如何在R中有效地使用Rprof ?

I would like to know if it is possible to get a profile from R-Code in a way that is similar to matlab's Profiler. That is, to get to know which line numbers are the one's that are especially slow.

我想知道是否有可能从R-Code中获得一个类似于matlab的分析器的概要文件。也就是说，要知道哪个行号是特别慢的。

What I acchieved so far is somehow not satisfactory. I used Rprof to make me a profile file. Using summaryRprof I get something like the following:

到目前为止，我所认为的不令人满意。我使用Rprof为我创建了一个概要文件。使用summary yrprof，我得到如下内容:

$by.self
                  self.time self.pct total.time total.pct
[.data.frame               0.72     10.1       1.84      25.8
inherits                   0.50      7.0       1.10      15.4
data.frame                 0.48      6.7       4.86      68.3
unique.default             0.44      6.2       0.48       6.7
deparse                    0.36      5.1       1.18      16.6
rbind                      0.30      4.2       2.22      31.2
match                      0.28      3.9       1.38      19.4
[<-.factor                 0.28      3.9       0.56       7.9
levels                     0.26      3.7       0.34       4.8
NextMethod                 0.22      3.1       0.82      11.5
...

and

和

$by.total
                      total.time total.pct self.time self.pct
data.frame                  4.86      68.3      0.48      6.7
rbind                       2.22      31.2      0.30      4.2
do.call                     2.22      31.2      0.00      0.0
[                           1.98      27.8      0.16      2.2
[.data.frame                1.84      25.8      0.72     10.1
match                       1.38      19.4      0.28      3.9
%in%                        1.26      17.7      0.14      2.0
is.factor                   1.20      16.9      0.10      1.4
deparse                     1.18      16.6      0.36      5.1
...

To be honest, from this output I don't get where my bottlenecks are because (a) I use data.frame pretty often and (b) I never use e.g., deparse. Furthermore, what is [?

坦率地说，从这个输出中，我找不到瓶颈所在，因为(a)我经常使用数据。此外,(是什么?

So I tried Hadley Wickham's profr, but it was not any more useful considering the following graph:

所以我尝试了哈德利·韦翰的教授，但考虑到下面的图表，这就没有任何用处了:

Is there a more convenient way to see which line numbers and particular function calls are slow?
Or, is there some literature that I should consult?

是否有一种更方便的方法来查看哪些行号和特定的函数调用比较慢?或者，我是否应该查阅一些文献?

Any hints appreciated.

任何暗示感激。

EDIT 1:
Based on Hadley's comment I will paste the code of my script below and the base graph version of the plot. But note, that my question is not related to this specific script. It is just a random script that I recently wrote. I am looking for a general way of how to find bottlenecks and speed up R-code.

编辑1:根据哈德利的评论，我将在下面粘贴我的脚本代码和基本图版本的情节。但是请注意，我的问题与这个特定的脚本无关。这只是我最近写的一个随机的脚本。我正在寻找一种找到瓶颈和加速R-code的通用方法。

The data (x) looks like this:

数据(x)如下:

type      word    response    N   Classification  classN
Abstract  ANGER   bitter      1   3a              3a
Abstract  ANGER   control     1   1a              1a
Abstract  ANGER   father      1   3a              3a
Abstract  ANGER   flushed     1   3a              3a
Abstract  ANGER   fury        1   1c              1c
Abstract  ANGER   hat         1   3a              3a
Abstract  ANGER   help        1   3a              3a
Abstract  ANGER   mad         13  3a              3a
Abstract  ANGER   management  2   1a              1a
... until row 1700

The script (with short explanations) is this:

剧本(简短的解释)是:

Rprof("profile1.out")

# A new dataset is produced with each line of x contained x$N times 
y <- vector('list',length(x[,1]))
for (i in 1:length(x[,1])) {
  y[[i]] <- data.frame(rep(x[i,1],x[i,"N"]),rep(x[i,2],x[i,"N"]),rep(x[i,3],x[i,"N"]),rep(x[i,4],x[i,"N"]),rep(x[i,5],x[i,"N"]),rep(x[i,6],x[i,"N"]))
}
all <- do.call('rbind',y)
colnames(all) <- colnames(x)

