2010/1/23 Alan G Isaac ais...@american.edu:
Suppose x and y are conformable 2d arrays.
I now want x to become a duplicate of y.
I could create a new array:
x = y.copy()
or I could assign the values of y to x:
x[:,:] = y
As expected the latter is faster (no array creation).
Are there
On 1/23/2010 5:01 PM, Anne Archibald wrote:
If both arrays are C contiguous, or more generally contiguous blocks
of memory with the same strided structure, you might get faster
copying by flattening them first, so that it can go in a single
memcpy().
I may misuderstand this. Did you just
On Sat, Jan 23, 2010 at 2:31 PM, Alan G Isaac ais...@american.edu wrote:
On 1/23/2010 5:01 PM, Anne Archibald wrote:
If both arrays are C contiguous, or more generally contiguous blocks
of memory with the same strided structure, you might get faster
copying by flattening them first, so that it
On Sat, Jan 23, 2010 at 4:00 PM, Keith Goodman kwgood...@gmail.com wrote:
On Sat, Jan 23, 2010 at 2:31 PM, Alan G Isaac ais...@american.edu wrote:
On 1/23/2010 5:01 PM, Anne Archibald wrote:
If both arrays are C contiguous, or more generally contiguous blocks
of memory with the same
2010/1/23 Alan G Isaac ais...@american.edu:
On 1/23/2010 5:01 PM, Anne Archibald wrote:
If both arrays are C contiguous, or more generally contiguous blocks
of memory with the same strided structure, you might get faster
copying by flattening them first, so that it can go in a single
On 1/23/2010 6:00 PM, Keith Goodman wrote:
x = y.view()
Thanks, but I'm not looking for a view.
And I need x to own its data.
Alan
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On 1/23/2010 7:29 PM, Anne Archibald wrote:
I had in mind accessing the underlying data through views
that were flat:
In [3]: x = np.random.random((1000,1000))
In [4]: y = np.random.random((1000,1000))
In [5]: xf = x.view()
In [6]: xf.shape = (-1,)
In [7]: yf = y.view()
In [8]:
