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该ggplot2软件包很容易成为我使用过的最好的绘图系统，除了对于较大的数据集（约 50k 点）性能不是很好。我正在考虑通过 Shiny 提供网络分析，ggplot2用作绘图后端，但我对性能并不满意，尤其是与基本图形相比。我的问题是是否有任何具体的方法来提高这种性能。

起点是以下代码示例：

library(ggplot2)

n = 86400 # a day in seconds
dat = data.frame(id = 1:n, val = sort(runif(n)))

dev.new()

gg_base = ggplot(dat, aes(x = id, y = val))
gg_point = gg_base + geom_point()
gg_line = gg_base + geom_line()
gg_both = gg_base + geom_point() + geom_line()

benchplot(gg_point)
benchplot(gg_line)
benchplot(gg_both)
system.time(plot(dat))
system.time(plot(dat, type = 'l'))

我在我的 MacPro 视网膜上得到以下时间：

> benchplot(gg_point)
       step user.self sys.self elapsed
1 construct     0.000    0.000   0.000
2     build     0.321    0.078   0.398
3    render     0.271    0.088   0.359
4      draw     2.013    0.018   2.218
5     TOTAL     2.605    0.184   2.975
> benchplot(gg_line)
       step user.self sys.self elapsed
1 construct     0.000    0.000   0.000
2     build     0.330    0.073   0.403
3    render     0.622    0.095   0.717
4      draw     2.078    0.009   2.266
5     TOTAL     3.030    0.177   3.386
> benchplot(gg_both)
       step user.self sys.self elapsed
1 construct     0.000    0.000   0.000
2     build     0.602    0.155   0.757
3    render     0.866    0.186   1.051
4      draw     4.020    0.030   4.238
5     TOTAL     5.488    0.371   6.046
> system.time(plot(dat))
   user  system elapsed 
  1.133   0.004   1.138 
# Note that the timing below depended heavily on wether or net the graphics device
# was in view or not. Not in view made performance much, much better.
> system.time(plot(dat, type = 'l'))
   user  system elapsed 
  1.230   0.003   1.233

有关我的设置的更多信息：

> sessionInfo()
R version 2.15.3 (2013-03-01)
Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit)

locale:
[1] C/UTF-8/C/C/C/C

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] ggplot2_0.9.3.1

loaded via a namespace (and not attached):
 [1] MASS_7.3-23        RColorBrewer_1.0-5 colorspace_1.2-1   dichromat_2.0-0   
 [5] digest_0.6.3       grid_2.15.3        gtable_0.1.2       labeling_0.1      
 [9] munsell_0.4        plyr_1.8           proto_0.3-10       reshape2_1.2.2    
[13] scales_0.2.3       stringr_0.6.2
4

1 回答 1

8

Hadley 在user2013 上对他的新软件包dplyr和ggvis进行了很酷的讨论。但他自己可能会更好地讲述这一点。

我不确定您的应用程序设计是什么样的，但我经常在将数据提供给 R 之前进行数据库内预处理。例如，如果您正在绘制时间序列，则实际上不需要显示每一秒X 轴上的日期。相反，您可能希望在例如一到五分钟的时间间隔内聚合并获得最小值/最大值/平均值。

下面是我多年前编写的一个函数示例，它在 SQL 中做了类似的事情。此特定示例使用模运算符，因为时间存储为纪元毫秒。但是如果 SQL 中的数据被正确地存储为日期/日期时间结构，SQL 有一些更优雅的本地方法来按时间段聚合。

#' @param table name of the table
#' @param start start time/date
#' @param end end time/date
#' @param aggregate one of "days", "hours", "mins" or "weeks"
#' @param group grouping variable
#' @param column name of the target column (y axis)
#' @export
minmaxdata <- function(table, start, end, aggregate=c("days", "hours", "mins", "weeks"), group=1, column){

  #dates
  start <- round(unclass(as.POSIXct(start))*1000);
  end <- round(unclass(as.POSIXct(end))*1000);

  #must aggregate
  aggregate <- match.arg(aggregate);

  #calcluate modulus
  mod <- switch(aggregate,
    "mins"   = 1000*60,
    "hours"  = 1000*60*60,
    "days"   = 1000*60*60*24,
    "weeks"  = 1000*60*60*24*7,
    stop("invalid aggregate value")
  );

  #we need to add the time differene between gmt and pst to make modulo work
  delta <- 1000 * 60 * 60 * (24 - unclass(as.POSIXct(format(Sys.time(), tz="GMT")) - Sys.time()));  

  #form query
  query <- paste("SELECT", group, "AS grouping, AVG(", column, ") AS yavg, MAX(", column, ") AS ymax, MIN(", column, ") AS ymin, ((CMilliseconds_g +", delta, ") DIV", mod, ") AS timediv FROM", table, "WHERE CMilliseconds_g BETWEEN", start, "AND", end, "GROUP BY", group, ", timediv;")
  mydata <- getquery(query);

  #data
  mydata$time <- structure(mod*mydata[["timediv"]]/1000 - delta/1000, class=c("POSIXct", "POSIXt"));
  mydata$grouping <- as.factor(mydata$grouping)

  #round timestamps
  if(aggregate %in% c("mins", "hours")){
    mydata$time <- round(mydata$time, aggregate)
  } else {
    mydata$time <- as.Date(mydata$time);
  }

  #return
  return(mydata)
}
于 2013-08-21T09:33:22.173 回答