...
and here is a maybe slightly neater approach using ?mapply (again with
the method column changed to character():
f <- function(meth,i,fr) do.call(meth,list((fr[i,])))
mapply(FUN=f,meth=input.df[,4],seq_len(nrow(input.df)),
MoreArgs = list(fr = input.df[,1:3]) )
Cheers,
Bert
Bert Gunter
"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sun, Mar 27, 2016 at 11:44 AM, Bert Gunter <bgunter.4...@gmail.com> wrote:
> OOPS! I forgot to tell you that I first changed the "method" column,
> which is a factor, to character, with
>
> input.df$method <- as.character(input.df$method)
>
> Then things will work properly.
>
> -- Bert
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Sun, Mar 27, 2016 at 11:35 AM, Bert Gunter <bgunter.4...@gmail.com> wrote:
>> 1. return() is not needed in R functions (it's harmless, however). You
>> might wish to go through an R function tutorial (many good ones are on
>> the web) to learn about what slick things you can do with functions in
>> R.
>>
>> 2. The following is just a brute force loop, so more elegant
>> approaches are likely possible, but this seems to do what you want:
>>
>> sapply(seq_len(nrow(input.df)),
>> function(i)do.call(input.df[i,4],list(x=unlist(input.df[i,1:3])))
>> )
>>
>> ## see ?do.call
>>
>> 3. I suspect that a change in your data structure might facilitate
>> your task, but of course, not knowing the task, I would not know what
>> changes would be useful. See ?switch for something that might be
>> related to what you are trying to do.
>>
>> Cheers,
>> Bert
>>
>>
>>
>>
>>
>> Bert Gunter
>>
>> "The trouble with having an open mind is that people keep coming along
>> and sticking things into it."
>> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>>
>>
>> On Sun, Mar 27, 2016 at 8:46 AM, Naresh Gurbuxani
>> <naresh_gurbux...@hotmail.com> wrote:
>>> I have a data frame with with several columns of parameters and one column
>>> of function name which should be used with these parameters. My goal is to
>>> call the appropriate function for each row and summarize the results in a
>>> list or data frame. I can partition the data frame according to function
>>> name, then call each function at a time. Is there a more elegant way to
>>> achieve this result with one call only?
>>>
>>> Below is an example. Here the functions are simple. It is possible to
>>> write a function that incorporates both the functions. It is not so easy
>>> in my actual task. My goal is to make one call and obtain res.df.
>>>
>>> Thanks,
>>> Naresh
>>>
>>>
>>>
>>> sum.sq <- function(x) {return(sum(x^2))}
>>>
>>> sum.cube <- function(x) {return(sum(x^3))}
>>>
>>> input.df <- data.frame(x1 = c(1,2,3,4), x2 = c(2,3,4,5), x3 = c(3,4,5,6),
>>> method = c(rep("sum.sq", 2), rep("sum.cube", 2)), case = c("a", "b", "c",
>>> "d"))
>>>
>>> library(plyr)
>>>
>>> res.df1 <- ddply(subset(input.df, method == "sum.sq"), c("case"),
>>> function(df) {
>>> x.vec <- c(df$x1, df$x2, df$x3)
>>> return(data.frame(sum.power = sum.sq(x.vec)))
>>> })
>>>
>>> res.df2 <- ddply(subset(input.df, method == "sum.cube"), c("case"),
>>> function(df) {
>>> x.vec <- c(df$x1, df$x2, df$x3)
>>> return(data.frame(sum.power = sum.cube(x.vec)))
>>> })
>>>
>>> res.df <- rbind(res.df1, res.df2)
>>> res.df <- merge(res.df, input.df[,c("method", "case")], by = "case")
>>>
>>>
>>>
>>>
>>> ______________________________________________
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>>> 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.
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