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Jul 24, 2013, 2:54:11 AM7/24/13

to eureqa...@googlegroups.com, Alison Reynolds

Yes your interpretation of what I was trying to do was correct

And the formula looks good.

Thank you again Alison

Nevi

---- Alison Reynolds <alison....@nutonian.com> wrote:

> Hi Nevi,

>

> First let me clarify exactly what you are looking for. Is the idea that

> given variable x1, x2, ... x8, you want the model produced to look like:

>

> z = (x1 + x2) / 2

> or

> z = (x1 + x2 + x3) / 3

>

> where the variables could be any subset of the 8 variables, and the

> denominator will be the number of variables? I.e. you expect that z will

> be the mean of some subset of the input variables?

>

> If my understanding is correct, then I think you can use a target

> expression that looks like this:

>

> y = (step(f0())*x1 + step(f1())*x2 + step(f2())*x3)/(step(f0()) +

> step(f1()) + step(f2()))

>

> (Mine is set up for 3 input variables; you'll need to update it for 8).

> The expressions step(fn()) will map to either 0 or 1. On the numerator

> that will either include or exclude a variable from being included in the

> sum of variables. On the denominator it will contribute to the count of

> variables. I've attached a sample project that shows that expression used

> in 2 searches. The solutions produced (once Eureqa has simplified them)

> are:

>

> y = 0.5*x1 + 0.5*x2

> and

> y = 0.3333*x1 + 0.3333*x2 + 0.3333*x3

>

> Those of course can be rewritten in the format you are looking for..

>

> y = (x1 + x2) / 2

> and

> y = (x1 + x2 + x3) / 3

>

> Let me know if that's what you were looking for.

> Alison

>

>

> On Mon, Jul 22, 2013 at 12:59 AM, aidabet <aid...@cox.net> wrote:

>

> > Another question

> >

> > Example Target Expression

> >

> > z = f(x1, x2, x3, x4, x5, x6, x7, x8)

> >

> > ****

> >

> > I’m looking for a way to Sum the number of variables being considered for

> > a Solution and divide the Sum of the Variables by the Number of Variables,

> > all included in the Final Solution.

> >

> > ****

> >

> > So the Final Solution would resemble

> >

> > z = f(( x1 + x2 ) / 2 ) if two variables used

> >

> > z = f(( x1 + x2 + x3 ) / 3 ) if three variables used

> >

> > ****

> >

> > If it’s possible your help is appreciated

> > Thanks

> >

> > --

> > You received this message because you are subscribed to the Google Groups

> > "formulize" group.

> > To unsubscribe from this group and stop receiving emails from it, send an

> > email to formulize+...@googlegroups.com.

> > For more options, visit https://groups.google.com/groups/opt_out.

> >

> >

> >

>

> --

> You received this message because you are subscribed to the Google Groups "Eureqa Group" group.

> To unsubscribe from this group and stop receiving emails from it, send an email to eureqa-group...@googlegroups.com.

> For more options, visit https://groups.google.com/groups/opt_out.

>

>

And the formula looks good.

Thank you again Alison

Nevi

---- Alison Reynolds <alison....@nutonian.com> wrote:

> Hi Nevi,

>

> First let me clarify exactly what you are looking for. Is the idea that

> given variable x1, x2, ... x8, you want the model produced to look like:

>

> z = (x1 + x2) / 2

> or

> z = (x1 + x2 + x3) / 3

>

> where the variables could be any subset of the 8 variables, and the

> denominator will be the number of variables? I.e. you expect that z will

> be the mean of some subset of the input variables?

>

> If my understanding is correct, then I think you can use a target

> expression that looks like this:

>

> y = (step(f0())*x1 + step(f1())*x2 + step(f2())*x3)/(step(f0()) +

> step(f1()) + step(f2()))

>

> (Mine is set up for 3 input variables; you'll need to update it for 8).

> The expressions step(fn()) will map to either 0 or 1. On the numerator

> that will either include or exclude a variable from being included in the

> sum of variables. On the denominator it will contribute to the count of

> variables. I've attached a sample project that shows that expression used

> in 2 searches. The solutions produced (once Eureqa has simplified them)

> are:

>

> y = 0.5*x1 + 0.5*x2

> and

> y = 0.3333*x1 + 0.3333*x2 + 0.3333*x3

>

> Those of course can be rewritten in the format you are looking for..

>

> y = (x1 + x2) / 2

> and

> y = (x1 + x2 + x3) / 3

>

> Let me know if that's what you were looking for.

> Alison

>

>

> On Mon, Jul 22, 2013 at 12:59 AM, aidabet <aid...@cox.net> wrote:

>

> > Another question

> >

> > Example Target Expression

> >

> > z = f(x1, x2, x3, x4, x5, x6, x7, x8)

> >

> > ****

> >

> > I’m looking for a way to Sum the number of variables being considered for

> > a Solution and divide the Sum of the Variables by the Number of Variables,

> > all included in the Final Solution.

> >

> > ****

> >

> > So the Final Solution would resemble

> >

> > z = f(( x1 + x2 ) / 2 ) if two variables used

> >

> > z = f(( x1 + x2 + x3 ) / 3 ) if three variables used

> >

> > ****

> >

> > If it’s possible your help is appreciated

> > Thanks

> >

> > --

> > You received this message because you are subscribed to the Google Groups

> > "formulize" group.

> > To unsubscribe from this group and stop receiving emails from it, send an

> > email to formulize+...@googlegroups.com.

> > For more options, visit https://groups.google.com/groups/opt_out.

> >

> >

> >

>

> --

> You received this message because you are subscribed to the Google Groups "Eureqa Group" group.

> To unsubscribe from this group and stop receiving emails from it, send an email to eureqa-group...@googlegroups.com.

> For more options, visit https://groups.google.com/groups/opt_out.

>

>

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