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Re: Piecewise and multiple values
*To*: mathgroup at smc.vnet.net
*Subject*: [mg90001] Re: Piecewise and multiple values
*From*: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
*Date*: Thu, 26 Jun 2008 04:41:43 -0400 (EDT)
*Organization*: The Open University, Milton Keynes, UK
*References*: <g3t6vs$imi$1@smc.vnet.net>
Giacomo Ciani wrote:
> Hi all,
>
> in brief my problem: I'm using mathematica to create the solutions for
> problems that actually came in few different versions (meaning that
> the problem in the same, but the initial data change). Suppose I ave
> version 1 in which m=3,k=1,..., version 2 in which m=5,k=2,... and so
> on... Typically, I use an approach like:
>
> data={m->{3,5,6},k->{1,2,7},...}
>
> Then I write and solve the general symbolic equations, and the replace
> the values with the above rule, like:
>
> sol = Solve[...]
> sol/.data
>
> obtaining a list of numerical solutions corresponding to the different
> set of data contained in the "data" list, like:
>
> {(num sol corresponding to m=3,k=5,...),(num sol corresponding to
> m=5,k=2,...)}
>
> This works very well for me, because it allows me to solve the problem
> once and check the numerical solution for all the different versions
> at a glance.
>
> The problem arises in the case I have to use a Piecewise function. In
> this case I define:
>
> | f x<z
> h = |
> | g x>z
>
> but when I evaluate
>
> f/.data
>
> I obtain something like:
>
> | {f1,f2,...} x<{z1,z2,...}
> h = |
> | {g1,g2,...} x>{z1,z2,...}
>
> that I don't know how to plot or "disentagle", i.e. obtain expressions
> like:
>
> | f1 x<z1
> h1 = |
> | g1 x>z1
>
>
> | f2 x<z2
> h2 = |
> | g2 x>z2
>
> and so on...
>
> Do you have any suggestion? I would prefer to retain my data
> replacement scheme, and just understand how to create "disentangled"
> piecewise functions...
Hi Giacomo,
I believe that something along the line
Thread /@ data // Transpose // ReplaceAll[f[x], #] &
should do what you are looking for (though I am not a hundred percent
sure to have fully grasped all of your requirements). For instance,
data = {m -> {3, 5, 6}, k -> {1, 2, 7}};
f[x_] := Piecewise[{{m + k, x < m}, {m - k, x >= m}}]
Thread /@ data // Transpose // ReplaceAll[f[x], #] &
(* returns *)
{\[Piecewise] {
{4, x < 3},
{2, x >= 3}
}, \[Piecewise] {
{7, x < 5},
{3, x >= 5}
}, \[Piecewise] {
{13, x < 6},
{-1, x >= 6}
}}
The following should illustrate how the above expression works:
Thread[data[[1]]]
{m -> 3, m -> 5, m -> 6}
Thread /@ data
{{m -> 3, m -> 5, m -> 6}, {k -> 1, k -> 2, k -> 7}}
Thread /@ data // Transpose
{{m -> 3, k -> 1}, {m -> 5, k -> 2}, {m -> 6, k -> 7}}
Thread /@ data // Transpose // ReplaceAll[m + k, #] &
{4, 7, 13}
Regards,
-- Jean-Marc
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