Re: Variable-length list?

• To: mathgroup at smc.vnet.net
• Subject: [mg83493] Re: Variable-length list?
• From: "Szabolcs HorvÃt" <szhorvat at gmail.com>
• Date: Wed, 21 Nov 2007 03:03:33 -0500 (EST)
• References: <fhu73f\$71m\$1@smc.vnet.net> <4742F526.5020201@gmail.com>

On Nov 21, 2007 3:32 AM, Hoa Bui <hoabui05 at gmail.com> wrote:
> Hi,
>
> Thanks for your comment, but this is not quite what I wanted to do.
> I tried to simplify the problem, but basically if I have a vector
> v={v[[1]],v[[2]],..v[[n]]}
> and each element v[[i]] has a range in which it can vary. I want to be
> able to search over all the possible combinations of the v[[i]]. (Each
> combination will return a vector, lets call it vi, and I want to use
> that vi to do something else.)
> The problem is that the number of dimensions (n) is not a fixed
> number, I want it to be a variable as well.
>
> Here's an example:
> If n=2, then v={a,b}. Say a={a1,a2} and b={b1,b2,b3}, then the combinations are:
> vi={{a1,b1}, {a1,b2}, {a1,b3}, {a2,b1}, {a2,b2}, {a2,b3}}
>
> If n=3, then v={a,b,c}. Say a={a1,a2}; b={b1,b2,b3}; c={c1,c2}, then I
> want to get:
> vi={{a1,b1,c1}, {a1,b2,c1}, {a1,b3,c1}, {a2,b1,c1}, {a2,b2,c1},
> {a2,b3,c1}, {a1,b1,c2}, {a1,b2,c2}, {a1,b3,c2}, {a2,b1,c2},
> {a2,b2,c2}, {a2,b3,c2}}
>
> If n=4, then I'll need another list d to represent my 4th dimension.

Well, why didn't you say so in the first place?  Take a look at Outer[]

In[1]:= f[v_] := Flatten[Outer[List, Sequence @@ v], Length[v] - 1]

In[2]:= f[{{a1, a2}, {b1, b2, b3}, {c1, c2}}]
Out[2]= {{a1, b1, c1}, {a1, b1, c2}, {a1, b2, c1}, {a1, b2, c2}, {a1,
b3, c1}, {a1, b3, c2}, {a2, b1, c1}, {a2, b1, c2}, {a2, b2,
c1}, {a2, b2, c2}, {a2, b3, c1}, {a2, b3, c2}}

>
> You see, it would be easy if n is a fixed number (because then I can
> just use Array[] with the specified dimensions {na, nb, nc,...}. But
> the trouble is, even the number n of dimensions is a variable.

That's why the second argument of Array was taken from the function
argument in my last example.  Did you try it?

In[3]:= g[{dd__Integer}]:=Array[Times,{dd}]

In[4]:= g[{3,4}]
Out[4]= {{1,2,3,4},{2,4,6,8},{3,6,9,12}}

In[5]:= g[{3,4,5}]
Out[5]= {{{1,2,3,4,5},{2,4,6,8,10},{3,6,9,12,15},{4,8,12,16,20}},{{2,4,6,8,10},{4,8,12,16,20},{6,12,18,24,30},{8,16,24,32,40}},{{3,6,9,12,15},{6,12,18,24,30},{9,18,27,36,45},{12,24,36,48,60}}}

> Obviously, I cannot use For[], because each For[] only solves 1
> dimension so I can only have a fixed number of dimensions searched.
>
> Is this doable?

Everything is doable in Mathematica :-)  But For[] is rarely a good
way to do things.

I hope this helps,
Szabolcs

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