The following does not raise an error:
a = np.arange(5) a[a>0] = a
although a.shape == (5,) while a[a>0].shape == (4,)
I get in on python2.6.5, numpy 1.4.1 on win32, and python 2.6.5, numpy 2.0.0.dev8469 on linux64.
Nadav.
On 7/29/2010 4:04 AM, Nadav Horesh wrote:
a = np.arange(5) a[a>0] = a
This has nothing to do with reusing ``a``::
>>> b = np.arange(50) >>> a[a>0] = b >>> a array([0, 0, 1, 2, 3])
Note however that reusing ``a`` is "unsafe". (You will get all zeros.)
fwiw, Alan Isaac
I was not aware that
a[b] = c
where b is an integer or boolean indexing array, is legal even if
a[b].shape != c.shape.
Nadav.
-----Original Message----- From: numpy-discussion-bounces@scipy.org on behalf of Alan G Isaac Sent: Thu 29-Jul-10 14:57 To: Discussion of Numerical Python Subject: Re: [Numpy-discussion] A bug in boolean indexing?
On 7/29/2010 4:04 AM, Nadav Horesh wrote:
a = np.arange(5) a[a>0] = a
This has nothing to do with reusing ``a``::
>>> b = np.arange(50) >>> a[a>0] = b >>> a array([0, 0, 1, 2, 3])
Note however that reusing ``a`` is "unsafe". (You will get all zeros.)
fwiw, Alan Isaac
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