[Numpy-discussion] Simple multi-arg wrapper for dot()
Bill Baxter
wbaxter@gmail....
Sat Mar 24 02:25:47 CDT 2007
On 3/24/07, Anne Archibald <peridot.faceted@gmail.com> wrote:
> On 24/03/07, Bill Baxter <wbaxter@gmail.com> wrote:
> > I mentioned in another thread Travis started on the scipy list that I
> > would find it useful if there were a function like dot() that could
> > multiply more than just two things.
> >
> > Here's a sample implementation called 'mdot'.
> >
> > mdot(a,b,c,d) ==> dot(dot(dot(a,b),c),d)
> > mdot(a,(b,c),d) ==> dot(dot(a,dot(b,c),d)
> > mdot(a,(b,(c,d))) ==> dot(a,dot(b,dot(c,d))
> >
> > ---
> > def mdot(*args):
> > """Multiply all the arguments using matrix product rules.
> > The output is equivalent to multiplying the arguments one by one
> > from left to right using dot().
> > Precedence can be controlled by creating tuples of arguments,
> > for instance mdot(a,((b,c),d)) multiplies a (a*((b*c)*d)).
> > Note that this means the output of dot(a,b) and mdot(a,b) will differ if
> > a or b is a pure tuple of numbers.
> > """
> > if len(args)==1:
> > return args[0]
> > elif len(args)==2:
> > return _mdot_r(args[0],args[1])
> > else:
> > return _mdot_r(args[:-1],args[-1])
> >
> > def _mdot_r(a,b):
> > """Recursive helper for mdot"""
> > if type(a)==types.TupleType:
> > if len(a)>1:
> > a = mdot(*a)
> > else:
> > a = a[0]
> > if type(b)==types.TupleType:
> > if len(b)>1:
> > b = mdot(*b)
> > else:
> > b = b[0]
> > return numpy.dot(a,b)
>
> You can do better:
> In [1]: from numpy import *
>
> In [2]: a = array([[0,-1],[1,0]])
>
> In [3]: reduce(dot,(a,a,a,a))
> Out[3]:
> array([[1, 0],
> [0, 1]])
>
> In [4]: def mdot(*args):
> ...: return reduce(dot,args)
> ...:
>
> In [5]: mdot(a,a,a,a)
> Out[5]:
> array([[1, 0],
> [0, 1]])
>
Nice, but how does that fare on things like mdot(a,(b,c),d) ? I'm
pretty sure it doesn't handle it.
I think an mdot that can only multiply things left to right comes up
short compared to an infix operator that can easily use parentheses to
control order.
--bb
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