NumPy 计算向量的范数 2 的平方

Question

I have vector a .
I want to calculate np.inner(a, a)
But I wonder whether there is prettier way to calc it.

[The disadvantage of this way, that if I want to calculate it for ab or a bit more complex expression, I have to do that with one more line. c = a - b and np.inner(c, c) instead of somewhat(a - b) ]

Answer 1

to calculate norm2

numpy.linalg.norm(x, ord=2) 

numpy.linalg.norm(x, ord=2)**2 for square

Answer 2

Honestly there's probably not going to be anything faster than np.inner or np.dot . If you find intermediate variables annoying, you could always create a lambda function:

sqeuclidean = lambda x: np.inner(x, x)

np.inner and np.dot leverage BLAS routines, and will almost certainly be faster than standard elementwise multiplication followed by summation.

In [1]: %%timeit -n 1 -r 100 a, b = np.random.randn(2, 1000000)
((a - b) ** 2).sum()
   ....: 
The slowest run took 36.13 times longer than the fastest. This could mean that an intermediate result is being cached 
1 loops, best of 100: 6.45 ms per loop

In [2]: %%timeit -n 1 -r 100 a, b = np.random.randn(2, 1000000)                                                                                                                                                                              
np.linalg.norm(a - b, ord=2) ** 2
   ....: 
1 loops, best of 100: 2.74 ms per loop

In [3]: %%timeit -n 1 -r 100 a, b = np.random.randn(2, 1000000)
sqeuclidean(a - b)
   ....: 
1 loops, best of 100: 2.64 ms per loop

np.linalg.norm(..., ord=2) uses np.dot internally, and gives very similar performance to using np.inner directly.

Answer 3

I don't know if the performance is any good, but (a**2).sum() calculates the right value and has the non-repeated argument you want. You can replace a with some complicated expression without binding it to a variable, just remember to use parentheses as necessary, since ** binds more tightly than most other operators: ((ab)**2).sum()

Answer 4

In case you end up here looking for a fast way to get the squared norm, these are some tests showing distances = np.sum((descriptors - desc[None])**2, axis=1) to be the quickest.

import timeit

setup_code = """
import numpy as np
descriptors = np.random.rand(3000, 512)
desc = np.random.rand(512)
"""

norm_code = """
np.linalg.norm(descriptors - desc[None], axis=-1)
"""
norm_time = timeit.timeit(stmt=norm_code, setup=setup_code, number=100, )

einsum_code = """
x = descriptors - desc[None]
sqrd_dist = np.einsum('ij,ij -> i', x, x)
"""
einsum_time = timeit.timeit(stmt=einsum_code, setup=setup_code, number=100, )

norm_sqrd_code = """
distances = np.sum((descriptors - desc[None])**2, axis=1)
"""
norm_sqrd_time = timeit.timeit(stmt=norm_sqrd_code, setup=setup_code, number=100, )

print(norm_time) # 0.7688689678907394
print(einsum_time) # 0.29194538854062557
print(norm_sqrd_time) # 0.274090813472867