oh i see thanks!
On Feb 10, 1:12 pm, John H Palmieri <[email protected]> wrote:
> On Feb 10, 11:58 am, green351 <[email protected]> wrote:
>
>
>
> > On Feb 10, 10:45 am, John H Palmieri <[email protected]> wrote:
>
> > > On Feb 10, 10:31 am, green351 <[email protected]> wrote:
>
> > > > Hi. I have the following problem: consider the following list of
> > > > compositions and its cartesian product with itself.
>
> > > > C83=Compositions(8, length=3);
> > > > cp=CartesianProduct(C83,C83)
>
> > > > I want to remove certain elements from this set/list and then to count
> > > > the elements left. For example I want to remove the element [[1,1,6],
> > > > [1,2,5]] and in general for [[a,b,c],[d,e,f]], I want to remove it if
> > > > a=d=1 OR b=e=1 Or c=f=1. There are other things I want to remove but
> > > > I think if I figure out the correct syntax for the above I could
> > > > probably figure the rest out. How would I do this?
>
> > > One way is to use "filter":
>
> > > Y = cp.filter(lambda x: x[0][0] != 1 or x[1][0] != 1)
>
> > > keeps pairs x=[[a,b,c], [d,e,f]] so that either a != 1 or d != 1, so
> > > you could do something like
>
> > > Y = cp.filter(lambda x: (x[0][0] != 1 or x[1][0] != 1) and (x[0][1] !=
> > > 1 or x[1][1] != 1) and
> > > (x[0][2] != 1 or x[1][2] != 1))
>
> > Thanks for the quick replies!
> > John,
> > So I tried the following using what you have above:
>
> > C22=Compositions(6, length=2);
> > cp2=CartesianProduct(C22,C22);
> > A= cp2.filter(lambda x: x[0][0] != 1 and x[1][1] != 1 and x[0][0] != 2
> > and x[1][1] != 2 and x[0][0] != 3 and x[1][1] != 3); A.list()
> > [[[4, 2], [1, 5]], [[4, 2], [2, 4]], [[5, 1], [1, 5]], [[5, 1], [2,
> > 4]]]
>
> > My goal here is to get rid of ALL [[a,b],[c,d]] with a=c or b=d. But
> > why does it also get rid of [[3,3],[2,4]]? Maybe I don't understand
> > the syntax. I'm guessing x[0][0] ! =1 refers to the element x= [[a,b],
> > [c,d]] with a=c=1. Is this correct?
>
> No: if x = [[a, b], [c,d]], then x[0] is [a,b], the 0th entry of x.
> Then x[0][0] is the 0th entry of [a,b], namely a. SImilarly, x[1] is
> [c,d]...
>
> If you want to get rid of all [[a,b], [c,d]] with a==c or b==d, then
> you want to keep all such terms with a != c and b != d. So you could
> say
>
> cp2.filter(lambda x: x[0][0] != x[1][0] and x[0][1] != x[1][1]).
>
> Does that clear anything up?
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