Re: MapThread and If
- To: mathgroup at smc.vnet.net
- Subject: [mg68039] Re: MapThread and If
- From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
- Date: Fri, 21 Jul 2006 05:37:31 -0400 (EDT)
- Organization: The Open University, Milton Keynes, UK
- References: <e9nkro$9u3$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Arkadiusz Majka wrote:
> Hi,
>
> Of course
>
> MapThread[If[#1 == #2, 0, tr] &, {{1, 2, 3}, {2, 2, 4}}]
>
> I get
>
> {tr, 0, tr}
>
> and for
>
> MapThread[If[#1 == #2, 0, tr] &, {{1, 2, 3}, {2, 2, {3, 4}}}]
>
> I get
>
> {tr, 0, If[3 == {3, 4}, 0, tr]}
>
> How I can build in above MapThread expression information that if 'a'
> is equal to any element from the list {a,b,c,d} the result is 0. In
> above example we have If[3=={3,4},0,tr] what I want to be 0, because 3
> is equal to an element belonging to {3,4}.
>
> Thx,
>
> Arek
>
Hi Arek,
What about the following expression?
MapThread[If[Length[Intersection[Flatten[{#1}], Flatten[{#2}]]] > 0, 0,
tr] & , {{1, 2, 3}, {2, 2, {3, 4}}}]
returns
{tr, 0, 0}
Since Intersection works only with lists, we transform each variable in
list to avoid atomic expression and then we Flatten them as one
dimensional lists.
HTH,
Jean-Marc