In the numpy library, one can pass a list into the numpy.searchsorted function, whereby it searched through a different list one element at a time and returns an array of the same sizes as the indices needed to preserve order. However, it seems to be wasting performance if both lists are sorted. For example:
m=[1,3,5,7,9]
n=[2,4,6,8,10]
numpy.searchsorted(m,n)
would return [1,2,3,4,5] which is the correct answer, but it looks like this would have complexity O(n ln(m)), whereby if one were to simply loop through m, and have some kind of pointer to n, it seems like the complexity is more like O(n+m)? Is there some kind of function in NumPy which does this?
AFAIK, this is not possible to do that in linear time only with Numpy without making additional assumptions on the inputs (eg. the integer are small and bounded). An alternative solution is to use Numba to do the merge manually:
import numba as nb
# Note: Numba requires a function signature with well defined array types
#nb.njit('int64[:](int64[::1], int64[::1])')
def search_both_sorted(a, b):
i, j = 0, 0
result = np.empty(b.size, np.int64)
while i < a.size and j < a.size:
if a[i] < b[j]:
i += 1
else:
result[j] = i
j += 1
for k in range(j, b.size):
result[k] = i
return result
a, b = np.cumsum(np.random.randint(0, 100, (2, 1000000)).astype(np.int64), axis=1)
result = search_both_sorted(a, b)
A faster implementation consists in using a branch-less approach so to remove the overhead of branch mis-prediction (especially on random/unpredictable inputs) when a and b are about the same size. Additionally, the O(n log m) algorithm can be faster when b is small so using np.searchsorted in that case is very efficient as pointed out by #MichaelSzczesny. Note that the Numba implementation of np.searchsorted can be a bit slower than the one of Numpy so it is better to pick the Numpy implementation. Here is the optimized version:
#nb.njit('int64[:](int64[::1], int64[::1])')
def search_both_sorted_opt_numba(a, b):
sa, sb = a.size, b.size
# Choose the best algorithm
if sb < sa * 0.15:
# Use a version with branches because `a[i] < b[j]`
# should be most of the time true.
i, j = 0, 0
result = np.empty(b.size, np.int64)
while i < a.size and j < b.size:
if a[i] < b[j]:
i += 1
else:
result[j] = i
j += 1
for k in range(j, b.size):
result[k] = i
else:
# Use a branchless approach to avoid miss-predictions
i, j = 0, 0
result = np.empty(b.size, np.int64)
while i < a.size and j < b.size:
tmp = a[i] < b[j]
result[j] = i
i += tmp
j += ~tmp
for k in range(j, b.size):
result[k] = i
return result
def search_both_sorted_opt(a, b):
sa, sb = a.size, b.size
# Choose the best algorithm
if 2 * sb * np.log2(sa) < sa + sb:
return np.searchsorted(a, b)
else:
return search_both_sorted_opt_numba(a, b)
searchsorted: 19.1 ms
snp_search: 11.8 ms
search_both_sorted: 6.5 ms
search_both_sorted_branchless: 4.3 ms
The optimized branchless Numba implementation is about 4.4 times faster than searchsorted which is pretty good considering that the code of searchsorted is already highly optimized. It can be even faster when a and b are huge because of cache locality.
You could use sortednp, unfortunately it does not give too much flexibility, In the code snippet below I used its merge tracking indices, but it produces three arrays, four times more memory than necessary is used, but it is faster than searchsorted.
import numpy as np
import sortednp as snp
a = np.cumsum(np.random.rand(1000000))
b = np.cumsum(np.random.rand(1000000))
def snp_search(a,b):
m, (ib, ia) = snp.merge(b, a, indices=True)
return ib - np.arange(len(ib))
assert(np.all(snp_search(a,b) == np.searchsorted(a,b)))
np.searchsorted(a, b); #58 ms
snp_search(a,b); # 22ms
np.searchsorted takes this into account already as can be seen from the source code:
/*
* Updating only one of the indices based on the previous key
* gives the search a big boost when keys are sorted, but slightly
* slows down things for purely random ones.
*/
if (cmp(last_key_val, key_val)) {
max_idx = arr_len;
}
else {
min_idx = 0;
max_idx = (max_idx < arr_len) ? (max_idx + 1) : arr_len;
}
Here min_idx, max_idx are used to perform binary search on the array. If last_key_val < key_val then only max_idx is reset to the array length, but min_idx remains at its current value, i.e. binary search starts at the same lower boundary as for the previous key.
