I have two points in 3D space:
a = (ax, ay, az) b = (bx, by, bz)
I want to calculate the distance between them:
dist = sqrt((ax-bx)^2 + (ay-by)^2 + (az-bz)^2)
How do I do this with NumPy? I have:
import numpy a = numpy.array((ax, ay, az)) b = numpy.array((bx, by, bz))
Another instance of this problem solving method:
def dist(x,y): return numpy.sqrt(numpy.sum((x-y)**2)) a = numpy.array((xa,ya,za)) b = numpy.array((xb,yb,zb)) dist_a_b = dist(a,b)
A nice one-liner:
dist = numpy.linalg.norm(a-b)
However, if speed is a concern I would recommend experimenting on your machine. I’ve found that using
sqrt with the
** operator for the square is much faster on my machine than the one-liner NumPy solution.
I ran my tests using this simple program:
#!/usr/bin/python import math import numpy from random import uniform def fastest_calc_dist(p1,p2): return math.sqrt((p2 - p1) ** 2 + (p2 - p1) ** 2 + (p2 - p1) ** 2) def math_calc_dist(p1,p2): return math.sqrt(math.pow((p2 - p1), 2) + math.pow((p2 - p1), 2) + math.pow((p2 - p1), 2)) def numpy_calc_dist(p1,p2): return numpy.linalg.norm(numpy.array(p1)-numpy.array(p2)) TOTAL_LOCATIONS = 1000 p1 = dict() p2 = dict() for i in range(0, TOTAL_LOCATIONS): p1[i] = (uniform(0,1000),uniform(0,1000),uniform(0,1000)) p2[i] = (uniform(0,1000),uniform(0,1000),uniform(0,1000)) total_dist = 0 for i in range(0, TOTAL_LOCATIONS): for j in range(0, TOTAL_LOCATIONS): dist = fastest_calc_dist(p1[i], p2[j]) #change this line for testing total_dist += dist print total_dist
On my machine,
math_calc_dist runs much faster than
numpy_calc_dist: 1.5 seconds versus 23.5 seconds.
To get a measurable difference between
math_calc_dist I had to up
TOTAL_LOCATIONS to 6000. Then
fastest_calc_dist takes ~50 seconds while
math_calc_dist takes ~60 seconds.
You can also experiment with
numpy.square though both were slower than the
math alternatives on my machine.
My tests were run with Python 2.6.6.
You can just subtract the vectors and then innerproduct.
Following your example,
a = numpy.array((xa, ya, za)) b = numpy.array((xb, yb, zb)) tmp = a - b sum_squared = numpy.dot(tmp.T, tmp) result = numpy.sqrt(sum_squared)
It can be done like the following. I don’t know how fast it is, but it’s not using NumPy.
from math import sqrt a = (1, 2, 3) # Data point 1 b = (4, 5, 6) # Data point 2 print sqrt(sum( (a - b)**2 for a, b in zip(a, b)))
I find a ‘dist’ function in matplotlib.mlab, but I don’t think it’s handy enough.
I’m posting it here just for reference.
import numpy as np import matplotlib as plt a = np.array([1, 2, 3]) b = np.array([2, 3, 4]) # Distance between a and b dis = plt.mlab.dist(a, b)
from scipy.spatial import distance a = (1, 2, 3) b = (4, 5, 6) dst = distance.euclidean(a, b)
Here’s some concise code for Euclidean distance in Python given two points represented as lists in Python.
def distance(v1,v2): return sum([(x-y)**2 for (x,y) in zip(v1,v2)])**(0.5)
np.dot (dot product):
a = numpy.array((xa,ya,za)) b = numpy.array((xb,yb,zb)) distance = (np.dot(a-b,a-b))**.5
b as you defined them, you can use also:
distance = np.sqrt(np.sum((a-b)**2))
Calculate the Euclidean distance for multidimensional space:
import math x = [1, 2, 6] y = [-2, 3, 2] dist = math.sqrt(sum([(xi-yi)**2 for xi,yi in zip(x, y)])) 5.0990195135927845
I want to expound on the simple answer with various performance notes. np.linalg.norm will do perhaps more than you need:
dist = numpy.linalg.norm(a-b)
Firstly – this function is designed to work over a list and return all of the values, e.g. to compare the distance from
pA to the set of points
sP = set(points) pA = point distances = np.linalg.norm(sP - pA, ord=2, axis=1.) # 'distances' is a list
Remember several things:
def distance(pointA, pointB): dist = np.linalg.norm(pointA - pointB) return dist
isn’t as innocent as it looks.
