Array Broadcasting without for loop

Question

I have the code

import numpy as np
import math

pos = np.array([[   1.72,   2.56],
                [   0.24,   5.67],
                [  -1.24,   5.45],
                [  -3.17,  -0.23],
                [   1.17,  -1.23],
                [   1.12,   1.08]])

ref = np.array([1.22, 1.18])

# Insert your solution below
d1 = math.sqrt((pos[0,0]-ref[0])**2 + (pos[0,1]-ref[1])**2)
d2 = math.sqrt((pos[1,0]-ref[0])**2 + (pos[1,1]-ref[1])**2)
d3 = math.sqrt((pos[2,0]-ref[0])**2 + (pos[2,1]-ref[1])**2)
d4 = math.sqrt((pos[3,0]-ref[0])**2 + (pos[3,1]-ref[1])**2)
d5 = math.sqrt((pos[4,0]-ref[0])**2 + (pos[4,1]-ref[1])**2)
d6 = math.sqrt((pos[5,0]-ref[0])**2 + (pos[5,1]-ref[1])**2)

The expected answer is

# [ 1.468,  4.596,  4.928 ,  4.611,  2.410,  0.141 ]

Is it possible to make my solution more efficient and short, preferably without the use of a for-loop. Thank you :D

@xxh sorry I don't really know what a SIMD is so can't answer your question, but apparently yes? stackoverflow.com/questions/44944367/… — Barry
– Barry, Commented Nov 10, 2020 at 1:09
The answers to that So question give an affirmative, "yes," and that means that the currently accepted answer is incredibly efficient. — user12492341
– user12492341, Commented Nov 10, 2020 at 1:11
@xxh - Only with modules like numpy or similar numeric modules. — Michael Szczesny
– Michael Szczesny, Commented Nov 10, 2020 at 1:11
@xxh - usually numpy is vectorized using a BLAS library and fast enough. But lacks gpu support and latest algorithmic optimizations. — Michael Szczesny
– Michael Szczesny, Commented Nov 10, 2020 at 1:16

Andy L. · Accepted Answer · 2020-11-10 01:17:22Z

3

Your equation is actually euclidean distance between pos and ref. You may simplify your equation further with np.linalg.norm

dist_arr = np.linalg.norm(pos-ref, axis=1)

Out[14]:
array([1.46778745, 4.59570452, 4.9279306 , 4.61087844, 2.41051862,
       0.14142136])

answered Nov 10, 2020 at 1:17

Andy L.

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1 Comment

Barry Over a year ago

wow, I've never really used np.linalg.norm. Haven't seen it before but thank you for letting me know. Looks like I should do a bit of research on it :D

Michael Szczesny · Accepted Answer · 2020-11-10 00:59:58Z

1

This does the same as your computations. Python's math module is not needed.

np.sqrt(((pos - ref)**2).sum(1))

Out:

[1.46778745, 4.59570452, 4.9279306 , 4.61087844, 2.41051862, 0.14142136]

answered Nov 10, 2020 at 0:59

Michael Szczesny

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1 Comment

Barry Over a year ago

Wow, just one line of code! Wow thank you so much, I'm still pretty new to python :D

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