Python NumPy 计算向量的范数 2 的平方
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NumPy calculate square of norm 2 of vector
提问by Dubon
I have vector a.
I want to calculate np.inner(a, a)
But I wonder whether there is prettier way to calc it.
我有向量a。
我想计算np.inner(a, a)
但我想知道是否有更漂亮的方法来计算它。
[The disadvantage of this way, that if I want to calculate it for a-bor a bit more complex expression, I have to do that with one more line. c = a - band np.inner(c, c)instead of somewhat(a - b)]
[这种方式的缺点是，如果我想计算它a-b或更复杂的表达式，我必须再用一行来完成。c = a - b而np.inner(c, c)不是somewhat(a - b)]
采纳答案by ali_m
Honestly there's probably not going to be anything faster than np.inneror np.dot. If you find intermediate variables annoying, you could always create a lambda function:
老实说，可能没有比np.inneror更快的了np.dot。如果你觉得中间变量很烦人，你总是可以创建一个 lambda 函数：
sqeuclidean = lambda x: np.inner(x, x)
np.innerand np.dotleverage BLAS routines, and will almost certainly be faster than standard elementwise multiplication followed by summation.
np.inner并np.dot利用 BLAS 例程，几乎肯定会比标准的元素乘法和求和更快。
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.dotinternally, and gives very similar performance to using np.innerdirectly.
np.linalg.norm(..., ord=2)在np.dot内部使用，并提供与np.inner直接使用非常相似的性能。
回答by Farseer
to calculate norm2
计算范数2
numpy.linalg.norm(x, ord=2)
numpy.linalg.norm(x, ord=2)**2for square
numpy.linalg.norm(x, ord=2)**2对于正方形
回答by Blckknght
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 awith some complicated expression without binding it to a variable, just remember to use parentheses as necessary, since **binds more tightly than most other operators: ((a-b)**2).sum()
我不知道性能是否有任何好处，但(a**2).sum()计算出正确的值并具有您想要的非重复参数。您可以a用一些复杂的表达式替换而不将其绑定到变量，只要记住必要时使用括号，因为**绑定比大多数其他运算符更紧密：((a-b)**2).sum()
