Jiajun Yang Jiajun Yang - 5 months ago 31
Python Question

Is "norm" equivalent to "Euclidean distance"?

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

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)

Answer

A norm is a function that takes a vector as an input and returns a scalar value that can be interpreted as the "size" or "length" of that vector. Norms have some other important mathematical properties:

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

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.


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.

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)

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

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

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

from scipy.spatial.distance import cdist

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

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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