# create a dataframe out of a word x class table
table_all <- table(all$word,all$classN)
dataf.all <- as.data.frame(table_all[,1:length(table_all[1,])])
dataf.all$words <- as.factor(rownames(dataf.all))
dataf.all$type <- "no"
# get type of the word.
words <- levels(dataf.all$words)
for (i in 1:length(words)) {
  dataf.all$type[i] <- as.character(all[pmatch(words[i],all$word),"type"])
}
dataf.all$type <- as.factor(dataf.all$type)
dataf.all$typeN <- as.numeric(dataf.all$type)

# aggregate response categories
dataf.all$c1 <- apply(dataf.all[,c("1a","1b","1c","1d","1e","1f")],1,sum)
dataf.all$c2 <- apply(dataf.all[,c("2a","2b","2c")],1,sum)
dataf.all$c3 <- apply(dataf.all[,c("3a","3b")],1,sum)

Rprof(NULL)

library(profr)
ggplot.profr(parse_rprof("profile1.out"))

Final data looks like this:

最终数据如下:

1a    1b  1c  1d  1e  1f  2a  2b  2c  3a  3b  pa  words   type    typeN   c1  c2  c3  pa
3 0   8   0   0   0   0   0   0   24  0   0   ANGER   Abstract    1   11  0   24  0
6 0   4   0   1   0   0   11  0   13  0   0   ANXIETY Abstract    1   11  11  13  0
2 11  1   0   0   0   0   4   0   17  0   0   ATTITUDE    Abstract    1   14  4   17  0
9 18  0   0   0   0   0   0   0   0   8   0   BARREL  Concrete    2   27  0   8   0
0 1   18  0   0   0   0   4   0   12  0   0   BELIEF  Abstract    1   19  4   12  0

The base graph plot:

基础图绘制:

Running the script today also changed the ggplot2 graph a little (basically only the labels), see here.

今天运行该脚本还稍微改变了ggplot2图(基本上只改变了标签)，请参见这里。

4 个解决方案

#1

Alert readers of yesterdays breaking news (R 3.0.0 is finally out) may have noticed something interesting that is directly relevant to this question:

提醒读者昨天的突发新闻(r3.0.0终于出来了)可能已经注意到一些与这个问题直接相关的有趣的事情:

• Profiling via Rprof() now optionally records information at the statement level, not just the function level.
• 通过Rprof()进行概要分析现在可以在语句级别(而不只是函数级别)记录信息。

And indeed, this new feature answers my question and I will show how.

的确，这个新特性回答了我的问题，我将展示如何解决这个问题。

---

Let's say, we want to compare whether vectorizing and pre-allocating are really better than good old for-loops and incremental building of data in calculating a summary statistic such as the mean. The, relatively stupid, code is the following:

比方说，我们想比较向量化和预分配在计算汇总统计量(如平均值)时是否真的比老的循环和增量数据构建更好。相对愚蠢的代码如下:

# create big data frame:
n <- 1000
x <- data.frame(group = sample(letters[1:4], n, replace=TRUE), condition = sample(LETTERS[1:10], n, replace = TRUE), data = rnorm(n))

# reasonable operations:
marginal.means.1 <- aggregate(data ~ group + condition, data = x, FUN=mean)

# unreasonable operations:
marginal.means.2 <- marginal.means.1[NULL,]

row.counter <- 1
for (condition in levels(x$condition)) {
  for (group in levels(x$group)) {  
    tmp.value <- 0
    tmp.length <- 0
    for (c in 1:nrow(x)) {
      if ((x[c,"group"] == group) & (x[c,"condition"] == condition)) {
        tmp.value <- tmp.value + x[c,"data"]
        tmp.length <- tmp.length + 1
      }
    }
    marginal.means.2[row.counter,"group"] <- group 
    marginal.means.2[row.counter,"condition"] <- condition
    marginal.means.2[row.counter,"data"] <- tmp.value / tmp.length
    row.counter <- row.counter + 1
  }
}

# does it produce the same results?
all.equal(marginal.means.1, marginal.means.2)

To use this code with Rprof, we need to parse it. That is, it needs to be saved in a file and then called from there. Hence, I uploaded it to pastebin, but it works exactly the same with local files.

