> Well what about A*(B*C)?
Point taken...
Johannes
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Hi,
in extension to the previous answers, I'd like to say that it is strongly
preferable to use dot(A,dot(B,C)) or dot(dot(A,B),C) instead of A*B*C.
The reason is that with dot(), you can control of which operation is performed
first, which can *massively* influence the time needed, depending o
On Thursday 26 October 2006 11:39, Vincent Schut wrote:
> > Of course I can easily loop over that axis, but if possible I'd like to
> > prevent array loops...
Depending on the problem's size, this could actually be the best solution --
if each particular sum is over enough elements, the overhead
Hi,
I absolutely do not know perl, so I do not know what the expression you posted
does.
However, the key is just to understand indexing in numpy:
if you have a matrix mat and index arrays index1, index2 with, lets say,
index1 = array([ 17, 19, 29])
index2 = array([ 12, 3, 9])
then the entri
> I'm just wondering if there is a way that i can increment all the values
> along a diagonal?
Assume you want to change mat.
# min() only necessary for non-square matrices
index = arange(min(mat.shape[0], mat.shape[1]))
# add 1 to each diagonal element
matrix[index, index] += 1
# add some other
Hi,
> > This seems like a rather common operation - I know I've needed it on
> > at least two occasions - is it worth creating some sort of C
> > implementation? What is the appropriate generalization?
>
> Some sort of indirect addressing infrastructure. But it looks like this
> could be tricky to
Hi,
one word in advance, instead of optimizing it is advisable to seek for a way
to refactorize the algorithm using smaller arrays, since this kind of
optimization almost certainly reduces readability. If you do it, comment
well. ;-)
If you have very large arrays and want to do some arithmetic
Am Mittwoch, 30. August 2006 19:20 schrieb Ghalib Suleiman:
> I'm somewhat new to both libraries...is there any way to create a 2D
> array of pixel values from an image object from the Python Image
> Library? I'd like to do some arithmetic on the values.
Yes.
To transport the data:
>>> import num
On Wednesday 19 July 2006 14:41,
[EMAIL PROTECTED] wrote:
> Your membership in the mailing list Numpy-discussion has been disabled
> due to excessive bounces The last bounce received from you was dated
> 19-Jul-2006. You will not get any more messages from this list until
> you re-enable your mem
On Wednesday 12 July 2006 02:07, Michael Sorich wrote:
> On 7/12/06, JJ <[EMAIL PROTECTED]> wrote:
> > 2) It would be very convienient to have some simple way to delete
> > selected columns of a matrix. For example, in matlab the command is
> > X[:,[3,5,7]]=[] to delete the three selected columns.
Hi,
> Opteron 64-bit, r2631 SVN.
>
> In [4]: depths_s2 = empty(shape=(5,),dtype=float)
> In [5]: depths_s2.fill(2.e5)
> In [6]: depths_s2
> Out[6]: array([ 20., 20., 20., 20., 20.])
>
> In [11]: depths_s2 = (empty(shape=(5,),dtype=float)).fill(2.e5)
> In [12]: print depths_
Hi,
> I'm not sure why bool arrays cannot be used as indices.
> The "natural" solution to the original problem seemed to be:
> M[:,V>0]
> but this is not allowed.
I started a thread on this earlier this year. Try searching the archive for
"boolean indexing" (if it comes back online somewhen).
T
Hi,
> ## Output:
> numpy.__version__: 0.9.8
> y: [ 0. 1. 2. 3. 4. 5. 6. 7. 8. 9.]
> y**2: [ 0. 1. 4. 9. 16. 25. 36. 49. 64. 81.]
> z: [ 0. 1. 2. 3. 4. 5. 6. 7. 8. 9.]
> z**2: [ 0.e+00 1.e+00 1.6000e+01 8.1000e+01
>2.5600e+0
Hi,
> def d4():
> d = zeros([4, 1000], dtype=float)
> for i in range(4):
> xy = A[i] - B
> d[i] = sqrt( sum(xy**2, axis=1) )
> return d
>
> Maybe there's another alternative to d4?
> Thanks again,
I think this is the fastest you can get. Maybe it would be nicer to use
Hi,
def dtest():
A = random( [4,2])
B = random( [1000,2])
# drawback: memory usage temporarily doubled
# solution see below
d = A[:, newaxis, :] - B[newaxis, :, :]
# written as 3 expressions for more clarity
d = sqrt((d**2).sum(axis=2))
return d
def dtest_lowmem(
> I've tried to send a message twice to scipy-user since friday without
> success (messages don't come back to me but I don't receive any message
> from scipy-user too and they don't appear in archives).
> Note that since friday there are no new messages from that list.
>
> Is scipy-user working?
Hi,
> I'm just starting with numpy (via scipy) and I'm wanting to perform
> adaptive thresholding
> (http://www.cee.hw.ac.uk/hipr/html/adpthrsh.html) on an image.
> Basically that means that I need to get a threshold for each pixel by
> examining the pixels around it. In numpy this translates to f
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