iterating over numpy arrays

Question

I am having a very difficult time vectoring, I can't seem to think about math in that way yet. I have this right now:

#!/usr/bin/env python

import numpy as np
import math

grid = np.zeros((2,2))
aList = np.arange(1,5).reshape(2,2)

i,j = np.indices((2,2))

iArray =  (i - aList[:,0:1]) 
jArray = (j - aList[:,1:2])

print np.power(np.power(iArray, 2) + np.power(jArray, 2), .5)

My print out looks like this:

[[ 2.23606798  1.41421356]
 [ 4.47213595  3.60555128]]

What I am trying to do is take a 2D array of pixel values, grid, and say how far each pixel is from a list of important pixels, aList.

# # @ 
# # #
* # *

An example is if the *s (0,2) and (2,2) are important pixels and I am currently at the @ (2,0) pixel, my value for the @ pixel would be:

[(0-2)^2 + (2-0)^2]^.5 + [(2-2)^2 + (0-2)^2]^.5

All grid does is hold pixel values so I need to get the index of each pixel value to associate distance. However my Alist array holds [x,y] coordinates, So that one is easy. I think I right now I have two issues: 1. I am not getting the indeces correctly 2. I am not looping over the coordinates in aList properly

Just a note: if your example is using array indices, then the coordinates of *s are (2,0), (2,2) and @ is (0, 2). — wflynny
– wflynny, Commented Jul 3, 2013 at 17:15
You could try to create a distance array for each of your "important points". dist2 = lambda index_array, i, j: (index_array[0]-i)**2 + (index_array[1]-j)**2, where index_array = np.indices((N,N)). This yields a mesh where each point is the distance^2 from the important point (i, j). Create one of these for each important point, then np.sqrt(sum(dist_arrays)) to get the combined distance from all important points. — wflynny
– wflynny, Commented Jul 3, 2013 at 17:28

Jaime · Accepted Answer · 2013-07-03 18:49:16Z

3

With a little help from broadcasting, I get this, with data based on your last example:

import numpy as np

grid = np.zeros((3, 3))
aList = np.array([[2, 0], [2, 2]])

important_rows, important_cols = aList.T
rows, cols  = np.indices(grid.shape)

dist = np.sqrt((important_rows - rows.ravel()[:, None])**2 +
               (important_cols - cols.ravel()[:, None])**2).sum(axis=-1)
dist = dist.reshape(grid.shape)

>>> dist
array([[ 4.82842712,  4.47213595,  4.82842712],
       [ 3.23606798,  2.82842712,  3.23606798],
       [ 2.        ,  2.        ,  2.        ]])

You can get more memory efficient by doing:

important_rows, important_cols = aList.T
rows, cols = np.meshgrid(np.arange(grid.shape[0]),
                         np.arange(grid.shape[1]),
                         sparse=True, indexing='ij')
dist2 = np.sqrt((rows[..., None] - important_rows)**2 +
                (cols[..., None] - important_cols)**2).sum(axis=-1)

edited Jul 3, 2013 at 18:49

answered Jul 3, 2013 at 18:41

Jaime

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Pablo Over a year ago

Very good answer! For an array of 3000x3000 is at least 1.73 times faster that my solution...

Pablo · Accepted Answer · 2013-07-03 18:37:59Z

1

My approach:

import numpy as np

n = 3

aList = np.zeros([n,n])
distance = np.zeros([n,n])

I,J = np.indices([n,n])

aList[2,2] = 1; aList[0,2] = 1   #Importan pixels
important = np.where(aList == 1) #Where the important pixels are

for i,j in zip(I[important],J[important]):   #This part could be improved...
    distance += np.sqrt((i-I)**2+(j-J)**2)

print distance

The last 'for' could be improved, but if you have only a few important pixels, the performance will be good...

Checking with:

import matplotlib.pyplot as plt

n = 500

...

aList[249+100,349] = 1; aList[249-100,349] = 1 ;aList[249,50] = 1

...

plt.plot(I[important],J[important],'rx',markersize=20)
plt.imshow(distance.T,origin='lower',
           cmap=plt.cm.gray)
plt.show()

The result is very comfortable:

enter image description here

edited Jul 3, 2013 at 18:37

answered Jul 3, 2013 at 18:20

Pablo

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