Calculating euclidean distance between consecutive points of an array with numpy

Question

I have an array which describes a polyline (ordered list of connected straight segments) as follows:

points = ((0,0),
          (1,2),
          (3,4),
          (6,5),
          (10,3),
          (15,4))
points = numpy.array(points, dtype=float)

Currently, I get a list of segment distances using the following loop:

segdists = []
for seg in xrange(points.shape[0]-1):
    seg = numpy.diff(points[seg:seg+2], axis=0)
    segdists.append(numpy.linalg.norm(seg))

I would like, instead, to apply a single function call, without loops, using some native Scipy/Numpy function.

The closest thing I could get is this:

from scipy.spatial.distance import pdist
segdists = pdist(points, metric='euclidean')

but in this later case, segdists provides EVERY distance, and I want to get only the distances between adjacent rows.

Also, I'd rather avoid creating custom functions (since I already have a working solution), but instead to use more "numpythonic" use of native functions.

Warren Weckesser · Accepted Answer · 2018-11-07 23:01:00Z

17

Here's one way:

Use the vectorized np.diff to compute the deltas:

d = np.diff(points, axis=0)

Then use np.hypot to compute the lengths:

segdists = np.hypot(d[:,0], d[:,1])

Or use a more explicit computation:

segdists = np.sqrt((d ** 2).sum(axis=1))

edited Nov 7, 2018 at 23:01

answered Nov 27, 2012 at 20:11

Warren Weckesser

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3 Comments

heltonbiker Over a year ago

After some dead-ends by myself, when you put it that way it's actually very straightforward. I've seen hypot being mentioned before, but googling "numpy hypot" doesn't return anything, I had to search at the numpy docs page. Thanks!

Varlor Over a year ago

Is this also possible in 3D ?

Warren Weckesser Over a year ago

@Varlor: Not with hypot, but the second version, segdists = np.sqrt((d ** 2).sum(axis=1)), works in 3D.

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