Oh yeah this could be. Floating point equality and bitwise equality are not the
same thing.
-----Original Message-----
From: "Jaime Fernández del Río" <jaime.f...@gmail.com>
Sent: 22-8-2014 16:22
To: "Discussion of Numerical Python" <numpy-discussion@scipy.org>
Subject: Re: [Numpy-discussion] np.unique with structured arrays
I can confirm, the issue seems to be in sorting:
>>> np.sort(V_)
array([([0.5, 0.0, 1.0],), ([0.5, 0.0, -1.0],), ([0.5, -0.0, 1.0],),
([0.5, -0.0, -1.0],)],
dtype=[('v', '<f4', (3,))])
These I think are handled by the generic sort functions, and it looks like the
comparison function being used is the one for a VOID dtype with no fields, so
it is being done byte-wise, hence the problems with 0.0 and -0.0. Not sure
where exactly the bug is, though...
Jaime
On Fri, Aug 22, 2014 at 6:20 AM, Nicolas P. Rougier <nicolas.roug...@inria.fr>
wrote:
Hello,
I've found a strange behavior or I'm missing something obvious (or np.unique is
not supposed to work with structured arrays).
I'm trying to extract unique values from a simple structured array but it does
not seem to work as expected.
Here is a minimal script showing the problem:
import numpy as np
V = np.zeros(4, dtype=[("v", np.float32, 3)])
V["v"] = [ [0.5, 0.0, 1.0],
[0.5, -1.e-16, 1.0], # [0.5, +1.e-16, 1.0] works
[0.5, 0.0, -1.0],
[0.5, -1.e-16, -1.0]] # [0.5, +1.e-16, -1.0]] works
V_ = np.zeros_like(V)
V_["v"][:,0] = V["v"][:,0].round(decimals=3)
V_["v"][:,1] = V["v"][:,1].round(decimals=3)
V_["v"][:,2] = V["v"][:,2].round(decimals=3)
print np.unique(V_)
[([0.5, 0.0, 1.0],) ([0.5, 0.0, -1.0],) ([0.5, -0.0, 1.0],) ([0.5, -0.0,
-1.0],)]
While I would have expected:
[([0.5, 0.0, 1.0],) ([0.5, 0.0, -1.0],)]
Can anyone confirm ?
Nicolas
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