是“规范"吗?相当于“欧几里德距离"? [英] Is "norm" equivalent to "Euclidean distance"?

问题描述

我不确定范数"和欧几里得距离"是否是同一个意思.请你能帮我解决这个区别吗?

I am not sure whether "norm" and "Euclidean distance" mean the same thing. Please could you help me with this distinction.

我有一个 n by m 数组 a，其中 m > 3. 我想计算第二个数据点 a[1,:] 到所有其他点(包括其自身)之间的欧几里得距离.所以我使用了 np.linalg.norm，它输出两个给定点的范数.但我不知道这是否是获得 ED 的正确方法.

I have an n by m array a, where m > 3. I want to calculate the Eculidean distance between the second data point a[1,:] to all the other points (including itself). So I used the np.linalg.norm, which outputs the norm of two given points. But I don't know if this is the right way of getting the EDs.

import numpy as np

a = np.array([[0, 0, 0 ,0 ], [1, 1 , 1, 1],[2,2, 2, 3], [3,5, 1, 5]])
N = a.shape[0] # number of row
pos = a[1,:] # pick out the second data point. 
dist = np.zeros((N,1), dtype=np.float64)

for i in range(N):
    dist[i]= np.linalg.norm(a[i,:] - pos)

推荐答案

A norm 是一个函数，它接受一个向量作为输入并返回一个标量值，该值可以解释为该向量的大小"、长度"或大小".更正式地，范数被定义为具有以下数学特性:

A norm is a function that takes a vector as an input and returns a scalar value that can be interpreted as the "size", "length" or "magnitude" of that vector. More formally, norms are defined as having the following mathematical properties:

• 它们按乘法缩放，即对于任何标量 Norm(a·v) = |a|·Norm(v)一个
• 它们满足三角不等式，即Norm(u + v) ≤ Norm(u) + Norm(>v)
• 向量的范数为零当且仅当它是零向量，即 Norm(v) = 0 ⇔ v = 0

• They scale multiplicatively, i.e. Norm(a·v) = |a|·Norm(v) for any scalar a
• They satisfy the triangle inequality, i.e. Norm(u + v) ≤ Norm(u) + Norm(v)
• The norm of a vector is zero if and only if it is the zero vector, i.e. Norm(v) = 0 ⇔ v = 0

欧几里德范数(也称为 L² 范数)只是众多不同范数中的一种 - 还有最大范数、曼哈顿范数等.单个向量的 L² 范数等价于到该点的欧几里德距离到原点，两个向量之差的L²范数等价于两点间的欧式距离.

The Euclidean norm (also known as the L² norm) is just one of many different norms - there is also the max norm, the Manhattan norm etc. The L² norm of a single vector is equivalent to the Euclidean distance from that point to the origin, and the L² norm of the difference between two vectors is equivalent to the Euclidean distance between the two points.

正如 @nobar 的回答所说，np.linalg.norm(x - y, ord=2)(或只是 np.linalg.norm(x - y)) 会给你向量 x 和 y 之间的欧几里得距离.

As @nobar's answer says, np.linalg.norm(x - y, ord=2) (or just np.linalg.norm(x - y)) will give you Euclidean distance between the vectors x and y.

由于您想计算 a[1, :] 与 a 中的其他每一行之间的欧几里德距离，您可以通过消除 a 来更快地完成此操作code>for 循环并广播 a 的行:

Since you want to compute the Euclidean distance between a[1, :] and every other row in a, you could do this a lot faster by eliminating the for loop and broadcasting over the rows of a:

dist = np.linalg.norm(a[1:2] - a, axis=1)

使用广播自己计算欧几里得距离也很容易:

It's also easy to compute the Euclidean distance yourself using broadcasting:

dist = np.sqrt(((a[1:2] - a) ** 2).sum(1))

最快的方法大概是scipy.spatial.distance.cdist:

The fastest method is probably scipy.spatial.distance.cdist:

from scipy.spatial.distance import cdist

dist = cdist(a[1:2], a)[0]

<小时>

(1000, 1000) 数组的一些时间:

---

Some timings for a (1000, 1000) array:

a = np.random.randn(1000, 1000)

%timeit np.linalg.norm(a[1:2] - a, axis=1)
# 100 loops, best of 3: 5.43 ms per loop

%timeit np.sqrt(((a[1:2] - a) ** 2).sum(1))
# 100 loops, best of 3: 5.5 ms per loop

%timeit cdist(a[1:2], a)[0]
# 1000 loops, best of 3: 1.38 ms per loop

# check that all 3 methods return the same result
d1 = np.linalg.norm(a[1:2] - a, axis=1)
d2 = np.sqrt(((a[1:2] - a) ** 2).sum(1))
d3 = cdist(a[1:2], a)[0]

assert np.allclose(d1, d2) and np.allclose(d1, d3)

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