python from 2D array to 3D coordinates array

Question

I have a mXn numpy array called a: I would like to write a function which returns an array with size (3, mxn) which contains for each couple (x,y) in the first array the correspondant value.

import numpy as np

m=5
n=10
a = np.random.random((m, n))    
x = np.random.random((m, 1))                          # x coordinates
y = np.random.random((1, n))                          # y coordinates


b = np.empty((3, m*n))                # array to store coordinates
k=0
for i in range (0,m):
    for j in range (0,n):
        b[0,k] = a[i,0]
        b[1,k] = a[0,j]
        b[2,k] = a[i,j]
        k=k+1

This seems to run ok, but is there a faster or better coded way to do this?

It looks like you're trying to build a meshgrid in your own way. Take a look at this: docs.scipy.org/doc/numpy/reference/generated/… — ma3oun
– ma3oun, Commented Jun 8, 2017 at 8:13
This is useful, but it only returns x,y coordinates, it doesn't return the third coordinate, from what I understand — Argentina
– Argentina, Commented Jun 8, 2017 at 8:33
I think you need to use range (0,m) and range (0,n) to cover all elements. — Divakar
– Divakar, Commented Jun 8, 2017 at 8:34
you did not use x,y at all? why bother put it in your code here? — Wang
– Wang, Commented Jun 8, 2017 at 8:48

Divakar · Accepted Answer · 2017-06-08 09:02:04Z

Steps :

Initialize a 3D array, such that m and n are separate ones. This lets us broadcast values.
Index into the first three elements along the first axis of output with appropriate elements off a and make sure that those shapes are broadcastable.
Reshape the output back to 2D.

That's all the play is about here! Here's the vectorized implementation -

b_out = np.empty((3, m,n),dtype=a.dtype)  # 1. Initialize 
b_out[0] = a[:,0,None]                    # 2. Assign
b_out[1] = a[0]
b_out[2] = a
b_out.shape = (3,m*n)                     # 3. Reshape back to 2D

Runtime test

Approaches -

def loopy_app(a):
    m,n = a.shape
    b = np.empty((3, m*n),dtype=a.dtype)
    k=0
    for i in range (0,m):
        for j in range (0,n):
            b[0,k] = a[i,0]
            b[1,k] = a[0,j]
            b[2,k] = a[i,j]
            k=k+1
    return b

def vectorized_app(a):
    b_out = np.empty((3, m,n),dtype=a.dtype)  
    b_out[0] = a[:,0,None]
    b_out[1] = a[0]
    b_out[2] = a
    b_out.shape = (3,m*n)
    return b_out

Timings -

In [194]: m=5
     ...: n=10
     ...: a = np.random.random((m, n))
     ...: 

In [195]: %timeit loopy_app(a)
     ...: %timeit vectorized_app(a)
     ...: 
10000 loops, best of 3: 28.2 µs per loop
100000 loops, best of 3: 2.48 µs per loop

In [196]: m=50
     ...: n=100
     ...: a = np.random.random((m, n))
     ...: 

In [197]: %timeit loopy_app(a)
     ...: %timeit vectorized_app(a)
     ...: 
100 loops, best of 3: 2.56 ms per loop
100000 loops, best of 3: 6.31 µs per loop

In [198]: 2560/6.31
Out[198]: 405.7052297939778

400x+ speedup on large datasets and more on larger ones!

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