>>> dis.dis(distance) 2 0 LOAD_GLOBAL 0 (np) 2 LOAD_ATTR 1 (linalg) 4 LOAD_ATTR 2 (norm) 6 LOAD_FAST 0 (pointA) 8 LOAD_FAST 1 (pointB) 10 BINARY_SUBTRACT 12 CALL_FUNCTION 1 14 STORE_FAST 2 (dist) 3 16 LOAD_FAST 2 (dist) 18 RETURN_VALUE
Firstly – every time we call it, we have to do a global lookup for "np", a scoped lookup for "linalg" and a scoped lookup for "norm", and the overhead of merely calling the function can equate to dozens of python instructions.
Lastly, we wasted two operations on to store the result and reload it for return…
First pass at improvement: make the lookup faster, skip the store
def distance(pointA, pointB, _norm=np.linalg.norm): return _norm(pointA - pointB)
We get the far more streamlined:
>>> dis.dis(distance) 2 0 LOAD_FAST 2 (_norm) 2 LOAD_FAST 0 (pointA) 4 LOAD_FAST 1 (pointB) 6 BINARY_SUBTRACT 8 CALL_FUNCTION 1 10 RETURN_VALUE
The function call overhead still amounts to some work, though. And you’ll want to do benchmarks to determine whether you might be better doing the math yourself:
def distance(pointA, pointB): return ( ((pointA.x - pointB.x) ** 2) + ((pointA.y - pointB.y) ** 2) + ((pointA.z - pointB.z) ** 2) ) ** 0.5 # fast sqrt
On some platforms,
**0.5 is faster than
math.sqrt. Your mileage may vary.
**** Advanced performance notes.
Why are you calculating distance? If the sole purpose is to display it,
print("The target is %.2fm away" % (distance(a, b)))
move along. But if you’re comparing distances, doing range checks, etc., I’d like to add some useful performance observations.
Let’s take two cases: sorting by distance or culling a list to items that meet a range constraint.
# Ultra naive implementations. Hold onto your hat. def sort_things_by_distance(origin, things): return things.sort(key=lambda thing: distance(origin, thing)) def in_range(origin, range, things): things_in_range =  for thing in things: if distance(origin, thing) <= range: things_in_range.append(thing)
The first thing we need to remember is that we are using Pythagoras to calculate the distance (
dist = sqrt(x^2 + y^2 + z^2)) so we’re making a lot of
sqrt calls. Math 101:
dist = root ( x^2 + y^2 + z^2 ) :. dist^2 = x^2 + y^2 + z^2 and sq(N) < sq(M) iff M > N and sq(N) > sq(M) iff N > M and sq(N) = sq(M) iff N == M
In short: until we actually require the distance in a unit of X rather than X^2, we can eliminate the hardest part of the calculations.
# Still naive, but much faster. def distance_sq(left, right): """ Returns the square of the distance between left and right. """ return ( ((left.x - right.x) ** 2) + ((left.y - right.y) ** 2) + ((left.z - right.z) ** 2) ) def sort_things_by_distance(origin, things): return things.sort(key=lambda thing: distance_sq(origin, thing)) def in_range(origin, range, things): things_in_range =  # Remember that sqrt(N)**2 == N, so if we square # range, we don't need to root the distances. range_sq = range**2 for thing in things: if distance_sq(origin, thing) <= range_sq: things_in_range.append(thing)
Great, both functions no-longer do any expensive square roots. That’ll be much faster, but before you go further, check yourself: why did sort_things_by_distance need a "naive" disclaimer both times above? Answer at the very bottom (*a1).
We can improve in_range by converting it to a generator:
def in_range(origin, range, things): range_sq = range**2 yield from (thing for thing in things if distance_sq(origin, thing) <= range_sq)
This especially has benefits if you are doing something like:
if any(in_range(origin, max_dist, things)): ...
But if the very next thing you are going to do requires a distance,
for nearby in in_range(origin, walking_distance, hotdog_stands): print("%s %.2fm" % (nearby.name, distance(origin, nearby)))
consider yielding tuples:
def in_range_with_dist_sq(origin, range, things): range_sq = range**2 for thing in things: dist_sq = distance_sq(origin, thing) if dist_sq <= range_sq: yield (thing, dist_sq)
This can be especially useful if you might chain range checks (‘find things that are near X and within Nm of Y’, since you don’t have to calculate the distance again).
