Hi Stefan,
Thanks for the kindly reply...
> Since both those objects have an __array_priority__ of 0.0, I guess it
> just takes whichever class comes first.
> In [15]: class A(np.ndarray):
> : __array_priority__ = -1.0
I think, playing more...
For np.multiply, it does not seem possib
2008/8/26 Anne Archibald <[EMAIL PROTECTED]>:
> 2008/8/22 Catherine Moroney <[EMAIL PROTECTED]>:
>> I'm looking for a way to acccomplish the following task without lots
>> of loops involved, which are really slowing down my code.
>>
>> I have a 128x512 array which I want to break down into 2x2 squa
Hi Neil
2008/8/26 Neil Crighton <[EMAIL PROTECTED]>:
>>
>> - Should we have a separate User manual and a Reference manual, or only
>> a single manual?
>>
>
> Are there still plans to write a 10 page 'Getting started with NumPy'
> document? I think this would be very useful. Ideally a 'getting st
Hey Matthew
2008/8/27 Matthew Brett <[EMAIL PROTECTED]>:
> In [148]: type(np.multiply(arr, obj)) # this is what I expected
> Out[148]:
>
> In [149]: type(np.multiply.outer(arr, obj)) # this is not - I expected
> class A again
> Out[149]:
Since both those objects have an __array_priority__ of 0.
sorry, 1D array
this is perfect, thanks.
On Aug 27, 10:18 pm, Gary Ruben <[EMAIL PROTECTED]> wrote:
> I don't know what you mean by a 1D vector, but for a 3-vector, you can
> do this (also works for N-dimensions)
>
> In [1]: a=r_[1.,2.,3.]
> In [2]: a
> Out[2]: array([ 1., 2., 3.])
> In [3]: b=
I don't know what you mean by a 1D vector, but for a 3-vector, you can
do this (also works for N-dimensions)
In [1]: a=r_[1.,2.,3.]
In [2]: a
Out[2]: array([ 1., 2., 3.])
In [3]: b=a/norm(a)
In [4]: b
Out[4]: array([ 0.26726124, 0.53452248, 0.80178373])
Gary R
> bit of a newb question, is t
bit of a newb question, is there a method for normalising a 1D vector
so it ends up with magnitude 1?
I can do it manually but I was hoping there was a neat numpy - or
scipy - trick. I've been web surfing but nothing really leaps out
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On Wed, Aug 27, 2008 at 12:01 PM, Claude Gouedard <[EMAIL PROTECTED]> wrote:
> Ok ,
> The same for asarray(1) ..
> The problem is that
> aa=asarray(1) is an numpy.array (right ? ) with a size 1 and a shape ( ) !
> No surprising ?
For me, this is not surprising at all :-) . Furthermore, if you try
Claude Gouedard wrote:
> Ok ,
> The same for asarray(1) ..
> The problem is that
> aa=asarray(1) is an numpy.array (right ? ) with a size 1 and a shape ( ) !
> No surprising ?
I think it is considered a 0-dimensional array (= a scalar).
--
Dag Sverre
_
Ok ,
The same for asarray(1) ..
The problem is that
aa=asarray(1) is an numpy.array (right ? ) with a size 1 and a shape ( ) !
No surprising ?
Claude
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Wed, 27 Aug 2008 16:48:48 +0200, Claude Gouedard wrote:
> Hi ,
>
> I'm just surprised by the behaviour of numpy.asarray on lists.
>
> Can someone comment this :
> =
> a=(1)
> aa=asarray(a)
> print aa.size , aa.shape
>>> 1 ( )
> =
() are not list delim
Manuel Metz wrote:
> Claude Gouedard wrote:
>> Hi ,
>>
>> I'm just surprised by the behaviour of numpy.asarray on lists.
>>
>> Can someone comment this :
>> =
>> a=(1)
>> aa=asarray(a)
>> print aa.size , aa.shape
1 ( )
>> =
>>
>> The shape doesnot r
Claude Gouedard wrote:
> Hi ,
>
> I'm just surprised by the behaviour of numpy.asarray on lists.
>
> Can someone comment this :
> =
> a=(1)
> aa=asarray(a)
> print aa.size , aa.shape
>>> 1 ( )
> =
>
> The shape doesnot reflect the actual size.
Becaus
Hi ,
I'm just surprised by the behaviour of numpy.asarray on lists.
Can someone comment this :
=
a=(1)
aa=asarray(a)
print aa.size , aa.shape
>> 1 ( )
=
The shape doesnot reflect the actual size.
If a=(1,2) there is no problem .
Hi Travis and team,
I am just writing some docs for subclassing, and ran into some
behavior I didn't understand:
In [143]: class A(np.ndarray): pass
In [144]: arr = np.arange(5)
In [145]: obj = arr.copy().view(A)
In [146]: type(obj)
Out[146]:
In [147]: obj.__array_priority__ # shouldn't thi
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