如何使用三个参数创建函数

时间:2013-10-24 09:34:07

标签: r

我正在尝试创建一个函数,它为我的作业带来了三个参数,并尝试了几次但没有运气!非常感谢一些建议或帮助:)

我已经创建了以下内容:

##### 1)
> raceIDs
[1] "GER" "SUI" "NZ2" "US1" "US2" "POR" "FRA" "AUS" "NZ1" "SWE"

##### 2)
#For each "raceIDs", there is a csv file which I have made a loop to read and created a list of data frames (assigned to the symbol "boatList")
#For example, if I select "NZ1" the output is:
> head(boatList[[9]]) #Only selected the first six lines as there is more than 30000 rows
  Boat       Date    Secs    LocalTime   SOG
1  NZ1 01:09:2013 38150.0 10:35:49.997 22.17
2  NZ1 01:09:2013 38150.2 10:35:50.197 22.19
3  NZ1 01:09:2013 38150.4 10:35:50.397 22.02
4  NZ1 01:09:2013 38150.6 10:35:50.597 21.90
5  NZ1 01:09:2013 38150.8 10:35:50.797 21.84
6  NZ1 01:09:2013 38151.0 10:35:50.997 21.95

##### 3)
# A matrix showing the race times for each raceIDs
> raceTimes
    start      finish    
GER "11:10:02" "11:35:05"
SUI "11:10:02" "11:35:22"
NZ2 "11:10:02" "11:34:12"
US1 "11:10:01" "11:33:29"
US2 "11:10:01" "11:36:05"
POR "11:10:02" "11:34:31"
FRA "11:10:02" "11:34:45"
AUS "11:10:03" "11:36:48"
NZ1 "11:10:01" "11:35:16"
SWE "11:10:03" "11:35:08"

我需要做的是我需要计算一艘船“在比赛时”的平均速度(SOG)(在开始和结束时间之间)。

基本上,我需要创建一个与此类似的函数:

> meanRaceSpeed("NZ1", boatList, raceTimes)
[1] 18.32  

> meanRaceSpeed("US1", boatList, raceTimes)
[1] 17.23

这是我作业的最后一个问题之一,我完全陷入困境:(

我甚至不知道从哪里开始。

有人可以请给我一些建议或支持吗?

2 个答案:

答案 0 :(得分:0)

由于这是一个家庭作业的问题,它可能不会很好地为任何人提供直接回答。

有关功能的更多信息,请参阅例如Quick-R / User-defined functions

具有三个参数的函数示例,该函数不执行任何操作,只返回它们的连接:

myfunc <- function(first, second, third) {
return(cat(first, second, third))
}

您可以将其称为myfunc("foo","bar","baz")并获取结果“foo bar baz”。

答案 1 :(得分:0)

我认为你不需要三个参数,因为你的三个参数是相互关联的,即当指定一艘船时,它也可以指定boatList中的位置,它的raceTimes可以也可以指定。

我没有测试以下功能,因为模拟随机数据有点痛苦,但我相信它可能会有所帮助:

meanRaceSpeed <- function(x)  #`x` is the boat's name
 {
  start_time <- raceTimes$start[rownames(raceTimes) == x] 
  finish_time <- raceTimes$finish[rownames(raceTimes) == x]

  #which `LocalTime` is the first with `start_time`
  start_LocalTime <- min(grep(start_time, boatList[[x]]$LocalTime))
  #which `LocalTime` is the last with `finish _time`
  #(if you want the first `finish_time` change `max` with `min`)
  finish_LocalTime <- max(grep(finish_time, boatList[[x]]$LocalTime))

  #which `SOG`s contain all the `LocalTimes` between start and finish
  #take their `mean`
  mean(boatList[[x]]$SOG[start_LocalTime : finish_LocalTime])
 }

#for all boats, something like:
sapply(raceIDs, meanRaceSpeed)

很抱歉,如果我误解了你的问题。