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    $\begingroup$ I think as soon as you start thinking about modifying a single element (setParam), your approach looses expressiveness, because within it this operation is not natural. If you go further down that road (I was actually doing this too, for a while), you end up with another emulation of mutable structures, and not the best one, because you are copying stuff (e.g. you have copy - internally - and traverse the entire parameter list to modify a single element - so your assignemnt is strictly speaking not constant time). $\endgroup$ Commented Jan 30, 2012 at 15:50
  • $\begingroup$ @Leonid, well, but these lists tend to be rather small, so performance is not an issue. At least not for the things I was using them for. This does not become slow if we have a large parameter value, only if we have a large number of parameters. It's like Append. $\endgroup$ Commented Jan 30, 2012 at 15:53
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    $\begingroup$ I don't think it should be called "quick and dirty". I think this is the best solution here. It has an added advantage that you can just add new rules modifying your parameters in a batch, and rule-application mechanism will take care of the rest. You can, if you wish, factor this into a separate function. But what I'd do within this approach, instead, is to define configurations. The key to expressiveness is composition. Here is what I'd do: config[newrules__]@config[rules__] := config[newrules, rules];. It's really that simple, and we can assign these to variables and get ... $\endgroup$ Commented Jan 30, 2012 at 16:37
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    $\begingroup$ Just one observation (which doesn't seem to have been made above): With the definition of params as list of rules you also can extract (but not set) a specific value using OptionValue[params, par1]. $\endgroup$ Commented May 25, 2012 at 12:54
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    $\begingroup$ @mcandril: I've just now seen your comment. In case it's still relevant for you: I think simply using //. instead of /. would solve that issue. For example, {a,b} /. { a->1, b->x+a } gives {1, a + x}, but {a,b} //. { a->1, b->x+a } gives {1, 1 + x}. $\endgroup$ Commented Jul 16, 2014 at 11:53