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如何为plyr创建函数
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Stack Overflow用户
提问于 2012-04-11 02:18:59
回答 1查看 281关注 0票数 1

我有一系列的足球成绩,我想知道一支球队在特定的比赛中得了多少分。

这是一个子集的头部,其中包含自最新结果以来在一个赛季内获得的累积积分

我因为没有使用dput而被打了几次手腕,所以忍住了。

代码语言:javascript
复制
allData <- structure(list(team = c("Arsenal", "Tottenham H", "Tottenham H", 
"Arsenal", "Arsenal", "Tottenham H"), venue = c("H", "A", "H", 
"A", "H", "A"), result = c("W", "D", "W", "L", "W", "D"), GF = c(1L, 
0L, 3L, 1L, 3L, 0L), GA = c(0L, 0L, 1L, 2L, 0L, 0L), gameDate = structure(c(1333868400, 
1333782000, 1333263600, 1333177200, 1332572400, 1332572400), class = c("POSIXct", 
"POSIXt"), tzone = ""), season = structure(c(2L, 2L, 2L, 2L, 
2L, 2L), .Label = c("2010/2011", "2011/2012"), class = "factor"), 
 points = c(3, 1, 3, 0, 3, 1), GD = c(1L, 0L, 2L, -1L, 3L, 
0L), cumpts = c(3, 1, 4, 3, 6, 5)), .Names = c("team", "venue", 
"result", "GF", "GA", "gameDate", "season", "points", "GD", "cumpts"
), row.names = c(NA, 6L), class = "data.frame")

这是一支球队在一个赛季的数据

代码语言:javascript
复制
spurs <- structure(list(team = c("Tottenham H", "Tottenham H", "Tottenham H", 
"Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H", 
"Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H", 
"Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H", 
"Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H", 
"Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H", 
"Tottenham H", "Tottenham H", "Tottenham H", "Tottenham H"), 
    venue = c("A", "H", "A", "H", "A", "H", "A", "H", "A", "H", 
    "A", "H", "H", "H", "A", "A", "H", "H", "A", "H", "A", "H", 
    "A", "H", "A", "A", "H", "A", "H", "A", "H", "A"), result = c("D", 
    "W", "D", "D", "L", "L", "L", "W", "D", "W", "L", "D", "W", 
    "W", "D", "W", "D", "W", "L", "W", "W", "W", "W", "W", "W", 
    "D", "W", "W", "W", "W", "L", "L"), GF = c(0L, 3L, 0L, 1L, 
    0L, 1L, 2L, 5L, 0L, 3L, 2L, 1L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 
    3L, 3L, 2L, 3L, 3L, 2L, 2L, 2L, 2L, 4L, 2L, 1L, 0L), GA = c(0L, 
    1L, 0L, 1L, 1L, 3L, 5L, 0L, 0L, 1L, 3L, 1L, 0L, 0L, 1L, 0L, 
    1L, 0L, 2L, 0L, 1L, 0L, 1L, 1L, 1L, 2L, 1L, 1L, 0L, 0L, 5L, 
    3L), gameDate = structure(c(1333782000, 1333263600, 1332572400, 
    1332313200, 1331366400, 1330848000, 1330243200, 1328947200, 
    1328515200, 1327996800, 1327219200, 1326528000, 1326268800, 
    1325577600, 1325318400, 1324972800, 1324540800, 1324281600, 
    1323590400, 1322899200, 1322294400, 1321862400, 1320562800, 
    1319958000, 1319353200, 1318748400, 1317538800, 1316847600, 
    1316329200, 1315638000, 1314514800, 1313996400), class = c("POSIXct", 
    "POSIXt"), tzone = ""), season = structure(c(2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
    2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("2010/2011", 
    "2011/2012"), class = "factor"), points = c(1, 3, 1, 1, 0, 
    0, 0, 3, 1, 3, 0, 1, 3, 3, 1, 3, 1, 3, 0, 3, 3, 3, 3, 3, 
    3, 1, 3, 3, 3, 3, 0, 0), GD = c(0L, 2L, 0L, 0L, -1L, -2L, 
    -3L, 5L, 0L, 2L, -1L, 0L, 2L, 1L, 0L, 2L, 0L, 1L, -1L, 3L, 
    2L, 2L, 2L, 2L, 1L, 0L, 1L, 1L, 4L, 2L, -4L, -3L), cumpts = c(1, 
    4, 5, 6, 6, 6, 6, 9, 10, 13, 13, 14, 17, 20, 21, 24, 25, 
    28, 28, 31, 34, 37, 40, 43, 46, 47, 50, 53, 56, 59, 59, 59
    )), .Names = c("team", "venue", "result", "GF", "GA", "gameDate", 
"season", "points", "GD", "cumpts"), row.names = c(NA, -32L), class = "data.frame")

然后我在马刺数据框上有这个代码来计算在特定游戏长度(这里是5)中的得分。

代码语言:javascript
复制
gameLength <- 5
seasonLength <- nrow(spurs)
cumPoints <- c()
cumPoints[1] <- spurs[gameLength,]$cumpts
for (i in gameLength+1:seasonLength) {
cumPoints[i-(gameLength-1)] <- ((spurs[i,]$cumpts)- 
 (spurs[i-gameLength,]$cumpts))
}
cumPoints <- cumPoints[!is.na(cumPoints)] # not sure why throws up NAs

这会产生正确的输出

代码语言:javascript
复制
 [1]  6  5  2  4  4  7  7  8  8 10  8 11 11 11  8 10 10 12 12 15 15
[22] 13 13 13 13 13 12  9

但是我需要能够转换allData,在数据框中包含每个赛季和球队的数据。

我假设我应该以某种方式使用ddply,除非有更好的替代方案。

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2012-04-11 03:19:54

要复制您的输出:

代码语言:javascript
复制
library(zoo)
rollapply(spurs$GD, gamelength, sum)

如果你有看起来像马刺data.frame的allData ...

代码语言:javascript
复制
rollsum <- function(df, gamelen=gamelength) {
  require(zoo)
  out <- rollapply(df$points, gamelen, sum)

  return(out)
}

ddply(allData, .(team), rollsum)
票数 1
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/10094226

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