Re: [Numpy-discussion] avoiding the matrix copy performance hit

[Tim H. <tim...@co...>]

Bill Baxter wrote:

> Is there anyway to get around this timing difference?
> *
>   >>> import timeit
> **  >>> t1 = timeit.Timer("a = zeros((1000,1000),'d'); a += 1.;",  
> 'from numpy import zeros,mat')
> **  >>> t2 = timeit.Timer("a = mat(zeros((1000,1000),'d')); a += 
> 1.;",  'from numpy import zeros,mat')
>   >>> **t1.timeit(100)
>   1.8391627591141742
>   >>> t2.timeit(100)
>   3.2988266117713465
>
> *Copying all the data of the input array seems wasteful when the array 
> is just going to go out of scope.  Or is this not something to be 
> concerned about? 

You could try using copy=False:

 >>> import timeit
 >>> t1 = timeit.Timer("a = zeros((1000,1000),'d'); a += 1.;",  'from 
numpy import zeros,mat')
 >>> t2 = timeit.Timer("a = mat(zeros((1000,1000),'d'), copy=False); a 
+= 1.;",  'from numpy import z
eros,mat')
 >>> t1.timeit(100)
3.6538127052460578
 >>> t2.timeit(100)
3.6567186611706237

I'd also like to point out that your computer appears to be much faster 
than mine.

-tim


>
> It seems like a copy-by-reference version of mat() would be useful.  
> Really I can't imagine any case when I'd want both a matrix and the 
> original version of the array both hanging around as separate copies.  
> I can imagine either 1) the array is just a temp and I won't ever need 
> it again or 2) temporarily wanting a "matrix view" on the array's data 
> to do some linalg, after which I'll go back to using the original (now 
> modified) array as an array again.
>
> --bill