I think that the only speedup you will get is defining an index only
once and reusing it.
2010/11/22 Ernest Adrogué :
> 22/11/10 @ 14:04 (-0600), thus spake Robert Kern:
>> > This way, I get the elements (0,1) and (1,1) which is what
>> > I wanted. The question is: is it possible to omit the [0,1]
22/11/10 @ 14:04 (-0600), thus spake Robert Kern:
> > This way, I get the elements (0,1) and (1,1) which is what
> > I wanted. The question is: is it possible to omit the [0,1]
> > in the index?
>
> No, but you can write generic code for it:
>
> t[np.arange(t.shape[0]), x, y]
Thank you. This i
22/11/10 @ 11:20 (-0800), thus spake John Salvatier:
> I didn't realize the x's and y's were varying the first time around.
> There's probably a way to omit it, but I think the conceptually
> simplest way is probably what you had to begin with. Build an index by
> saying i = numpy.arange(0, t.shape
2010/11/21 Ernest Adrogué :
> Hi,
>
> Suppose an array of shape (N,2,2), that is N arrays of
> shape (2,2). I want to select an element (x,y) from each one
> of the subarrays, so I get a 1-dimensional array of length
> N. For instance:
>
> In [228]: t=np.arange(8).reshape(2,2,2)
>
> In [229]: t
> O
22/11/10 @ 11:08 (-0800), thus spake Christopher Barker:
> On 11/21/10 11:37 AM, Ernest Adrogué wrote:
> >>so you want
> >>
> >>t[:,x,y]
> >
> >I tried that, but it's not the same:
> >
> >In [307]: t[[0,1],x,y]
> >Out[307]: array([1, 7])
> >
> >In [308]: t[:,x,y]
> >Out[308]:
> >array([[1, 3],
> >
I didn't realize the x's and y's were varying the first time around.
There's probably a way to omit it, but I think the conceptually
simplest way is probably what you had to begin with. Build an index by
saying i = numpy.arange(0, t.shape[0])
then you can do t[i, x,y]
On Mon, Nov 22, 2010 at 11:0
On 11/21/10 11:37 AM, Ernest Adrogué wrote:
>> so you want
>>
>> t[:,x,y]
>
> I tried that, but it's not the same:
>
> In [307]: t[[0,1],x,y]
> Out[307]: array([1, 7])
>
> In [308]: t[:,x,y]
> Out[308]:
> array([[1, 3],
> [5, 7]])
what is your t? Here's my example, which I think matches wh
Hi,
21/11/10 @ 11:28 (-0800), thus spake John Salvatier:
> yes use the symbol ':'
>
> so you want
>
> t[:,x,y]
I tried that, but it's not the same:
In [307]: t[[0,1],x,y]
Out[307]: array([1, 7])
In [308]: t[:,x,y]
Out[308]:
array([[1, 3],
[5, 7]])
No?
--
Ernest
yes use the symbol ':'
so you want
t[:,x,y]
2010/11/21 Ernest Adrogué :
> Hi,
>
> Suppose an array of shape (N,2,2), that is N arrays of
> shape (2,2). I want to select an element (x,y) from each one
> of the subarrays, so I get a 1-dimensional array of length
> N. For instance:
>
> In [228]: t=
read about basic slicing :
http://docs.scipy.org/doc/numpy/reference/arrays.indexing.html
On Sun, Nov 21, 2010 at 11:28 AM, John Salvatier
wrote:
> yes use the symbol ':'
>
> so you want
>
> t[:,x,y]
>
> 2010/11/21 Ernest Adrogué :
>> Hi,
>>
>> Suppose an array of shape (N,2,2), that is N arrays
Hi,
Suppose an array of shape (N,2,2), that is N arrays of
shape (2,2). I want to select an element (x,y) from each one
of the subarrays, so I get a 1-dimensional array of length
N. For instance:
In [228]: t=np.arange(8).reshape(2,2,2)
In [229]: t
Out[229]:
array([[[0, 1],
[2, 3]],
On Mon, Jun 21, 2010 at 7:10 PM, Robert Kern wrote:
> On Mon, Jun 21, 2010 at 17:42, Neal Becker wrote:
>> Robert Kern wrote:
>>
>>> On Mon, Jun 21, 2010 at 14:01, Neal Becker wrote:
Can I find an efficient way to do this?
I have a 2d array, A, 80 rows by 880 columns.
I
On Mon, Jun 21, 2010 at 17:42, Neal Becker wrote:
> Robert Kern wrote:
>
>> On Mon, Jun 21, 2010 at 14:01, Neal Becker wrote:
>>> Can I find an efficient way to do this?
>>>
>>> I have a 2d array, A, 80 rows by 880 columns.
>>>
>>> I have a vector, B, of length 80, with scalar indexes.
>>
>> I as
Robert Kern wrote:
> On Mon, Jun 21, 2010 at 14:01, Neal Becker wrote:
>> Can I find an efficient way to do this?
>>
>> I have a 2d array, A, 80 rows by 880 columns.
>>
>> I have a vector, B, of length 80, with scalar indexes.
>
> I assume you mean 880.
>
>> I want a vector output C where
>> C[
On Mon, Jun 21, 2010 at 14:01, Neal Becker wrote:
> Can I find an efficient way to do this?
