Dealing with zeros in numpy array normalization

Question

I have a numpy array of 2D vectors, which I am trying to normalize as below. The array can have vectors with magnitude zero.

x = np.array([[0.0, 0.0], [1.0, 0.0]])
norms = np.array([np.linalg.norm(a) for a in x])

>>> x/norms
array([[ nan,   0.],
       [ inf,   0.]])

>>> nonzero = norms > 0.0
>>> nonzero
array([False,  True], dtype=bool)

Can I somehow use nonzero to apply the division only to x[i] such that nonzero[i] is True? (I can write a loop for this - just wondering if there's a numpy way of doing this)

Or is there a better way of normalizing the array of vectors, skipping all zero vectors in the process?

np.linalp.norm has a second argument axis you can use to increase speed, as discussed here: stackoverflow.com/a/19794741/1959808 — 0 _
– 0 _, Commented Nov 18, 2013 at 9:10

Jaime · Accepted Answer · 2013-07-21 15:17:03Z

8

If you can do the normalization in place, you can use your boolean indexing array like this:

nonzero = norms > 0
x[nonzero] /= norms[nonzero]

edited Jul 21, 2013 at 15:17

answered Jul 21, 2013 at 5:53

Jaime

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IanH · Accepted Answer · 2013-07-21 05:51:57Z

2

Here's one possible way of doing this

norms = np.sqrt((x**2).sum(axis=1,keepdims=True))
x[:] = np.where(norms!=0,x/norms,0.)

This uses np.where to do the substitution you need.

Note: in this case x is modified in place.

answered Jul 21, 2013 at 5:51

IanH

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2 Comments

Jaime Over a year ago

+1 Beat me toit by a few seconds! To do it in place I think it is better to use a boolean indexing array: idx = (norms != 0); x[idx] /= norms[idx]

M-V Over a year ago

@Jaime - thanks! Please post it as an answer so I can give you credit.

tom10 · Accepted Answer · 2013-07-21 05:51:05Z

1

It's probably easiest just to do the calculation and then modify the results to be what you'd like:

y = x/norms
y[np.isnan(y) | np.isinf(y)]=0

#y = array([[ 0.,  0.],
#       [ 0.,  0.]])

answered Jul 21, 2013 at 5:51

tom10

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