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))
Use numpy.linalg.norm:
dist = numpy.linalg.norm(a-b)
This works because the Euclidean distance is the l2 norm, and the default value of the ord parameter in numpy.linalg.norm is 2.
For more theory, see Introduction to Data Mining:
Use scipy.spatial.distance.euclidean:
from scipy.spatial import distance
a = (1, 2, 3)
b = (4, 5, 6)
dst = distance.euclidean(a, b)
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 sqrt-sum with axis=0, linalg.norm with axis=0, or
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[0]
return numpy.linalg.norm(a - b, axis=1)
def linalg_norm_T(data):
a, b = data[1]
return numpy.linalg.norm(a - b, axis=0)
def sqrt_sum(data):
a, b = data[0]
return numpy.sqrt(numpy.sum((a - b) ** 2, axis=1))
def sqrt_sum_T(data):
a, b = data[1]
return numpy.sqrt(numpy.sum((a - b) ** 2, axis=0))
def scipy_distance(data):
a, b = data[0]
return list(map(distance.euclidean, a, b))
def sqrt_einsum(data):
a, b = data[0]
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[1]
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")
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:
sP = set(points)
pA = point
distances = np.linalg.norm(sP - pA, ord=2, axis=1.) # 'distances' is a list
Remember several things:
Python function calls are expensive.
[Regular] Python doesn't cache name lookups.
So
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...)
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)
Starting Python 3.8, the math module directly provides the dist function, which returns the euclidean distance between two points (given as tuples or lists of coordinates):
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
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)))
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 math library's 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[0] - p1[0]) ** 2 +
(p2[1] - p1[1]) ** 2 +
(p2[2] - p1[2]) ** 2)
def math_calc_dist(p1,p2):
return math.sqrt(math.pow((p2[0] - p1[0]), 2) +
math.pow((p2[1] - p1[1]), 2) +
math.pow((p2[2] - p1[2]), 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 fastest_calc_dist and 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.sqrt and numpy.square though both were slower than the math alternatives on my machine.
My tests were run with Python 2.6.6.
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)
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)
I like np.dot (dot product):
a = numpy.array((xa,ya,za))
b = numpy.array((xb,yb,zb))
distance = (np.dot(a-b,a-b))**.5
With Python 3.8, it's very easy.
https://docs.python.org/3/library/math.html#math.dist
math.dist(p, q)
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)))
Having a and b as you defined them, you can use also:
distance = np.sqrt(np.sum((a-b)**2))
Since Python 3.8
Since Python 3.8 the math module includes the function math.dist().
See here https://docs.python.org/3.8/library/math.html#math.dist.
math.dist(p1, p2)
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
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)
import math
dist = math.hypot(math.hypot(xa-xb, ya-yb), za-zb)
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
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)
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.
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: print(np.linalg.norm(np.subtract(a,b)))
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([1], dtype='uint8')
>>> a2 = np.array([2], dtype='uint8')
>>> a1 - a2
array([255], 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)
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
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
Speed wise math.hypot look better
%%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)
Underflow
i, j = 1e-200, 1e-200
np.sqrt(i**2+j**2)
# 0.0
Overflow
i, j = 1e+200, 1e+200
np.sqrt(i**2+j**2)
# inf
No Underflow
i, j = 1e-200, 1e-200
np.hypot(i, j)
# 1.414213562373095e-200
No Overflow
i, j = 1e+200, 1e+200
np.hypot(i, j)
# 1.414213562373095e+200
Refer
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:
np.sqrt(np.sum((a-b)**2))
Even with arrays of 10_000_000 elements this still runs at 0.1s on my machine.
I can implement the error function, erf, myself, but I'd prefer not to. Is there a python package with no external dependencies that contains an implementation of this function? I have found this but this seems to be part of some much larger package (and it's not even clear which one!).
Since v.2.7. the standard math module contains erf function. This should be the easiest way.
http://docs.python.org/2/library/math.html#math.erf
I recommend SciPy for numerical functions in Python, but if you want something with no dependencies, here is a function with an error error is less than 1.5 * 10-7 for all inputs.
def erf(x):
# save the sign of x
sign = 1 if x >= 0 else -1
x = abs(x)
# constants
a1 = 0.254829592
a2 = -0.284496736
a3 = 1.421413741
a4 = -1.453152027
a5 = 1.061405429
p = 0.3275911
# A&S formula 7.1.26
t = 1.0/(1.0 + p*x)
y = 1.0 - (((((a5*t + a4)*t) + a3)*t + a2)*t + a1)*t*math.exp(-x*x)
return sign*y # erf(-x) = -erf(x)
The algorithm comes from Handbook of Mathematical Functions, formula 7.1.26.
I would recommend you download numpy (to have efficiant matrix in python) and scipy (a Matlab toolbox substitute, which uses numpy). The erf function lies in scipy.
>>>from scipy.special import erf
>>>help(erf)
You can also use the erf function defined in pylab, but this is more intended at plotting the results of the things you compute with numpy and scipy. If you want an all-in-one
installation of these software you can use directly the Python Enthought distribution.
