Re: Coding an inverse permutation
- To: mathgroup at smc.vnet.net
- Subject: [mg79116] Re: Coding an inverse permutation
- From: Peter Pein <petsie at dordos.net>
- Date: Thu, 19 Jul 2007 03:26:20 -0400 (EDT)
- References: <f7kdht$49n$1@smc.vnet.net>
Diana schrieb:
> Folks,
>
> I have the following list:
>
> aa={1, 2, 5, 3, 4, 8, 9, 7, 11, 6, 13, 17, 10, 16, 19, 15, 14, 20, 21,
> 23, 25, 12, 29, 31, 18, 22,
> 37, 27, 26, 28, 33, 35, 32, 24, 41, 43, 30, 34, 47, 39, 38, 40, 45,
> 49, 44, 36, 53, 55, 46, 52,
> 59, 51, 50, 56, 57, 61, 62, 42, 67, 71, 48, 58, 65, 63, 64, 68, 69,
> 73, 74, 54, 77, 79, 60, 76,
> 83, 75, 80, 70, 81, 89, 85, 66, 95, 97, 72, 82, 91, 87}
>
> I want to figure out a clean way to code its inverse permutation.
>
> The inverse permutation list would start as follows:
>
> bb={1,2,4,5,3,10,8,6, ...}
>
> Since "3" is in position 4 of aa, position 3 of bb will be "4".
> Since "5" is in position 3 of aa, position 5 of bb will be "3".
>
> Can someone give me a suggestion as to how to code this?
>
> Thanks, Diana
>
>
Hi Diana,
try Ordering[]:
In[2]:=
bb = Ordering[aa]
Out[2]=
{1, 2, 4, 5, 3, 10, 8, 6, 7, 13, 9,
22, 11, 17, 16, 14, 12, 25, 15, 18,
19, 26, 20, 34, 21, 29, 28, 30, 23,
37, 24, 33, 31, 38, 32, 46, 27, 41,
40, 42, 35, 58, 36, 45, 43, 49, 39,
61, 44, 53, 52, 50, 47, 70, 48, 54,
55, 62, 51, 73, 56, 57, 64, 65, 63,
82, 59, 66, 67, 78, 60, 85, 68, 69,
76, 74, 71, 72, 77, 79, 86, 75, 81,
88, 80, 87, 83, 84}
Peter