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Re: Did something change?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg58122] Re: [mg58109] Did something change?
  • From: Andrzej Kozlowski <andrzej at akikoz.net>
  • Date: Sun, 19 Jun 2005 03:43:25 -0400 (EDT)
  • References: <200506181008.GAA08878@smc.vnet.net> <A5BD3CF8-FDED-4DAA-8E0F-A0CC0B8ADD02@akikoz.net>
  • Sender: owner-wri-mathgroup at wolfram.com

On 18 Jun 2005, at 22:00, Andrzej Kozlowski wrote:

>
> On 18 Jun 2005, at 19:08, Ronald Bruck wrote:
>
>
>>
>> Recently I created a list of several matrices, something like
>>
>>    Do[F[i] = IdentityMatrix[8]; F[i][[1,3]] = 7, {i,10}]
>>
>> and was surprised to get the message that F[i] wasn't a symbol.  I'm
>> pretty sure code similar to that worked prior to Mathematica 5.  Did
>> something change?
>>
>> I've also tried Symbolize[F[i]] in that loop, but it doesn't help;
>> Mathematica refuses to access part [[1,3]] of F[i], claiming F[i]  
>> isn't
>> a symbol.
>>
>> How are you supposed to set the individual values of a matrix if you
>> can't do this?  This is an EXTREMELY unhelpful "feature".
>>
>> --Ron Bruck
>>
>>
>>
>
>
> This is the way Set and Part work in Mathematica work and always  
> have done. One way to deal with this type of problem is to use  
> ReplacePart instead of Part:
>
> Do[F[i] = IdentityMatrix[8]; F[i] = ReplacePart[F[i], 7, {1, 3}],  
> {i, 10}]
>
> or, if you really want to use Set and Part as in your original  
> example you have to introduce a temporary symbol, e.g.:
>
> Do[F[i] = IdentityMatrix[8]; Block[{B = F[i]}, B[[1, 3]] =
>    7; F[i] = B], {i, 10}]
>
>
> Andrzej Kozlowski
>
> Chiba,Japan
>
>


I forgot to say that a much better approach than the above is to give  
up using Do altogether and instead use a functional approach:

F = Table[ IdentityMatrix[8], {i, 10}];

F[[All, 1, 3]] = 7;


Now you have


F[[1]]


{{1, 0, 7, 0, 0, 0, 0, 0}, {0, 1, 0, 0, 0, 0, 0, 0},
   {0, 0, 1, 0, 0, 0, 0, 0}, {0, 0, 0, 1, 0, 0, 0, 0},
   {0, 0, 0, 0, 1, 0, 0, 0}, {0, 0, 0, 0, 0, 1, 0, 0},
   {0, 0, 0, 0, 0, 0, 1, 0}, {0, 0, 0, 0, 0, 0, 0, 1}}

and so on.

ANdrzej


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