On Mon, Sep 14, 2009 at 4:46 AM, Ruben Salvador wrote:
> Hi there!
>
> It's a time since I'm asking this question to myself, and still don't know
> a Pythonic way to solve it. I want to create a 2D array where each *row* is
> a copy of an already existing 1D array. For example:
>
> In [21]: a = np
Perfect, that's exactly what I need! Somehow I missed this routine when
checking documentation :S
In [58]: a
Out[58]: array([1, 2, 3])
In [59]: np.tile(a, (5,1))
Out[59]:
array([[1, 2, 3],
[1, 2, 3],
[1, 2, 3],
[1, 2, 3],
[1, 2, 3]])
Thanks a lot!
On Mon, Sep 14, 2009 at 1:07 PM, Citi, Luca wr
if I get your question correctly, np.tile could be what you need
___
NumPy-Discussion mailing list
NumPy-Discussion@scipy.org
http://mail.scipy.org/mailman/listinfo/numpy-discussion
Ruben Salvador skrev:
> [...] I want to create a 2D array where each
> *row* is a copy of an already existing 1D array. For example:
> In [25]: a
> Out[25]: array([1, 2, 3])
> [...]
> In [30]: b
> Out[30]:
> array([[1, 2, 3],
> [1, 2, 3],
> [1, 2, 3],
> [1, 2, 3],
> [1, 2, 3]])
>
Without unders
Hi there!
It's a time since I'm asking this question to myself, and still don't know a
Pythonic way to solve it. I want to create a 2D array where each *row* is a
copy of an already existing 1D array. For example:
In [21]: a = np.array[1, 2, 3]
In [25]: a
Out[25]: array([1, 2, 3])
To create a 2D
