Norm of a arrays of vectors in python

Question

I have this array ``` A = array([[-0.49740509, -0.48618909, -0.49145315], [-0.48959259, -0.48618909, -0.49145315], [-0.49740509, -0.47837659, -0.49145315], ..., [ 0.03079315, -0.01194593, -0.06872366], [ 0.03054901, -0.01170179, -0.06872366], [ 0.03079315, -0.01170179, -0.06872366]]) ``` which is a collection of 3D vector. I was wondering if I could use a vectorial operation to get an array with the norm of each of my vector. I tried with `norm(A)` but it didn't work.

Accepted Answer

Doing it manually might be fastest (although there's always some neat trick someone posts I didn't think of):

```
In [75]: from numpy import random, array

In [76]: from numpy.linalg import norm

In [77]:

In [77]: A = random.rand(1000,3)

In [78]: timeit normedA_0 = array([norm(v) for v in A])
100 loops, best of 3: 16.5 ms per loop

In [79]: timeit normedA_1 = array(map(norm, A))
100 loops, best of 3: 16.9 ms per loop

In [80]: timeit normedA_2 = map(norm, A)
100 loops, best of 3: 16.7 ms per loop

In [81]: timeit normedA_4 = (A*A).sum(axis=1)**0.5
10000 loops, best of 3: 46.2 us per loop
```

This assumes everything's real. Could multiply by the conjugate instead if that's not true.

Update: Eric's suggestion of using `math.sqrt` won't work -- it doesn't handle numpy arrays -- but the idea of using sqrt instead of `**0.5` is a good one, so let's test it.

```
In [114]: timeit normedA_4 = (A*A).sum(axis=1)**0.5
10000 loops, best of 3: 46.2 us per loop

In [115]: from numpy import sqrt

In [116]: timeit normedA_4 = sqrt((A*A).sum(axis=1))
10000 loops, best of 3: 45.8 us per loop
```

I tried it a few times, and this was the largest difference I saw.

Norm of a arrays of vectors in python

Solution

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