Question

我已经编写了一个非常长的R函数，我确信它可以在2行或更少的行中重新创建，但我还是无法处理它。

我向我的函数传递一个greyhound结果数据的数据框和一个表示距离运行的整数。然后它返回该距离上每个箱号（总共8个箱子）的胜率。

目前：

boxPercents <- function(dist, data) {

## get each box num of wins
x1 <- data$position == 1 & data$distance == dist & data$box == 1
x2 <- data$position == 1 & data$distance == dist & data$box == 2
x3 <- data$position == 1 & data$distance == dist & data$box == 3
...
x8 <- data$position == 1 & data$distance == dist & data$box == 8

## count the total num of races at that 
numRaces <- data$position == 1 & data$distance == dist

## print out the winning percent for each box
print(sum(x1) / sum(numRaces))
print(sum(x2) / sum(numRaces))
print(sum(x3) / sum(numRaces))
...
print(sum(x8) / sum(numRaces))

}

我的输出如下所示，然后转换为矢量：

[1] 0.2452107
[1] 0.1340996
[1] 0.09961686
[1] 0.1034483
[1] 0.08045977
[1] 0.1034483
[1] 0.09961686
[1] 0.1340996

我很确定其中一个应用函数是我应该使用的，但所有的努力都没有结果。

编辑：这是数据的标题：

       track      date race position box             name   sp  fave distance
 Warrnambool 02 Jan 14    1        1   1       TOP SECRET  1.7  true      450
 Warrnambool 02 Jan 14    1        2   4     FLASH WILSON  4.7 false      450
 Warrnambool 02 Jan 14    1        3   8 HEAPS OF ABILITY 11.8 false      450
 Warrnambool 02 Jan 14    1        4   7   OCCUPATION LAD 24.1 false      450
 Warrnambool 02 Jan 14    1        5   2   HE'S A VILLIAN 19.3 false      450
 Warrnambool 02 Jan 14    1        6   5 ZAC'S A SIXPENCE  9.7 false      450

Answer 1

您可以将其设为功能并使用相应的dist

dist <- 450
vapply(1:8, function(i) sum(with(data, 
           position==1 & distance==dist & box==i))/sum(with(data, 
               position==1 & distance==dist)), numeric(1L))

或

sapply(1:8, function(i) sum(with(data,
     position==1 & distance==dist & box==i))/sum(with(data,
                                 position==1 & distance==dist)))

因为position和distance中的numerator和denominator相同，我会这样做

sapply(1:8, function(i) {indx <- with(data, position==1 & distance==dist)
                      sum(indx & data$box==i)/sum(indx)} )

更新

big datasets的更快选项将使用data.table

library(data.table)
 setDT(data)[position==1 & distance==dist, c(.SD,numRaces= .N)][,
                list(percentage=unique(.N/numRaces)), by=box]

或者可以缩短上述内容（由@Arun评论）

 setDT(data)[position==1 & distance==dist, .N, by=box][, N := N/sum(N)]

使用prop.table

的选项

 as.data.frame(prop.table(table(subset(data,
        position==1 & distance==dist, select=c(position, box)))))

Answer 2

使用dplyr的另一个选项可能比大数据集的sapply方法更快：

更新

library(dplyr)
boxPercents <- function(dist, data) {
  data <- data %>% filter(position == 1 & distance == dist) %>% select(box)
  data %>% count(box) %>% transmute(percentage = n / sum(n))
}

原件：

boxPercents <- function(dist, data) {
  data <- data %>% filter(position == 1 & distance == dist) %>% select(box)
  numRaces <- nrow(data)
  data %>% 
    group_by(box) %>%
    summarise(percentage = n() / numRaces) 
}

使用该功能（注意我修改了输入数据 - 请参阅下面的dput）：

boxPercents(450, data)
#Source: local data frame [2 x 2]
#
#  box percentage
#1   1  0.6666667
#2   5  0.3333333

数据

data <- structure(list(track = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "Warrnambool", class = "factor"), 
    date = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "02 Jan 14", class = "factor"), 
    race = c(1L, 1L, 1L, 1L, 1L, 1L), position = c(1, 2, 3, 4, 
    1, 1), box = c(1, 4, 8, 7, 1, 5), name = structure(c(5L, 
    1L, 2L, 4L, 3L, 6L), .Label = c("FLASH WILSON", "HEAPS OF ABILITY", 
    "HES A VILLIAN", "OCCUPATION LAD", "TOP SECRET", "ZACS A SIXPENCE"
    ), class = "factor"), sp = c(1.7, 4.7, 11.8, 24.1, 19.3, 
    9.7), fave = structure(c(2L, 1L, 1L, 1L, 1L, 1L), .Label = c("false", 
    "true"), class = "factor"), distance = c(450L, 450L, 450L, 
    450L, 450L, 450L)), .Names = c("track", "date", "race", "position", 
"box", "name", "sp", "fave", "distance"), row.names = c(NA, -6L
), class = "data.frame")

简单的R功能问题

2 个答案:

更新

数据