Re: Multiple use of Set on lists
- To: mathgroup at smc.vnet.net
- Subject: [mg119761] Re: Multiple use of Set on lists
- From: Oleksandr Rasputinov <oleksandr_rasputinov at hmamail.com>
- Date: Tue, 21 Jun 2011 05:51:38 -0400 (EDT)
- References: <itm0of$143$1@smc.vnet.net> <201106201204.IAA07658@smc.vnet.net> <itolks$erm$1@smc.vnet.net>
The problem is due to Set being HoldFirst but MapThread not having any
corresponding Hold attributes. While the c's are being substituted for their values too early, this is not an issue with MapThread per se; rather it is a consequence of how attributes affect the order of evaluation in Mathematica.
An ugly hack which solves the problem as stated is the following:
SubValues[c] = Block[{c},
MapThread[Set, {cs, rands}];
SubValues[c]
];
However, it may be better to reformulate the problem in such a way
that this construct is not needed.
On Jun 21, 12:38 am, Jonathan Frazer <J.Fra... at sussex.ac.uk> wrote:
> Hello,
>
> I have a largish list of awkwardly "indexed variables", of the form cs =
> {c[1/2][-(1/2)][1/2][-(1/2)][-1], c[1/2][-(1/2)][1/2][1/2][-1]......etc....
> ....} which need to be assigned random values multiple times inside a Do loop. I was hoping I could generate a new list of random numbers each time then do something like
>
> MapThread[Set, {cs, rands}]
>
> Where "rands" would be the list of random numbers. I'm clearly misunderstanding how MapThread works though as this doesn't work. Rather than getting "c[1/2][-(1/2)][1/2][-(1/2)][-1]=new random number", instead "old random number=new random number" is what seems to be happening.
>
> As a last resort I thought I would just clear the variables at the beginning of the do loop but you can't use Clear to do this and I don't see how you can apply "=." (Unset) over a list either.
>
> Any suggestions?
>
> Many thanks,
>
> Jonny
- References:
- Re: How to find one expression in terms of another expression?
- From: David Bailey <dave@removedbailey.co.uk>
- Re: How to find one expression in terms of another expression?