On 10/02/2008, Matthew Brett <[EMAIL PROTECTED]> wrote:
> > Ah, I see. You definitely do not want to reassign the .data buffer in
> > this case. An out= parameter does not reassign the memory location
> > that the array object points to. It should use the allocated memory
> > that was already there
On Feb 8, 2008 5:29 AM, Francesc Altet <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I'm a bit confused that the sort method of a string character doesn't
> allow a mergesort:
>
> >>> s = numpy.empty(10, "S10")
> >>> s.sort(kind="merge")
> TypeError: desired sort not supported for this type
>
> However, by
Matthew Brett wrote:
>>> import numpy as np
>>> a = np.arange(10)
>>> b = np.arange(10)+1
>>> a.data = b.data # raises error, but I hope you see what I mean
>>>
>>> ?
>> Not really, no. Can you describe your use case in more detail?
>
> Yes - I am just writing the new median implementation. To al
On Feb 10, 2008 7:17 PM, Matthew Brett <[EMAIL PROTECTED]> wrote:
> > Ah, I see. You definitely do not want to reassign the .data buffer in
> > this case. An out= parameter does not reassign the memory location
> > that the array object points to. It should use the allocated memory
> > that was alr
On Feb 10, 2008 7:43 PM, Brad Malone <[EMAIL PROTECTED]> wrote:
> Hi, I am receiving a "AttributeError: incompatible shape for a
> non-contiguous array" error. A quick illustration of the type of code
> that gives me the error is shown below:
>
> from n
Hi, I am receiving a "AttributeError: incompatible shape for a
non-contiguous array" error. A quick illustration of the type of code
that gives me the error is shown below:
from numpy import *
list=[i for i in range(0,27)]
c=array(list)
c.shape=(3,3,3)
> Ah, I see. You definitely do not want to reassign the .data buffer in
> this case. An out= parameter does not reassign the memory location
> that the array object points to. It should use the allocated memory
> that was already there. It shouldn't "copy" anything at all;
> otherwise, "median(x, o
On Feb 10, 2008 6:48 PM, Matthew Brett <[EMAIL PROTECTED]> wrote:
> > > import numpy as np
> > > a = np.arange(10)
> > > b = np.arange(10)+1
> > > a.data = b.data # raises error, but I hope you see what I mean
> > >
> > > ?
> >
> > Not really, no. Can you describe your use case in more detail?
>
>
> > import numpy as np
> > a = np.arange(10)
> > b = np.arange(10)+1
> > a.data = b.data # raises error, but I hope you see what I mean
> >
> > ?
>
> Not really, no. Can you describe your use case in more detail?
Yes - I am just writing the new median implementation. To allow
future optimization
On Feb 10, 2008 5:15 PM, Matthew Brett <[EMAIL PROTECTED]> wrote:
> Hi,
>
> I am sorry if I have missed something obvious, but is there any way in
> python of doing this:
>
> import numpy as np
> a = np.arange(10)
> b = np.arange(10)+1
> a.data = b.data # raises error, but I hope you see what I mea
Hi,
I am sorry if I have missed something obvious, but is there any way in
python of doing this:
import numpy as np
a = np.arange(10)
b = np.arange(10)+1
a.data = b.data # raises error, but I hope you see what I mean
?
Thanks a lot for any pointers.
Matthew
Hi,
I just noticed this:
>From the sort method docstring:
axis : integer
Axis to be sorted along. None indicates that the flattened array
should be used. Default is -1.
In [40]: import numpy as N
In [41]: a = N.arange(10)
In [42]: N.sort(a, None)
Out[42]: array([0, 1, 2, 3
I need just a single number "in avarage".
I have committed some changes to NLP/NSP ralg solver from
scikits.openopt, for non-noisy funcs it works better, but for noisy
funcs vise versa, hence now my examples/nssolveVSfsolve.py doesn't work
as it should be, so I need to implement "noise" paramete
On Sun, Feb 10, 2008 at 4:23 AM, dmitrey <[EMAIL PROTECTED]> wrote:
> hi all,
> I need a good estimation of noise value for simple calculations.
>
> I.e. when I calculate something like sin(15)+cos(80) I get a solution
> with precision, for example, 1e-11.
>
> I guess the precision depends on syst
hi all,
I need a good estimation of noise value for simple calculations.
I.e. when I calculate something like sin(15)+cos(80) I get a solution
with precision, for example, 1e-11.
I guess the precision depends on system arch, isn't it?
So what's the best way to estimate the value?
I guess here
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