>
> I have a 2d array, A, 80 rows by 880 columns.
>
> I have a vector, B, of length 80, with scalar indexes.
I assume you mean 880.
> I want a vector output C where
> C[i] = A[b[i],i] (i=0,879)
C = A[b,
Can I find an efficient way to do this?
I have a 2d array, A, 80 rows by 880 columns.
I have a vector, B, of length 80, with scalar indexes.
I want a vector output C where
C[i] = A[b[i],i] (i=0,879)
___
NumPy-Discussion mailing list
NumPy-Discussion@s
On Tue, Mar 30, 2010 at 10:13 AM, Tom K. wrote:
>
> This one bit me again, and I am trying to understand it better so I can
> anticipate when it will happen.
>
> What I want to do is get rid of singleton dimensions, and index into the
> last dimension with an array.
>
> In [1]: import numpy as np
On Tue, Mar 30, 2010 at 09:46, Charles R Harris
wrote:
>
>
> On Tue, Mar 30, 2010 at 8:13 AM, Tom K. wrote:
>>
>> This one bit me again, and I am trying to understand it better so I can
>> anticipate when it will happen.
>>
>> What I want to do is get rid of singleton dimensions, and index into t
On Tue, Mar 30, 2010 at 8:13 AM, Tom K. wrote:
>
> This one bit me again, and I am trying to understand it better so I can
> anticipate when it will happen.
>
> What I want to do is get rid of singleton dimensions, and index into the
> last dimension with an array.
>
> In [1]: import numpy as np
On 3/30/2010 10:13 AM, Tom K. wrote:
> What I want to do is get rid of singleton dimensions, and index into the
> last dimension with an array.
>>> x=np.zeros((10,1,1,1,14,1024))
>>> np.squeeze(x).shape
(10, 14, 1024)
hth,
Alan Isaac
___
NumPy-Discussio
This one bit me again, and I am trying to understand it better so I can
anticipate when it will happen.
What I want to do is get rid of singleton dimensions, and index into the
last dimension with an array.
In [1]: import numpy as np
In [2]: x=np.zeros((10,1,1,1,14,1024))
In [3]: x[:,0,0,0,:
On Sun, Dec 20, 2009 at 8:58 PM, Alan G Isaac wrote:
> Why is s3 F_CONTIGUOUS, and perhaps equivalently,
> why is its C_CONTIGUOUS data in s3.base (below)?
> Thanks,
> Alan Isaac
>
a3
> array([[ 0, 1, 2, 3, 4, 5],
> [ 6, 7, 8, 9, 10, 11]])
a3.flags
> C_CONTIGUOUS : True
Why is s3 F_CONTIGUOUS, and perhaps equivalently,
why is its C_CONTIGUOUS data in s3.base (below)?
Thanks,
Alan Isaac
>>> a3
array([[ 0, 1, 2, 3, 4, 5],
[ 6, 7, 8, 9, 10, 11]])
>>> a3.flags
C_CONTIGUOUS : True
F_CONTIGUOUS : False
OWNDATA : True
WRITEABLE : True
ALIG
On Thu, Mar 5, 2009 at 9:15 PM, Stéfan van der Walt wrote:
> Hi Robin
>
> 2009/3/5 Robin :
>> On Thu, Mar 5, 2009 at 10:57 AM, Robin wrote:
>>> On Thu, Mar 5, 2009 at 10:40 AM, Robin wrote:
Hi,
I have an indexing problem, and I know it's a bit lazy to ask the
list, sometime w
Hi Robin
2009/3/5 Robin :
> On Thu, Mar 5, 2009 at 10:57 AM, Robin wrote:
>> On Thu, Mar 5, 2009 at 10:40 AM, Robin wrote:
>>> Hi,
>>>
>>> I have an indexing problem, and I know it's a bit lazy to ask the
>>> list, sometime when people do interesting tricks come up so I hope no
>>> one minds!
>>
On Thu, Mar 5, 2009 at 10:57 AM, Robin wrote:
> On Thu, Mar 5, 2009 at 10:40 AM, Robin wrote:
>> Hi,
>>
>> I have an indexing problem, and I know it's a bit lazy to ask the
>> list, sometime when people do interesting tricks come up so I hope no
>> one minds!
>>
>> I have a 2D array X.shape = (a,
On Thu, Mar 5, 2009 at 10:40 AM, Robin wrote:
> Hi,
>
> I have an indexing problem, and I know it's a bit lazy to ask the
> list, sometime when people do interesting tricks come up so I hope no
> one minds!
>
> I have a 2D array X.shape = (a,b)
>
> and I want to change it into new array which is s
On Thu, Mar 5, 2009 at 10:40 AM, Robin wrote:
> Hi,
>
> I have an indexing problem, and I know it's a bit lazy to ask the
> list, sometime when people do interesting tricks come up so I hope no
> one minds!
>
> I have a 2D array X.shape = (a,b)
>
> and I want to change it into new array which is s
Hi,
I have an indexing problem, and I know it's a bit lazy to ask the
list, sometime when people do interesting tricks come up so I hope no
one minds!
I have a 2D array X.shape = (a,b)
and I want to change it into new array which is shape (2,(a*b)) which
has the following form:
[ X[0,0], X[0,1]
29 matches
Mail list logo
