ,  2.23606798,  2.82842712],
   [ 1.41421356,  1.        ,  1.41421356,  2.23606798],
   [ 1.        ,  0.        ,  1.        ,  2.        ],
   [ 1.41421356,  1.        ,  1.41421356,  2.23606798],
   [ 2.23606798,  2.        ,  2.23606798,  2.82842712]])

Isn't there a better way to do this (without the 2 loops)?

Upvotes: 2

Views: 12211

Answers (2)

jakevdp
jakevdp

Reputation: 86513

Use scipy.spatial.distance.cdist. It requires 2D inputs, so you can do something like this:

from scipy.spatial import distance
dist_matrix = distance.cdist(l_arr.reshape(-1, 2), [pos_goal]).reshape(l_arr.shape[:2])

This is quite succinct, and for large arrays will be faster than a manual approach based on looping or broadcasting.

Upvotes: 4

juanpa.arrivillaga
juanpa.arrivillaga

Reputation: 96277

The np.linalg.norm function takes an axis argument, so you want:

In [6]: np.linalg.norm(l_arr - pos_goal, axis=2)
Out[6]:
array([[ 2.23606798,  2.        ,  2.23606798,  2.82842712],
       [ 1.41421356,  1.        ,  1.41421356,  2.23606798],
       [ 1.        ,  0.        ,  1.        ,  2.        ],
       [ 1.41421356,  1.        ,  1.41421356,  2.23606798],
       [ 2.23606798,  2.        ,  2.23606798,  2.82842712]])

You can just use -1 for "the last" axis:

In [7]: np.linalg.norm(l_arr - pos_goal, axis=-1)
Out[7]:
array([[ 2.23606798,  2.        ,  2.23606798,  2.82842712],
       [ 1.41421356,  1.        ,  1.41421356,  2.23606798],
       [ 1.        ,  0.        ,  1.        ,  2.        ],
       [ 1.41421356,  1.        ,  1.41421356,  2.23606798],
       [ 2.23606798,  2.        ,  2.23606798,  2.82842712]])

Note, I used array-broadcasting to get the differences:

In [11]: l_arr - pos_goal
Out[11]:
array([[[-1, -2],
        [ 0, -2],
        [ 1, -2],
        [ 2, -2]],

       [[-1, -1],
        [ 0, -1],
        [ 1, -1],
        [ 2, -1]],

       [[-1,  0],
        [ 0,  0],
        [ 1,  0],
        [ 2,  0]],

       [[-1,  1],
        [ 0,  1],
        [ 1,  1],
        [ 2,  1]],

       [[-1,  2],
        [ 0,  2],
        [ 1,  2],
        [ 2,  2]]])

Generally, learning how to use broadcasting in combination with built-in numpy/scipy vectorized functions is the way to achieve substantial speed improvements.

Upvotes: 2

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