How can I make a function equal to 0?

Gabriel Genellina gagsl-py2 at yahoo.com.ar
Fri Mar 21 22:51:49 CET 2008


On 21 mar, 17:43, Martin Manns <mma... at gmx.net> wrote:

> Basically I have a lot of functions inside a numpy array,
> of which most return "" and actually are derived as follows:
>
> from numpy import *
>
> grid_size = 1000000
>
> class F(object):
>     def __cmp__(self, other): return 0
>     def __call__(self, dummy): return ""
> f = F()
> myarray = array([f] * grid_size, dtype="O")
>
> # I re-define an element in the array
>
> myarray[34424] = lambda x: "23"
>
> # Now, I would like to loop over all re-defined elements:
>
> for i in itertools.izip(*nonzero(myarray)):
>     print myarray[i]()
>
> If I fill the array with 0 instead of the functions that
> return "" this works really fast.
>
> However, I would like to be able call the content of each
> cell in the array as a function.

If you drop this condition then you could fill the array with zeroes
as before, replace only the interesting ones with actual functions,
and write:

for item in myarray[nonzero(myarray)]:
    print item() if callable(item) else item

> As I said, the callable object approach works but is
> about as slow as iterating through the whole array.
> Perhaps I should add a __call__ function to the built-in
> 0? But I doubt that this helps.

You can't - anyway, being Python code, would not help with the speed.

--
Gabriel Genellina



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