But what about if we’re searching a really large list of
things and we anticipate a lot of them not being worth consideration?
There is actually a very simple optimization:
def in_range_all_the_things(origin, range, things): range_sq = range**2 for thing in things: dist_sq = (origin.x - thing.x) ** 2 if dist_sq <= range_sq: dist_sq += (origin.y - thing.y) ** 2 if dist_sq <= range_sq: dist_sq += (origin.z - thing.z) ** 2 if dist_sq <= range_sq: yield thing
Whether this is useful will depend on the size of ‘things’.
def in_range_all_the_things(origin, range, things): range_sq = range**2 if len(things) >= 4096: for thing in things: dist_sq = (origin.x - thing.x) ** 2 if dist_sq <= range_sq: dist_sq += (origin.y - thing.y) ** 2 if dist_sq <= range_sq: dist_sq += (origin.z - thing.z) ** 2 if dist_sq <= range_sq: yield thing elif len(things) > 32: for things in things: dist_sq = (origin.x - thing.x) ** 2 if dist_sq <= range_sq: dist_sq += (origin.y - thing.y) ** 2 + (origin.z - thing.z) ** 2 if dist_sq <= range_sq: yield thing else: ... just calculate distance and range-check it ...
And again, consider yielding the dist_sq. Our hotdog example then becomes:
# Chaining generators info = in_range_with_dist_sq(origin, walking_distance, hotdog_stands) info = (stand, dist_sq**0.5 for stand, dist_sq in info) for stand, dist in info: print("%s %.2fm" % (stand, dist))
(*a1: sort_things_by_distance’s sort key calls distance_sq for every single item, and that innocent looking key is a lambda, which is a second function that has to be invoked…)
For anyone interested in computing multiple distances at once, I’ve done a little comparison using perfplot (a small project of mine).
The first advice is to organize your data such that the arrays have dimension
(3, n) (and are C-contiguous obviously). If adding happens in the contiguous first dimension, things are faster, and it doesn’t matter too much if you use
a_min_b = a - b numpy.sqrt(numpy.einsum('ij,ij->j', a_min_b, a_min_b))
which is, by a slight margin, the fastest variant. (That actually holds true for just one row as well.)
The variants where you sum up over the second axis,
axis=1, are all substantially slower.
Code to reproduce the plot:
import numpy import perfplot from scipy.spatial import distance def linalg_norm(data): a, b = data return numpy.linalg.norm(a - b, axis=1) def linalg_norm_T(data): a, b = data return numpy.linalg.norm(a - b, axis=0) def sqrt_sum(data): a, b = data return numpy.sqrt(numpy.sum((a - b) ** 2, axis=1)) def sqrt_sum_T(data): a, b = data return numpy.sqrt(numpy.sum((a - b) ** 2, axis=0)) def scipy_distance(data): a, b = data return list(map(distance.euclidean, a, b)) def sqrt_einsum(data): a, b = data a_min_b = a - b return numpy.sqrt(numpy.einsum("ij,ij->i", a_min_b, a_min_b)) def sqrt_einsum_T(data): a, b = data a_min_b = a - b return numpy.sqrt(numpy.einsum("ij,ij->j", a_min_b, a_min_b)) def setup(n): a = numpy.random.rand(n, 3) b = numpy.random.rand(n, 3) out0 = numpy.array([a, b]) out1 = numpy.array([a.T, b.T]) return out0, out1 b = perfplot.bench( setup=setup, n_range=[2 ** k for k in range(22)], kernels=[ linalg_norm, linalg_norm_T, scipy_distance, sqrt_sum, sqrt_sum_T, sqrt_einsum, sqrt_einsum_T, ], xlabel="len(x), len(y)", ) b.save("norm.png")
import numpy as np from scipy.spatial import distance input_arr = np.array([[0,3,0],[2,0,0],[0,1,3],[0,1,2],[-1,0,1],[1,1,1]]) test_case = np.array([0,0,0]) dst= for i in range(0,6): temp = distance.euclidean(test_case,input_arr[i]) dst.append(temp) print(dst)
import math dist = math.hypot(math.hypot(xa-xb, ya-yb), za-zb)
You can easily use the formula
distance = np.sqrt(np.sum(np.square(a-b)))
which does actually nothing more than using Pythagoras’ theorem to calculate the distance, by adding the squares of Δx, Δy and Δz and rooting the result.
