On 8/27/06, Carlos Pita <[EMAIL PROTECTED]> wrote:
Maybe something like this:
I n [12]: a = empty((3,3), dtype=int)
In [13]: a.fill(11)
In [14]: a
Out[14]:
array([[11, 11, 11],
[11, 11, 11],
[11, 11, 11]])
I haven't timed it, so don't know how fast it is. Looking at this makes me think fill should return the array so that one could do something like: a = empty((3,3), dtype=int).fill(10)
Chuck
Hi all!
Is there a more efficient way of creating a constant K-valued array of size N than:
zeros(N) + K
?
Maybe something like this:
I n [12]: a = empty((3,3), dtype=int)
In [13]: a.fill(11)
In [14]: a
Out[14]:
array([[11, 11, 11],
[11, 11, 11],
[11, 11, 11]])
I haven't timed it, so don't know how fast it is. Looking at this makes me think fill should return the array so that one could do something like: a = empty((3,3), dtype=int).fill(10)
Chuck
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