Hi Berend,
Your method is really much better. Thank you very much. (Yes I also forgot
to add the $root at the end.)
Best,
Prakasit
On Tue, Feb 19, 2013 at 10:51 PM, Berend Hasselman <b...@xs4all.nl> wrote:
>
> On 19-02-2013, at 09:55, Prakasit Singkateera <asltjoey.rs...@gmail.com>
> wrote:
>
> > Hi everyone,
> >
> > From your helps of giving me several ideas, eventually I can solve the
> posted problem. Here is the R code. It can be done by applying the
> uniroot.all to the data frame together with the proper form of equation
> (slightly modification of the original equation).
> >
> > #Generate the sample data frame
> > customer.name = c("John","Mike","Peter")
> > product = c("Toothpaste","Toothpaste","Toothpaste")
> > cost = c(30,45,40)
> > mydata = data.frame(customer.name,product,cost)
> >
> > #Original cost function - not used
> > #fcost = function(orders) 3.40 + (1.20 * orders^2)
> >
> > #Slightly modification of the cost function to be a proper form for root
> finding
> > #This is basically to set ==> 3.40 + (1.20 * orders^2) - fcost = 0
> > f.to.findroot = function(orders,fcost) 3.40 + (1.20 * orders^2) - fcost
> >
> > #Using rootSolve package which contains uniroot.all function
> > library(rootSolve)
> >
> > #Using plyr package which contains adply function
> > library(plyr)
> >
> > #Use uniroot function to find the 'orders' variable (from the
> f.to.findroot function) for each customer and put it into no.of.orders
> column in mysolution data frame
> > #Replace 'fcost' with 'cost' column from mydata
> > #Interval of 0 to 1,000 is to make the f.to.findroot function have both
> negative and positive sign, otherwise uniroot.all will give an error
> > mysolution = data.frame(adply(mydata, 1, summarize, no.of.orders =
> uniroot.all(f.to.findroot,interval = c(0,1000),fcost=cost)))
> > mysolution
> >
> > #Remove the redundant mydata as mysolution it is an extended version of
> mydata
> > rm(mydata)
> >
> > #Note uniroot.all can be used for both linear (e.g.orders^1) and
> non-linear (e.g.orders^2) equations.
>
>
> 1. You don't need rootSolve. uniroot is sufficient in your case. You don't
> have multiple roots for each element of cost.
>
> 2. You are now storing more information than you require into the
> resulting dataframe. Use uniroot(
)$root to store only the root of the
> equation.
>
> 3. you don't need plyr. You can do it like this
>
> mysolution <- within(mydata,
> no.of.orders <- sapply(seq_len(length(cost)),function(k)
> uniroot(f.to.findroot,interval = c(0,1000),fcost=cost[k])$root )
> )
> # for printing the dataframe
>
> mysolution
>
> Berend
>
>
[[alternative HTML version deleted]]
______________________________________________
R-help@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.
