On 2010-08-06 13:11, Martin Spacek wrote:
> Josef, I'd forgotten you could use None to increase the dimensionality of an
> array. Neat. And, somehow, it's almost twice as fast as the Cython version!:
>
> >>> timeit a[np.arange(a.shape[0])[:, None], i]
> 10 loops, best of 3: 5.76 us per loop
On 2010-08-06 06:57, Keith Goodman wrote:
> You can speed it up by getting rid of two copies:
>
> idx = np.arange(a.shape[0])
> idx *= a.shape[1]
> idx += i
Keith, you're right of course. I'd forgotten about your earlier suggestion
about
operating in-place. Here's my new version:
def rowta
On Fri, Aug 6, 2010 at 3:01 AM, Martin Spacek wrote:
> Keith Goodman wrote:
> > Here's one way:
> >
> >>> a.flat[i + a.shape[1] * np.arange(a.shape[0])]
> > array([0, 3, 5, 6, 9])
>
>
> I'm afraid I made my example a little too simple. In retrospect, what I really
> want is to be able to u
On Fri, Aug 6, 2010 at 6:01 AM, Martin Spacek wrote:
> Keith Goodman wrote:
> > Here's one way:
> >
> >>> a.flat[i + a.shape[1] * np.arange(a.shape[0])]
> > array([0, 3, 5, 6, 9])
>
>
> I'm afraid I made my example a little too simple. In retrospect, what I really
> want is to be able to u
Keith Goodman wrote:
> Here's one way:
>
>>> a.flat[i + a.shape[1] * np.arange(a.shape[0])]
> array([0, 3, 5, 6, 9])
I'm afraid I made my example a little too simple. In retrospect, what I really
want is to be able to use a 2D index array "i", like this:
>>> a = np.array([[ 0, 1, 2,
choose might be slower if you weren't doing an "arange(N)" each time.
On Thu, Aug 5, 2010 at 1:51 PM, Keith Goodman wrote:
> On Thu, Aug 5, 2010 at 1:32 PM, wrote:
> > On Thu, Aug 5, 2010 at 4:07 PM, Martin Spacek
> wrote:
> >> josef.pkt wrote:
> > a = np.array([[0, 1],
> >>
On Thu, Aug 5, 2010 at 1:32 PM, wrote:
> On Thu, Aug 5, 2010 at 4:07 PM, Martin Spacek wrote:
>> josef.pkt wrote:
> a = np.array([[0, 1],
>> [2, 3],
>> [4, 5],
>> [6, 7],
>> [8, 9]])
> i = np.array([0, 1, 1, 0, 1])
>
On 08/05/2010 03:07 PM, Martin Spacek wrote:
> josef.pkt wrote:
a = np.array([[0, 1],
> [2, 3],
> [4, 5],
> [6, 7],
> [8, 9]])
i = np.array([0, 1, 1, 0, 1])
a[range(a.shape[0]), i]
> array([0, 3, 5, 6,
On Thu, Aug 5, 2010 at 4:07 PM, Martin Spacek wrote:
> josef.pkt wrote:
a = np.array([[0, 1],
> [2, 3],
> [4, 5],
> [6, 7],
> [8, 9]])
i = np.array([0, 1, 1, 0, 1])
a[range(a.shape[0]), i]
> array([0, 3, 5, 6, 9
josef.pkt wrote:
>>> a = np.array([[0, 1],
[2, 3],
[4, 5],
[6, 7],
[8, 9]])
>>> i = np.array([0, 1, 1, 0, 1])
>>> a[range(a.shape[0]), i]
array([0, 3, 5, 6, 9])
>>> a[np.arange(a.shape[0]), i]
array([0, 3, 5, 6, 9])
Thank
You may also use the choose function:
http://docs.scipy.org/doc/numpy/reference/generated/numpy.choose.html
choose(i, (a[:,0], a[:,1])
On Thu, Aug 5, 2010 at 10:31 AM, Keith Goodman wrote:
> On Thu, Aug 5, 2010 at 10:26 AM, wrote:
> > On Thu, Aug 5, 2010 at 1:12 PM, Martin Spacek
> wrote:
>
On Thu, Aug 5, 2010 at 10:26 AM, wrote:
> On Thu, Aug 5, 2010 at 1:12 PM, Martin Spacek wrote:
>> I want to take an n x m array "a" and index into it using an integer index
>> array
>> "i" of length n that will pull out the value at the designated column from
>> each
>> corresponding row of "a
On Thu, Aug 5, 2010 at 10:12 AM, Martin Spacek wrote:
> I want to take an n x m array "a" and index into it using an integer index
> array
> "i" of length n that will pull out the value at the designated column from
> each
> corresponding row of "a".
>
a = np.arange(10)
a.shape = 5, 2
On Thu, Aug 5, 2010 at 1:12 PM, Martin Spacek wrote:
> I want to take an n x m array "a" and index into it using an integer index
> array
> "i" of length n that will pull out the value at the designated column from
> each
> corresponding row of "a".
>
a = np.arange(10)
a.shape = 5, 2
>
I want to take an n x m array "a" and index into it using an integer index
array
"i" of length n that will pull out the value at the designated column from each
corresponding row of "a".
>>> a = np.arange(10)
>>> a.shape = 5, 2
>>> a
array([[0, 1],
[2, 3],
[4, 5],
[6, 7]
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