Question

我几天前仍在处理一个问题，并希望收到有关如何创建功能的反馈/支持。非常感谢您的专业知识。

我创建了以下内容：

##### 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"

我需要做的是通过创建一个名为meanRaceSpeed并具有的功能，我需要计算船“在比赛期间”（在开始和结束时间之间）的平均速度（SOG）三个论点：

到目前为止，我所尝试的是创建一个包含3个参数的函数（这里有一些专家的帮助）：

meanRaceSpeed <- function(raceIDs, boatList, raceTimes)  
 {
  #Probably need to compare times, and thought it might be useful to convert character values into `DateTime` values but not to sure how to use it
  #DateTime <- as.POSIXct(paste(boatList$Date, boatList$Time), format="%Y%m%d %H%M%S")

  #To get the times for each boat
  start_time <- raceTimes$start[rownames(raceTimes) = raceIDs] 
  finish_time <- raceTimes$finish[rownames(raceTimes) = raceIDs]
  start_LocalTime <- min(grep(start_time, boatList$LocalTime))
  finish_LocalTime <- max(grep(finish_time, boatList$LocalTime))

  #which `SOG`s contain all the `LocalTimes` between start and finish
  #take their `mean`
  mean(boatList$SOG[start_LocalTime : finish_LocalTime])
 }
 ### Obviously, my code does not work :( and I don't know where.

所以基本上，我需要用三个参数创建一个函数，预期的结果是：

#e.g For NZ1
> meanRaceSpeed("NZ1", boatList, raceTimes)   
[1] 18.32   #Mean speed for NZ1 between 11:10:01 - 11:35:16

#e.g for US1
> meanRaceSpeed("US1", boatList, raceTimes)
[1] 17.23    #Mean speed for US1 between 11:10:01 - 11:33:29

我可能出错的任何帮助？非常感谢您的帮助。

Answer 1

我将为R提供一些一般性建议，但我也会帮助您解决具体问题。每当我在R中遇到问题时，我通常会发现它有助于使事情更明确。

如果函数不能使用这些方法（是函数中的数据框还是矩阵？）那么你应该尝试另一种方法。如果这些表操作方法不起作用，请尝试不同的方法。怎么样？

您可以使用以下几种不同的方法来测试您的功能，以及一些可能会让您感到兴奋的建议。（我不想为你解决整件事，因为这是你的功课，而是让你继续前进。）

1）为什么不尝试使用循环而不是括号？

start_time <- raceTimes$start[rownames(raceTimes) = raceIDs]

将其变为for循环。做起来并不难。

2）调试你的功能。 R中内置了很多工具，您可以在包中添加。既然你很可能没有时间做功课。我建议这样做。拆开该功能并将其中的每个部分应用于您想要的变量。它们的长度合适吗？它们是正确的数据类型吗？在你把它们放在一起之前，他们是否得到了正确的答案？确保这一点。

3）如果一切都失败了，不要害怕功能和代码是否优雅。 R并不总是优雅的语言。（实际上，它很少是一种优雅的语言。）特别是当你是初学者时，你的代码可能会很难看。只要确保它有效。

Answer 2

由于我已经拥有您的数据经验，所以我坐下来做一个完整的例子。

首先，看起来像你的数据：

raceIDs <- c("GER", "SUI", "NZ2", "US1", "US2", "POR", "FRA", "AUS", "NZ1", "SWE")

raceTimes <- as.matrix(read.table(text = '    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"', header = T))

#turn matrix to data.frame or, else, `$` won't work
raceTimes <- as.data.frame(raceTimes, stringsAsFactors = F)

blDF <- data.frame(Boat = rep(raceIDs, 3), 
  LocalTime = c(raceTimes$start, rep("11:20:25", length(raceIDs)), raceTimes$finish),
       SOG = runif(3 * length(raceIDs), 15, 25), stringsAsFactors = F)

boatList <- split(blDF, blDF$Boat)
#remove `names` to create them from scratch
names(boatList) <- NULL

然后：

#create `names` by searching each element of 
#`boatList` of what `boat` it contains
names(boatList) <- unlist(lapply(boatList, function(x) unique(x$Boat)))

#the function
meanRaceSpeed <- function(ID, boatList, raceTimes)  
 {              #named the first argument `ID` instead of `raceIDs`
  start_time <- raceTimes$start[rownames(raceTimes) == ID] 
  finish_time <- raceTimes$finish[rownames(raceTimes) == ID]

  start_LocalTime <- min(grep(start_time, boatList[[ID]]$LocalTime))
  finish_LocalTime <- max(grep(finish_time, boatList[[ID]]$LocalTime))

  mean(boatList[[ID]]$SOG[start_LocalTime : finish_LocalTime])
 }

测试：

  meanRaceSpeed("US1", boatList, raceTimes)
#[1] 19.7063
  meanRaceSpeed("NZ1", boatList, raceTimes)
#[1] 21.74729
  mean(boatList$NZ1$SOG) #to test function
#[1] 21.74729
  mean(boatList$US1$SOG) #to test function 
#[1] 19.7063

通过少量参数和计算来创建函数

2 个答案: