[Numpy-discussion] from_function

Yakov Keselman yakov.keselman@gmail....
Tue Feb 10 15:11:36 CST 2009


Perhaps you can do something along the following lines to get around
this limitation:

#################

# parameterizes the original function by delta, size.
def parameterized_function(delta, size, function):
    center = (size-1)/2.0
    return lambda i: function( (i-center)*delta )


# the function to which we want to apply "fromfunction".
def square(x): return x*x

# test script
from numpy import fromfunction
Delta = 0.1
Size = 9
print fromfunction( parameterized_function(Delta, Size, square), (Size,) )

###########

[ 0.16  0.09  0.04  0.01  0.    0.01  0.04  0.09  0.16]


On 2/3/09, Neal Becker <ndbecker2@gmail.com> wrote:
> I've been using something I wrote:
>
> coef_from_function (function, delta, size)
> which does (c++ code):
>
>   double center = double(size-1)/2;
>   for (int i = 0; i < size; ++i)
>     coef[i] = call<value_t> (func, double(i - center) * delta);
>
> I thought to translate this to np.fromfunction.  It seems fromfunction is
> not as flexible, it uses only a fixed integer grid?
>
>
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