[Numpy-discussion] "Extended" Outer Product
Geoffrey Zhu
zyzhu2000 at gmail.com
Tue Aug 21 15:46:28 EDT 2007
On 8/21/07, Timothy Hochberg <tim.hochberg at ieee.org> wrote:
>
>
>
> On 8/21/07, Charles R Harris <charlesr.harris at gmail.com> wrote:
> >
> >
> >
> > On 8/20/07, Geoffrey Zhu < zyzhu2000 at gmail.com> wrote:
> > > Hi Everyone,
> > >
> > > I am wondering if there is an "extended" outer product. Take the
> > > example in "Guide to Numpy." Instead of doing an multiplication, I
> > > want to call a custom function for each pair.
> > >
> > > >>> print outer([1,2,3],[10,100,1000])
> > >
> > > [[ 10 100 1000]
> > > [ 20 200 2000]
> > > [ 30 300 3000]]
> > >
> > >
> > > So I want:
> > >
> > > [
> > > [f(1,10), f(1,100), f(1,1000)],
> > > [f(2,10), f(2, 100), f(2, 1000)],
> > > [f(3,10), f(3, 100), f(3,1000)]
> > > ]
> >
> >
> > Maybe something like
> >
> > In [15]: f = lambda x,y : x*sin(y)
> >
> > In [16]: a = array([[f(i,j) for i in range(3)] for j in range(3)])
> >
> > In [17]: a
> > Out[17]:
> > array([[ 0. , 0. , 0. ],
> > [ 0. , 0.84147098, 1.68294197],
> > [ 0. , 0.90929743, 1.81859485]])
> >
> > I don't know if nested list comprehensions are faster than two nested
> loops, but at least they avoid array indexing.
>
> This is just a general comment on recent threads of this type and not
> directed specifically at Chuck or anyone else.
>
> IMO, the emphasis on avoiding FOR loops at all costs is misplaced. It is
> often more memory friendly and thus faster to vectorize only the inner loop
> and leave outer loops alone. Everything varies with the specific case of
> course, but trying to avoid FOR loops on principle is not a good strategy.
>
I agree. My original post asked for solutions without using two nested
for loops because I already know the two for loop solution. Besides, I
was hoping that some version of 'outer' will take in a function
reference and call the function instead of doing multiplifcation.
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