我在3D中有两点:
(xa, ya, za)
(xb, yb, zb)
我想计算距离:
dist = sqrt((xa-xb)^2 + (ya-yb)^2 + (za-zb)^2)
使用NumPy或Python的最佳方法是什么?我有:
a = numpy.array((xa ,ya, za))
b = numpy.array((xb, yb, zb))
dist = numpy.linalg.norm(a-b)
我喜欢np.dot
(点积):
a = numpy.array((xa,ya,za))
b = numpy.array((xb,yb,zb))
distance = (np.dot(a-b,a-b))**.5
在定义它们时使用a
和b
,您还可以使用:
distance = np.sqrt(np.sum((a-b)**2))
一个不错的单线程:
dist = numpy.linalg.norm(a-b)
但是,如果考虑到速度,我建议您在机器上进行试验。我发现在我的机器上使用math
库的sqrt
和**
算子比使用单线NumPy解决方案要快得多。
我使用这个简单的程序运行我的测试:
#!/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
在我的机器上,math_calc_dist
跑得比numpy_calc_dist
快得多:1.5秒对23.5秒。
为了获得fastest_calc_dist
和math_calc_dist
之间可测量的差异,我必须将TOTAL_LOCATIONS
提高到6000.然后fastest_calc_dist
需要约50秒而math_calc_dist
需要约60秒。
您也可以尝试使用numpy.sqrt
和numpy.square
,尽管两者都比我机器上的math
替代品慢。
我的测试是使用Python 2.6.6运行的。
下面是Python中欧几里德距离的一些简洁代码,给出了在Python中表示为列表的两个点。
def distance(v1,v2):
return sum([(x-y)**2 for (x,y) in zip(v1,v2)])**(0.5)
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)
您可以轻松使用该公式
distance = np.sqrt(np.sum(np.square(a-b)))
这实际上只不过是使用毕达哥拉斯定理来计算距离,通过加上Δx,Δy和Δz的平方并使结果生根。
计算多维空间的欧几里德距离:
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
首先找出两个矩阵的差异。然后,使用numpy的multiply命令应用元素乘法。之后,找到元素的总和乘以新矩阵。最后,找到求和的平方根。
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
SciPy中有一个功能。它被称为Euclidean。
例:
from scipy.spatial import distance
a = (1, 2, 3)
b = (4, 5, 6)
dst = distance.euclidean(a, b)
对于有兴趣一次计算多个距离的人,我使用perfplot(我的一个小项目)进行了一些比较。
第一个建议是组织您的数据,使得数组具有维度(3, n)
(并且显然是C连续的)。如果在连续的第一维中发生添加,事情会更快,如果你使用sqrt-sum
与axis=0
,linalg.norm
与axis=0
,或者
a_min_b = a - b
numpy.sqrt(numpy.einsum('ij,ij->j', a_min_b, a_min_b))
这是最快的变种。 (这实际上只适用于一行。)
你在第二轴上总结的变体axis=1
都慢得多。
重现情节的代码:
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
perfplot.save(
"norm.png",
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,
],
logx=True,
logy=True,
xlabel="len(x), len(y)",
)
this problem solving method的另一个例子:
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)
我想用各种性能说明来阐述简单的答案。 np.linalg.norm可能比您需要的更多:
dist = numpy.linalg.norm(a-b)
首先 - 此功能旨在处理列表并返回所有值,例如比较从pA
到点sP
的距离:
sP = set(points)
pA = point
distances = np.linalg.norm(sP - pA, ord=2, axis=1.) # 'distances' is a list
记住几件事:
所以
def distance(pointA, pointB):
dist = np.linalg.norm(pointA - pointB)
return dist
并不像看起来那样无辜。
>>> 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
首先 - 每次调用它时,我们必须对“np”进行全局查找,对“linalg”进行范围查找,对“norm”进行范围查找,仅调用函数的开销可以等同于几十个python说明。
最后,我们浪费了两个操作来存储结果并重新加载它以便返回...
第一步改进:使查找更快,跳过商店
def distance(pointA, pointB, _norm=np.linalg.norm):
return _norm(pointA - pointB)
我们得到了更加精简:
>>> 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
但是,函数调用开销仍然相当于一些工作。你会想做基准来确定你自己是否可以更好地做数学:
def distance(pointA, pointB):
return (
((pointA.x - pointB.x) ** 2) +
((pointA.y - pointB.y) ** 2) +
((pointA.z - pointB.z) ** 2)
) ** 0.5 # fast sqrt
在某些平台上,**0.5
比math.sqrt
更快。你的旅费可能会改变。
****高级演奏说明。
你为什么计算距离?如果唯一的目的是显示它,
print("The target is %.2fm away" % (distance(a, b)))
向前走。但是,如果您要比较距离,进行范围检查等,我想添加一些有用的性能观察。
我们来看两种情况:按距离排序或剔除列表到满足范围约束的项目。
# 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)
我们需要记住的第一件事是我们使用Pythagoras来计算距离(dist = sqrt(x^2 + y^2 + z^2)
)所以我们正在进行大量的sqrt
调用。数学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
简而言之:在我们实际需要以X为单位而不是X ^ 2的距离之前,我们可以消除计算中最难的部分。
# 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)
太棒了,这两个功能不再做任何昂贵的平方根。那会更快。我们还可以通过将in_range转换为生成器来改进in_range:
def in_range(origin, range, things):
range_sq = range**2
yield from (thing for thing in things
if distance_sq(origin, thing) <= range_sq)
如果您正在执行以下操作,这尤其有益:
if any(in_range(origin, max_dist, things)):
...
但如果你要做的下一件事需要距离,
for nearby in in_range(origin, walking_distance, hotdog_stands):
print("%s %.2fm" % (nearby.name, distance(origin, nearby)))
考虑产生元组:
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)
如果您可能链接范围检查('找到靠近X且在Nm内的东西',因为您不必再次计算距离),这可能特别有用。
但是,如果我们正在搜索一个非常大的things
列表,我们预计其中很多都不值得考虑呢?
实际上有一个非常简单的优化:
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
这是否有用将取决于“事物”的大小。
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 ...
再次,考虑让dist_sq产生。然后我们的热狗示例变为:
# 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))
它可以像下面这样完成。我不知道它有多快,但它不使用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)))
我在matplotlib.mlab中找到了'dist'函数,但我认为它不够方便。
我在这里发帖仅供参考。
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)
你可以减去向量然后减去内积。
按照你的例子,
a = numpy.array((xa, ya, za))
b = numpy.array((xb, yb, zb))
tmp = a - b
sum_squared = numpy.dot(tmp.T, tmp)
result sqrt(sum_squared)
它代码简单,易于理解。