Re: Select from list

• To: mathgroup at smc.vnet.net
• Subject: [mg87749] Re: [mg87729] Select from list
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Wed, 16 Apr 2008 06:50:45 -0400 (EDT)
• References: <200804160904.FAA23961@smc.vnet.net>

On 16 Apr 2008, at 18:04, Steve Gray wrote:
> I have a list like this:
>
> ptX=
> {{1, 2, 3, 4}, {1, 2, 3, 5}, {1, 2, 4, 5}, {1, 3, 4, 5}, {2, 3, 4, 5}}
>
> and I want a list pointing to all the sublists above that contain both
> a 2 and a 3. In this example I would get {1,2,5}. The best solution I
> have, with more generality, is:
>
> va = 2;
> vb = 3;
> za = Map[Cases[#1, va]&, ptX] /. {} -> {0}
> zb = Map[Cases[#1, vb]&, ptX] /. {} -> {0}
> Flatten[Position[za*zb, {va*vb}]]
>
> which gives
>
> {{2}, {2}, {2}, {0}, {2}}
> {{3}, {3}, {0}, {3}, {3}}
> {1, 2, 5}.
>
> (This doesn't work if va or vb is zero. That's ok.)
> There's probably a better way. Anyone? Thank you.
>
> Steve Gray
>

Flatten[Position[ptX, {___, 2, ___, 3, ___}]]
{1, 2, 5}

That's assuming that 3 comes always after 2 in your list. If not:

Flatten[Position[ptX, {___, 2, ___, 3, ___} | {___, 3, ___, 2, ___}]]

Andrzej Kozlowski

• Prev by Date: DifferenitalD vs CapitalDifferenitalD
• Next by Date: Re: Select from list
• Previous by thread: Select from list
• Next by thread: Re: Select from list