[R] Making a function and applying it over a list(?)

[Christoph Häni]

You could store your first approach in a function and lapply it to
your by_hour variable:

  df <- data.frame(
hour = factor(rep(1:5,4)),
  id       = factor(rep(c("supply", "demand"), each = 10)),
  price    = c(5,7,9,11,13,15,17,19,21,23,
20,18,16,14,12,10,8,6,4,2 ),
  quantity = c(3,5,7,13,19,31,37,53,61,67,6,18,20,24,40,46,66,70,76,78)
)

myfu <- function(x){

df <- x # for simplicity

quantity_points <- with(
  df,
  seq(min(quantity), max(quantity), length.out = 500)
)

by_id <- split(df[, c("price", "quantity")], df$id)

interpolated_price <- lapply(
  by_id,
  function(x)
  {
    with(
      x,
      approx(
        quantity,
        price,
        xout = quantity_points
      )
    )$y
  }
)

index_of_equality <- with(interpolated_price, which.min(abs(supply - demand)))
quantity_points[index_of_equality]

}

by_hour <- split(df,df$hour)

lapply(by_hour,myfu)


Was that what you were looking for?

Cheers,
Christoph


2013/10/24 Lasse Thorst <lath at energidanmark.dk>:
> Hi All
>
> I've gotten some awesome help getting a formular that finds the intersection of two vectors. This works brilliantly, but I can't figure out how to make it run over another factor. A simple example looks likes this:
>
>   df <- data.frame(
>   id       = factor(rep(c("supply", "demand"), each = 10)),
>   price    = c(5,7,9,11,13,15,17,19,21,23,20,18,16,14,12,10,8,6,4,2 ),
>   quantity = c(3,5,7,13,19,31,37,53,61,67,6,18,20,24,40,46,66,70,76,78)
> )
>
> quantity_points <- with(
>   df,
>   seq(min(quantity), max(quantity), length.out = 500)
> )
>
> by_id <- split(df[, c("price", "quantity")], df$id)
>
> interpolated_price <- lapply(
>   by_id,
>   function(x)
>   {
>     with(
>       x,
>       approx(
>         quantity,
>         price,
>         xout = quantity_points
>       )
>     )$y
>   }
> )
>
> index_of_equality <- with(interpolated_price, which.min(abs(supply - demand)))
> quantity_points[index_of_equality]
>
> Question: I need to run this over a larger data frame, where I have the same data, but also a new factor variable (called hour). So if you have the original data frame and add:
>
>   df <- data.frame(
> hour = factor(seq(1:20)),
>   id       = factor(rep(c("supply", "demand"), each = 10)),
>   price    = c(5,7,9,11,13,15,17,19,21,23,20,18,16,14,12,10,8,6,4,2 ),
>   quantity = c(3,5,7,13,19,31,37,53,61,67,6,18,20,24,40,46,66,70,76,78)
> )
>
> How can I run it for each hour? I tried using:
> by_hour <- split(df[, c("price", "quantity")], df$hour)
> mapply(fx, by_hour)
>
> And gathering the above into a fx <- function(){"the neat code"}, but I can't get it to work.
>
> Kind Regards,
> Lasse
>
>         [[alternative HTML version deleted]]
>
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