[R] LM with summation function

Tue May 22 18:48:54 CEST 2012

Ahh.... sorry -- I didn't understand that x was supposed to be an
index so I was using the row number an index for the summation -- yes,
my proposal probably won't work without further assumptions....[I.e.,
you could assume linear growth between observations, but that will
bias something some direction...(not sure which)]

I'll ponder it some more and get back to you if I come up with anything

Michael

On Tue, May 22, 2012 at 12:43 PM, Robbie Edwards
<robbie.edwards at gmail.com> wrote:
> I don't think I can.
>
> For the sample data
>
>  d <- data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
>
> when x = 4, s = 1200.  However, that s4 is sum of y1 + y2 + y3 + y4.
>  Wouldn't I have to know the y for x = 2 and x = 3 to get the value of y
> for x = 4?
>
> In the previous message, I created two sample data frames.  d is what I'm
> trying to use to create df.  I only know what's in d, df is just used to
> illustrate what I'm trying to get from d.
>
> robbie
>
>
>
>
>
> On Tue, May 22, 2012 at 12:30 PM, R. Michael Weylandt <
> michael.weylandt at gmail.com> wrote:
>
>> But if I understand your problem correctly, you can get the y values
>> from the s values. I'm relying on your statement that "s is sum of the
>> current y and all previous y (s3 = y1 + y2 + y3)." E.g.,
>>
>> y <- c(1, 4, 6, 9, 3, 7)
>>
>> s1 = 1
>> s2 = 4 + s1 = 5
>> s3 = 6 + s2 = 11
>>
>> more generally
>>
>> s <- cumsum(y)
>>
>> Then if we only see s, we can get back the y vector by doing
>>
>> c(s[1], diff(s))
>>
>> which is identical to y.
>>
>> So for your data, the underlying y must have been c(109, 1091, 4125,
>> 2891) right?
>>
>> Or have I completely misunderstood your problem?
>>
>> Michael
>>
>> On Tue, May 22, 2012 at 12:25 PM, Robbie Edwards
>> <robbie.edwards at gmail.com> wrote:
>> > Actually, I can't.  I don't know the y values.  Only the s and only for a
>> > subset of the data.
>> >
>> > Like this.
>> >
>> > d <- data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
>> >
>> >
>> >
>> > On Tue, May 22, 2012 at 11:57 AM, R. Michael Weylandt
>> > <michael.weylandt at gmail.com> wrote:
>> >>
>> >> You can reconstruct the y values by taking first-differences of the s
>> >> vector, no? Then it sounds like you're good to go
>> >>
>> >> Best, Michael
>> >>
>> >> On Tue, May 22, 2012 at 11:40 AM, Robbie Edwards
>> >> <robbie.edwards at gmail.com> wrote:
>> >> > Hi all,
>> >> >
>> >> > Thanks for the replies, but I realize I've done a bad job explaining
>> my
>> >> > problem.  To help, I've created some sample data to explain the
>> problem.
>> >> >
>> >> > df <- data.frame(x=c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), y=c(109,
>> >> > 232,
>> >> > 363, 496, 625, 744, 847, 928, 981, 1000, 979, 912), s=c(109, 341, 704,
>> >> > 1200, 1825, 2569, 3416, 4344, 5325, 6325, 7304, 8216))
>> >> >
>> >> > In this data frame, y results from y = x * b1 + x^2 * b2 + x^3 * b3
>> and
>> >> > s
>> >> > is sum of the current y and all previous y (s3 = y1 + y2 + y3).
>> >> >
>> >> > I know I can find b1, b2 and b3 using:
>> >> > lm(y ~ 0 + x + I(x^2) + I(x^3), data=df)
>> >> >
>> >> > yielding...
>> >> > Coefficients:
>> >> >     x  I(x^2)  I(x^3)
>> >> >   100      10      -1
>> >> >
>> >> > However, I need to find b1, b2 and b3 using the s column.  The reason
>> >> > being, I don't actually know the values of y in the actual data set.
>> >> >  And
>> >> > in the actual data, I only have a few of the values.  Imagine this
>> data
>> >> > is
>> >> > being used a reward schedule for like a loyalty points program.  y
>> >> > represents the number of points needed for each level while s is the
>> >> > total
>> >> > number of points to reach that level.  In the real problem, my data
>> >> > looks
>> >> > more like this:
>> >> >
>> >> > d <- data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
>> >> >
>> >> > Where I need to use a few sample points to help define the parameters
>> of
>> >> > the curve.
>> >> >
>> >> > thanks again and hopefully this makes the problem a bit clearer.
>> >> >
>> >> > robbie
>> >> >
>> >> >
>> >> >
>> >> > On Fri, May 18, 2012 at 7:40 PM, David Winsemius
>> >> > <dwinsemius at comcast.net>wrote:
>> >> >
>> >> >>
>> >> >> On May 18, 2012, at 1:44 PM, Robbie Edwards wrote:
>> >> >>
>> >> >>  Hi all,
>> >> >>>
>> >> >>> I'm trying to model some data where the y is defined by
>> >> >>>
>> >> >>> y = summation[1 to 50] B1 * x + B2 * x^2 + B3 * x^3
>> >> >>>
>> >> >>> Hopefully that reads clearly for email.
>> >> >>>
>> >> >>>
>> >> >> cumsum( rowSums( cbind(B1 * x,  B2 * x^2, B3 * x^3)))
>> >> >>
>> >> >>
>> >> >>
>> >> >>  Anyway, if it wasn't for the summation, I know I would do it like
>> this
>> >> >>>
>> >> >>> lm(y ~ x + x2 + x3)
>> >> >>>
>> >> >>> Where x2 and x3 are x^2 and x^3.
>> >> >>>
>> >> >>> However, since each value of x is related to the previous values of
>> x,
>> >> >>> I
>> >> >>> don't know how to do this.  Any help is greatly appreciated.
>> >> >>>
>> >> >>>
>> >> >>>
>> >> >>
>> >> >> David Winsemius, MD
>> >> >> West Hartford, CT
>> >> >>
>> >> >>
>> >> >
>> >> >        [[alternative HTML version deleted]]
>> >> >
>> >> > ______________________________________________
>> >> > R-help at r-project.org mailing list
>> >> > https://stat.ethz.ch/mailman/listinfo/r-help
>> >> > PLEASE do read the posting guide
>> >> > http://www.R-project.org/posting-guide.html
>> >> > and provide commented, minimal, self-contained, reproducible code.
>> >
>> >
>>
>
>        [[alternative HTML version deleted]]
>
> ______________________________________________
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