Find difference of two matrices first. Then, apply element wise multiplication with numpy’s multiply command. After then, find summation of the element wise multiplied new matrix. Finally, find square root of the summation.
def findEuclideanDistance(a, b): euclidean_distance = a - b euclidean_distance = np.sum(np.multiply(euclidean_distance, euclidean_distance)) euclidean_distance = np.sqrt(euclidean_distance) return euclidean_distance
from math import dist dist((1, 2, 6), (-2, 3, 2)) # 5.0990195135927845
And if you’re working with lists:
dist([1, 2, 6], [-2, 3, 2]) # 5.0990195135927845
Since Python 3.8 the
math module includes the function
See here https://docs.python.org/3.8/library/math.html#math.dist.
Return the Euclidean distance between two points p1 and p2,
each given as a sequence (or iterable) of coordinates.
import math print( math.dist( (0,0), (1,1) )) # sqrt(2) -> 1.4142 print( math.dist( (0,0,0), (1,1,1) )) # sqrt(3) -> 1.7321
With Python 3.8, it’s very easy.
Return the Euclidean distance between two points p and q, each given
as a sequence (or iterable) of coordinates. The two points must have
the same dimension.
Roughly equivalent to:
sqrt(sum((px - qx) ** 2.0 for px, qx in zip(p, q)))
import numpy as np # any two python array as two points a = [0, 0] b = [3, 4]
You first change list to numpy array and do like this:
print(np.linalg.norm(np.array(a) - np.array(b))). Second method directly from python list as:
The other answers work for floating point numbers, but do not correctly compute the distance for integer dtypes which are subject to overflow and underflow. Note that even
scipy.distance.euclidean has this issue:
>>> a1 = np.array(, dtype='uint8') >>> a2 = np.array(, dtype='uint8') >>> a1 - a2 array(, dtype=uint8) >>> np.linalg.norm(a1 - a2) 255.0 >>> from scipy.spatial import distance >>> distance.euclidean(a1, a2) 255.0
This is common, since many image libraries represent an image as an ndarray with dtype="uint8". This means that if you have a greyscale image which consists of very dark grey pixels (say all the pixels have color
#000001) and you’re diffing it against black image (
#000000), you can end up with
x-y consisting of
255 in all cells, which registers as the two images being very far apart from each other. For unsigned integer types (e.g. uint8), you can safely compute the distance in numpy as:
np.linalg.norm(np.maximum(x, y) - np.minimum(x, y))
For signed integer types, you can cast to a float first:
np.linalg.norm(x.astype("float") - y.astype("float"))
For image data specifically, you can use opencv’s norm method:
import cv2 cv2.norm(x, y, cv2.NORM_L2)
What’s the best way to do this with NumPy, or with Python in general? I have:
Well best way would be safest and also the fastest
I would suggest hypot usage for reliable results for chances of underflow and overflow are very little compared to writing own sqroot calculator
Lets see math.hypot, np.hypot vs vanilla
np.sqrt(np.sum((np.array([i, j, k])) ** 2, axis=1))
i, j, k = 1e+200, 1e+200, 1e+200 math.hypot(i, j, k) # 1.7320508075688773e+200
np.sqrt(np.sum((np.array([i, j, k])) ** 2)) # RuntimeWarning: overflow encountered in square
%%timeit math.hypot(i, j, k) # 100 ns ± 1.05 ns per loop (mean ± std. dev. of 7 runs, 10000000 loops each)
%%timeit np.sqrt(np.sum((np.array([i, j, k])) ** 2)) # 6.41 µs ± 33.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
i, j = 1e-200, 1e-200 np.sqrt(i**2+j**2) # 0.0
i, j = 1e+200, 1e+200 np.sqrt(i**2+j**2) # inf
i, j = 1e-200, 1e-200 np.hypot(i, j) # 1.414213562373095e-200
i, j = 1e+200, 1e+200 np.hypot(i, j) # 1.414213562373095e+200
The fastest solution I could come up with for large number of distances is using numexpr. On my machine it is faster than using numpy einsum:
import numexpr as ne import numpy as np np.sqrt(ne.evaluate("sum((a_min_b)**2,axis=1)"))
If you want something more explicit you can easily write the formula like this:
Even with arrays of 10_000_000 elements this still runs at 0.1s on my machine.