How can I do something like the following?
a = empty((5,7), dtype=4 element array of floats)
c = a[:,-1] # last column of 4 element arrays
a[0,0] = 2.0
print a[0,0]
[2. 2. 2. 2.]
a[1,0] = 3.0
a[0,1] = a[0,0] * a[1,0]
print a[0,1]
[6. 6. 6. 6.]
etc.
As always, thanks for the help!
Alex
Thanks Ben,
I submitted a patch with your fix.
Ticket 189 http://projects.scipy.org/scipy/numpy/ticket/189
David
2007/3/17, Ben Granett [EMAIL PROTECTED]:
Hi,
There seems to be a problem with histogramdd (numpy 1.0.1). Depending on
the
number of bins in each axis, it may not produce an
Stefan van der Walt wrote:
On Thu, Mar 22, 2007 at 04:33:53PM -0700, Travis Oliphant wrote:
I would rather opt for changing the spline fitting algorithm than for
padding with zeros.
From what I understand, the splines used in ndimage have the implicit
mirror-symmetric boundary
James Turner wrote:
By the way, ringing at sharp edges is an intrinsic feature of higher-
order spline interpolation, right? I believe this kind of interpolant
is really intended for smooth (band-limited) data. I'm not sure why
the pre-filtering makes a difference though; I don't yet understand
On 3/22/07, Bill Baxter [EMAIL PROTECTED] wrote:
On 3/23/07, Eric Firing [EMAIL PROTECTED] wrote:
Sebastian Haase wrote:
On 3/22/07, Stefan van der Walt [EMAIL PROTECTED] wrote:
On Thu, Mar 22, 2007 at 08:13:22PM -0400, Brian Blais wrote:
Hello,
I'd like to concatenate a couple
Hello all,
On Mar 23, 2007, at 3:04 AM, Stefan van der Walt wrote:
On Thu, Mar 22, 2007 at 11:20:37PM -0700, Zachary Pincus wrote:
The actual transform operators then use these coefficients to
(properly) compute pixel values at different locations. I just
assumed that the pre-filtering was
On Fri, Mar 23, 2007 at 11:09:03AM -0400, Robert Pyle wrote:
In [65]:concatenate((a.reshape(10,1),b.reshape(10,1)),axis=1)
Out[65]:
array([[ 0, -10],
[ 1, -9],
[ 2, -8],
[ 3, -7],
[ 4, -6],
[ 5, -5],
[ 6, -4],
On 3/24/07, Sebastian Haase [EMAIL PROTECTED] wrote:
Then of course, there's r_ and c_:
c = numpy.c_[a,b]
c = numpy.r_[a[None],b[None]].T
--bb
So,
None is the same as newaxis - right?
Yes, newaxis is None. None is newaxis. Same thing. I just don't see
much advantage in
Hi,
this is regarding the svn commit by wfspotz.
Author: [EMAIL PROTECTED]
Date: 2007-03-23 13:04:37 -0500 (Fri, 23 Mar 2007)
New Revision: 3593
Modified:
trunk/numpy/doc/swig/numpy.i
Log:
Added typecheck typemaps to INPLACE_ARRAY typemap suites
Hi wfspotz,
I was just wondering if you
On 3/23/07, Alexander Michael [EMAIL PROTECTED] wrote:
On 3/23/07, Nadav Horesh [EMAIL PROTECTED] wrote:
How about
a = empty((5,7,4))
c = a[...,-1]
Solely because I want to use the array with code that assumes it is
working with two-dimensional arrays but yet only performs operations
On 3/23/07, Sebastian Haase [EMAIL PROTECTED] wrote:
Hold on -- aren't the ... at the *end* always implicit:
I you have
a.shape = (6,5,4,3)
a[3,2] is the same as a[3,2,:,:] is the same as a[3,2,...]
only if you wanted a[...,3,2] you would have to change your code !?
Am I confused !?
Ha!
Anyone,
What is the easiest way to detect in python/C if an object is a subclass of
ndarray?
Chuck
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I mentioned in another thread Travis started on the scipy list that I
would find it useful if there were a function like dot() that could
multiply more than just two things.
Here's a sample implementation called 'mdot'.
mdot(a,b,c,d) == dot(dot(dot(a,b),c),d)
mdot(a,(b,c),d) ==
On 24/03/07, Bill Baxter [EMAIL PROTECTED] wrote:
I mentioned in another thread Travis started on the scipy list that I
would find it useful if there were a function like dot() that could
multiply more than just two things.
Here's a sample implementation called 'mdot'.
mdot(a,b,c,d) ==
On 23/03/07, Charles R Harris [EMAIL PROTECTED] wrote:
Anyone,
What is the easiest way to detect in python/C if an object is a subclass of
ndarray?
Um, how about isinstance or issubclass? (if you want strictness you
can look at whether x.__class__ is zeros(1).__class__)
Anne
Hi Zach,
Hmm, this is worrisome. There really shouldn't be ringing on
continuous-tone images like Lena -- right? (And at no step in an
image like that should gaussian filtering be necessary if you're
doing spline interpolation -- also right?)
That's hard to say. Just because it's mainly a
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