简单的 R 函数问题

Question

我写了一个非常冗长的 R 函数，我确信它可以在 2 行或更少的代码中重新创建，但我现在还不知道如何处理。

我向我的函数传递了一个灰狗结果数据的数据帧和一个表示距离的整数运行。然后它 returns 在该距离内每个框数（总共 8 个框）的获胜百分比。

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，然后我将其转换为向量：

[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 header：

       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

你可以把它变成一个函数并使用合适的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 在 numerator 和 denominator 中是相同的，所以我会做

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

更新

big datasets 的更快选择是使用 data.table

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

或者上面的内容可以缩短（如@Arun 所评论）

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

或使用 prop.table

的选项

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

Answer 2

另一个使用 dplyr 的选项，对于大型数据集可能比 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))
}

原文：

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

使用函数（注意我修改了输入数据 - 见下面的 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 函数问题

Simple R function issue

r

更新

数据