I wanted to generate 1 or -1 in Python as a step to randomizing between non-negative and non-positive numbers or to randomly changing sign of an already existing integer. What would be the best way to generate 1 or -1 in Python? Assuming even distribution I know I could use:
import random
#method1
my_number = random.choice((-1, 1))
#method2
my_number = (-1)**random.randrange(2)
#method3
# if I understand correctly random.random() should never return exactly 1
# so I use "<", not "<="
if random.random() < 0.5:
my_number = 1
else:
my_number = -1
#method4
my_number = random.randint(0,1)*2-1
Using timeit module I got the following results:
#method1
s = "my_number = random.choice((-1, 1))"
timeit.timeit(stmt = s, setup = "import random")
>2.814896769857569
#method2
s = "my_number = (-1)**random.randrange(2)"
timeit.timeit(stmt = s, setup = "import random")
>3.521280517518562
#method3
s = """
if random.random() < 0.5: my_number = 1
else: my_number = -1"""
timeit.timeit(stmt = s, setup = "import random")
>0.25321546903273884
#method4
s = "random.randint(0,1)*2-1"
timeit.timeit(stmt = s, setup = "import random")
>4.526625442240402
So unexpectedly method 3 is the fastest. My bet was on method 1 to be the fastest as it is also shortest. Also both method 1 (since Python 3.6 I think?) and 3 give the possibility to introduce uneven distributions. Although method 1 is shortest (main advantege) for now I would choose method 3:
def positive_or_negative():
if random.random() < 0.5:
return 1
else:
return -1
Testing:
s = """
import random
def positive_or_negative():
if random.random() < 0.5:
return 1
else:
return -1
"""
timeit.timeit(stmt = "my_number = positive_or_negative()", setup = s)
>0.3916183138621818
Any better (faster or shorter) method to randomly generate -1 or 1 in Python? Any reason why would you choose method 1 over method 3 or vice versa?
A one liner variation of #3:
return 1 if random.random() < 0.5 else -1
It's fast(er) than the 'math' variants, because it doesn't involve additional arithmetic.
Here's another one-liner that my timings show to be faster than the if/else comparison to 0.5:
[-1,1][random.randrange(2)]
not sure what your application is exactly, but I needed something similar for a large vectorized array.
Here's a good way to get a sign array:
(2*np.random.randint(0,2,size=(your_size))-1)
The result is an array, for example:
array([-1, -1, -1, 1, 1, 1, -1, -1, 1, 1, 1, -1, -1, 1, -1])
and you can use the reshape command to get the above to the size of your matrix:
(2*np.random.randint(0,2,size=(m*n))-1).reshape(m,n)
Then you can multiply a matrix by the above and get all of the members with random signs.
A= np.array([[1, 2, 3],
[4, 5, 6]])
B = A*(2*np.random.randint(0,2,size=(2*3))-1).reshape(2,3)
Then you get something like :
B = array([[ 1, 2, -3],[ 4, 5, -6]])
Pretty quick, if your data is vectorized.
Maths made simple:
Generate random number: 0 or 1
Get them mutiplied by 2: 0 or 2
Substract 1: -1 or 1
Adapt that to any programming code. No need for test functions.
print(random.randint(0,1)*2-1)
works also without randint
print(int(random.random()*2)*2-1)
The fastest way to generate random numbers if you're going to be doing lots of them is by using numpy:
In [1]: import numpy as np
In [2]: import random
In [3]: %timeit [random.choice([-1,1]) for i in range(100000)]
10 loops, best of 3: 88.9 ms per loop
In [4]: %timeit [(-1)**random.randrange(2) for i in range(100000)]
10 loops, best of 3: 110 ms per loop
In [5]: %timeit [1 if random.random() < 0.5 else -1 for i in range(100000)]
100 loops, best of 3: 18.4 ms per loop
In [6]: %timeit [random.randint(0,1)*2-1 for i in range(100000)]
1 loop, best of 3: 180 ms per loop
In [7]: %timeit np.random.choice([-1,1],size=100000)
1000 loops, best of 3: 1.52 ms per loop
If you need single bits (one per call), you already did your benchmark and other answers provide additional info.
If you need many bits or can pre-calculate bit-arrays for later consumption, numpy's methods might shine.
Here is some more demo-approach using numpy (which surprisingly does not have a method dedicated for this job exactly):
import numpy as np
import random
def sample_bits(N):
assert N % 8 == 0 # demo only
n_bytes = N // 8
rbytes = np.random.randint(0, 255, dtype=np.uint8, size=n_bytes)
return np.unpackbits(rbytes)
def alt(N):
return np.random.choice([-1,1],size=N)
def alt2(N):
return [1 if random.random() < 0.5 else -1 for i in range(N)]
if __name__ == '__main__':
import timeit
print(timeit.timeit("sample_bits(1024)", setup="from __main__ import sample_bits", number=10000))
print(timeit.timeit("alt(1024)", setup="from __main__ import alt", number=10000))
print(timeit.timeit("alt2(1024)", setup="from __main__ import alt2", number=10000))
Output:
0.06640421246836543
0.352129537507486
1.5522800431775592
The general idea is:
use numpy to generate many uint8's in one step
(there might be something better using internal functions without the randint-API)
unpack uint8's to 8 bits
uniformity follows from randint's uniformity guarantees
Again, this is only a demo:
for one specific case
not caring about different result-types of these functions
not caring about -1 vs. 0 (might be important in your use-case)
(not even optimal compared to much more low-level approaches; MT used internally can be used as a bit-source, which does not need fp-math, like many other PRNGs!)
My code is as
vals = array("i", [-1, 1])
def my_rnd():
return vals[randint(0, 7) % 2]
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.