A pure python implementation can be found in the mpmath module (http://code.google.com/p/mpmath/)
From the doc string:
>>> from mpmath import *
>>> mp.dps = 15
>>> print erf(0)
0.0
>>> print erf(1)
0.842700792949715
>>> print erf(-1)
-0.842700792949715
>>> print erf(inf)
1.0
>>> print erf(-inf)
-1.0
For large real x, \mathrm{erf}(x) approaches 1 very
rapidly::
>>> print erf(3)
0.999977909503001
>>> print erf(5)
0.999999999998463
The error function is an odd function::
>>> nprint(chop(taylor(erf, 0, 5)))
[0.0, 1.12838, 0.0, -0.376126, 0.0, 0.112838]
:func:erf implements arbitrary-precision evaluation and
supports complex numbers::
>>> mp.dps = 50
>>> print erf(0.5)
0.52049987781304653768274665389196452873645157575796
>>> mp.dps = 25
>>> print erf(1+j)
(1.316151281697947644880271 + 0.1904534692378346862841089j)
Related functions
See also :func:erfc, which is more accurate for large x,
and :func:erfi which gives the antiderivative of
\exp(t^2).
The Fresnel integrals :func:fresnels and :func:fresnelc
are also related to the error function.
To answer my own question, I have ended up using the following code, adapted from a Java version I found elsewhere on the web:
# from: http://www.cs.princeton.edu/introcs/21function/ErrorFunction.java.html
# Implements the Gauss error function.
# erf(z) = 2 / sqrt(pi) * integral(exp(-t*t), t = 0..z)
#
# fractional error in math formula less than 1.2 * 10 ^ -7.
# although subject to catastrophic cancellation when z in very close to 0
# from Chebyshev fitting formula for erf(z) from Numerical Recipes, 6.2
def erf(z):
t = 1.0 / (1.0 + 0.5 * abs(z))
# use Horner's method
ans = 1 - t * math.exp( -z*z - 1.26551223 +
t * ( 1.00002368 +
t * ( 0.37409196 +
t * ( 0.09678418 +
t * (-0.18628806 +
t * ( 0.27886807 +
t * (-1.13520398 +
t * ( 1.48851587 +
t * (-0.82215223 +
t * ( 0.17087277))))))))))
if z >= 0.0:
return ans
else:
return -ans
I have a function which does 10^5 erf calls. On my machine...
scipy.special.erf makes it time at 6.1s
erf Handbook of Mathematical Functions takes 8.3s
erf Numerical Recipes 6.2 takes 9.5s
(three-run averages, code taken from above posters).
One note for those aiming for higher performance: vectorize, if possible.
import numpy as np
from scipy.special import erf
def vectorized(n):
x = np.random.randn(n)
return erf(x)
def loopstyle(n):
x = np.random.randn(n)
return [erf(v) for v in x]
%timeit vectorized(10e5)
%timeit loopstyle(10e5)
gives results
# vectorized
10 loops, best of 3: 108 ms per loop
# loops
1 loops, best of 3: 2.34 s per loop
SciPy has an implementation of the erf function, see scipy.special.erf.
From Python's math.erf function documentation, it uses up to 50 terms in the approximation:
Implementations of the error function erf(x) and the complementary error
function erfc(x).
Method: we use a series approximation for erf for small x, and a continued
fraction approximation for erfc(x) for larger x;
combined with the relations erf(-x) = -erf(x) and erfc(x) = 1.0 - erf(x),
this gives us erf(x) and erfc(x) for all x.
The series expansion used is:
erf(x) = x*exp(-x*x)/sqrt(pi) * [
2/1 + 4/3 x**2 + 8/15 x**4 + 16/105 x**6 + ...]
The coefficient of x**(2k-2) here is 4**k*factorial(k)/factorial(2*k).
This series converges well for smallish x, but slowly for larger x.
The continued fraction expansion used is:
erfc(x) = x*exp(-x*x)/sqrt(pi) * [1/(0.5 + x**2 -) 0.5/(2.5 + x**2 - )
3.0/(4.5 + x**2 - ) 7.5/(6.5 + x**2 - ) ...]
after the first term, the general term has the form:
k*(k-0.5)/(2*k+0.5 + x**2 - ...).
This expansion converges fast for larger x, but convergence becomes
infinitely slow as x approaches 0.0. The (somewhat naive) continued
fraction evaluation algorithm used below also risks overflow for large x;
but for large x, erfc(x) == 0.0 to within machine precision. (For
example, erfc(30.0) is approximately 2.56e-393).
Parameters: use series expansion for abs(x) < ERF_SERIES_CUTOFF and
continued fraction expansion for ERF_SERIES_CUTOFF <= abs(x) <
ERFC_CONTFRAC_CUTOFF. ERFC_SERIES_TERMS and ERFC_CONTFRAC_TERMS are the
numbers of terms to use for the relevant expansions.
#define ERF_SERIES_CUTOFF 1.5
#define ERF_SERIES_TERMS 25
#define ERFC_CONTFRAC_CUTOFF 30.0
#define ERFC_CONTFRAC_TERMS 50
Error function, via power series.
Given a finite float x, return an approximation to erf(x).
Converges reasonably fast for small x.