import numpy as np
# any two python array as two points
a = [0, 0]
b = [3, 4]

首先将list更改为numpy数组，并像这样做:print(np.linalg.norm(np.array(a) - np.array(b)))。第二种方法直接从python列表as: print(np.linalg.norm(np.subtract(a,b)))

一个很好的句子:

dist = numpy.linalg.norm(a-b)

---

但是，如果速度是一个问题，我建议在您的机器上进行试验。我发现，在我的机器上，使用数学库的√rt和**运算符对平方进行运算要比使用一行NumPy解决方案快得多。

我用这个简单的程序进行了测试:

#!/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运行得比numpy_calc_dist快得多:1.5秒对23.5秒。

为了在fastst_calc_dist和math_calc_dist之间获得一个可测量的差异，我必须将TOTAL_LOCATIONS增加到6000。然后，fastst_calc_dist耗时约50秒，math_calc_dist耗时约60秒。

您也可以尝试使用numpy。SQRT和numpy。不过这两个运算都比我机器上的数学运算要慢。

我的测试使用Python 2.6.6运行。

使用numpy.linalg.norm:

dist = numpy.linalg.norm(a-b)

这是因为欧氏距离是l2范数，而numpy.linalg.norm中ord参数的默认值是2。 要了解更多理论，请参阅数据挖掘介绍:

我想用各种表演笔记来阐述这个简单的答案。Np.linalg.norm可能会做的比你需要的更多:

dist = numpy.linalg.norm(a-b)

首先，这个函数被设计用于处理一个列表并返回所有的值，例如比较pA到点集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:

Python函数调用的开销很大。 [常规]Python不缓存名称查找。

So

def distance(pointA, pointB):
    dist = np.linalg.norm(pointA - pointB)
    return dist

并不像看上去那么无辜。

>>> 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，每次我们调用它时，我们都必须对“np”进行全局查找，对“linalg”进行范围查找，对“norm”进行范围查找，而仅仅调用这个函数的开销就相当于几十条python指令。

最后，我们浪费了两个操作来存储结果并重新加载它以返回…

改进的第一步:使查找更快，跳过存储

def distance(pointA, pointB, _norm=np.linalg.norm):
    return _norm(pointA - pointB)

我们得到了更精简的:

>>> 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

不过，函数调用开销仍然需要一些工作。你会想要做基准测试，以确定你自己做数学是否会更好:

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快。你的里程可能会有所不同。

****高级性能说明。

你为什么要计算距离?如果唯一的目的是展示它，

 print("The target is %.2fm away" % (distance(a, b)))

沿着。但是如果你在比较距离，进行范围检查等等，我想添加一些有用的性能观察。

让我们以两种情况为例:按距离排序或将列表剔除到满足范围约束的项。

# 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，我们使用毕达哥拉斯来计算距离(dist =根号(x²+ y²+ z²))，所以我们做了很多根号调用。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:直到我们真正需要以X为单位的距离，而不是X^2，我们才能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)

很好，这两个函数都不再做昂贵的平方根了。这样会快得多，但在进一步讨论之前，请检查自己:为什么sort_things_by_distance两次都需要一个“天真”的免责声明?在最下面回答(*a1)。

我们可以通过将in_range转换为生成器来改进它:

def in_range(origin, range, things):
    range_sq = range**2
    yield from (thing for thing in things
                if distance_sq(origin, thing) <= range_sq)

如果你在做以下事情，这尤其有好处:

if any(in_range(origin, max_dist, things)):
    ...

但如果你接下来要做的事需要一段距离，

for nearby in in_range(origin, walking_distance, hotdog_stands):
    print("%s %.2fm" % (nearby.name, distance(origin, nearby)))

考虑生成元组:

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)

如果你可能要进行连锁范围检查(“找到X附近和Y的Nm范围内的东西”，因为你不需要再次计算距离)，这可能特别有用。

但如果我们在搜索一个很大的列表，我们预计其中有很多不值得考虑呢?

其实有一个很简单的优化:

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 ...

同样，考虑生成dist_sq。热狗的例子就变成了:

# 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的排序键为每一项调用distance_sq，而那个看起来无辜的键是一个lambda，这是必须调用的第二个函数…)

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)

首先求两个矩阵的差。然后，使用numpy的multiply命令应用元素乘法。然后，求元素与新矩阵相乘的和。最后，求求和的平方根。

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