要使用Rprof代码，我们需要对其进行解析。也就是说，它需要保存在一个文件中，然后从那里调用。因此，我将它上传到了pastebin，但它与本地文件完全相同。

Now, we

现在,我们

• simply create a profile file and indicate that we want to save the line number,
• 只需创建一个概要文件并指示要保存行号，
• source the code with the incredible combination eval(parse(..., keep.source = TRUE)) (seemingly the infamous fortune(106) does not apply here, as I haven't found another way)
• 使用不可思议的组合eval(解析(…),保持。(似乎《臭名昭著的财富》(106)在这里并不适用，因为我还没有找到另一种方法)
• stop the profiling and indicate that we want the output based on the line numbers.
• 停止分析并表明我们需要基于行号的输出。

The code is:

的代码是:

Rprof("profile1.out", line.profiling=TRUE)
eval(parse(file = "http://pastebin.com/download.php?i=KjdkSVZq", keep.source=TRUE))
Rprof(NULL)

summaryRprof("profile1.out", lines = "show")

Which gives:

这使:

$by.self
                           self.time self.pct total.time total.pct
download.php?i=KjdkSVZq#17      8.04    64.11       8.04     64.11
<no location>                   4.38    34.93       4.38     34.93
download.php?i=KjdkSVZq#16      0.06     0.48       0.06      0.48
download.php?i=KjdkSVZq#18      0.02     0.16       0.02      0.16
download.php?i=KjdkSVZq#23      0.02     0.16       0.02      0.16
download.php?i=KjdkSVZq#6       0.02     0.16       0.02      0.16

$by.total
                           total.time total.pct self.time self.pct
download.php?i=KjdkSVZq#17       8.04     64.11      8.04    64.11
<no location>                    4.38     34.93      4.38    34.93
download.php?i=KjdkSVZq#16       0.06      0.48      0.06     0.48
download.php?i=KjdkSVZq#18       0.02      0.16      0.02     0.16
download.php?i=KjdkSVZq#23       0.02      0.16      0.02     0.16
download.php?i=KjdkSVZq#6        0.02      0.16      0.02     0.16

$by.line
                           self.time self.pct total.time total.pct
<no location>                   4.38    34.93       4.38     34.93
download.php?i=KjdkSVZq#6       0.02     0.16       0.02      0.16
download.php?i=KjdkSVZq#16      0.06     0.48       0.06      0.48
download.php?i=KjdkSVZq#17      8.04    64.11       8.04     64.11
download.php?i=KjdkSVZq#18      0.02     0.16       0.02      0.16
download.php?i=KjdkSVZq#23      0.02     0.16       0.02      0.16

$sample.interval
[1] 0.02

$sampling.time
[1] 12.54

Checking the source code tells us that the problematic line (#17) is indeed the stupid if-statement in the for-loop. Compared with basically no time for calculating the same using vectorized code (line #6).

检查源代码告诉我们，有问题的行(#17)实际上是for循环中愚蠢的if语句。与基本上没有时间使用矢量化代码计算相同的数据相比(第6行)。

I haven't tried it with any graphical output, but I am already very impressed by what I got so far.

我还没有尝试过任何图形输出，但是到目前为止，我已经对我所得到的印象非常深刻。

#2

Update: This function has been re-written to deal with line numbers. It's on github here.

更新:这个函数被重写以处理行号。这里在github上。

I wrote this function to parse the file from Rprof and output a table of somewhat clearer results than summaryRprof. It displays the full stack of functions (and line numbers if line.profiling=TRUE), and their relative contribution to run time:

我编写这个函数来解析来自Rprof的文件，并输出一个比summary更清晰的结果表。它显示完整的函数堆栈(如果是line.profiling=TRUE)，以及它们对运行时间的相对贡献:

proftable <- function(file, lines=10) {
# require(plyr)
  interval <- as.numeric(strsplit(readLines(file, 1), "=")[[1L]][2L])/1e+06
  profdata <- read.table(file, header=FALSE, sep=" ", comment.char = "",
                         colClasses="character", skip=1, fill=TRUE,
                         na.strings="")
  filelines <- grep("#File", profdata[,1])
  files <- aaply(as.matrix(profdata[filelines,]), 1, function(x) {
                        paste(na.omit(x), collapse = " ") })
  profdata <- profdata[-filelines,]
  total.time <- interval*nrow(profdata)
  profdata <- as.matrix(profdata[,ncol(profdata):1])
  profdata <- aaply(profdata, 1, function(x) {
                      c(x[(sum(is.na(x))+1):length(x)],
                        x[seq(from=1,by=1,length=sum(is.na(x)))])
              })
  stringtable <- table(apply(profdata, 1, paste, collapse=" "))
  uniquerows <- strsplit(names(stringtable), " ")
  uniquerows <- llply(uniquerows, function(x) replace(x, which(x=="NA"), NA))
  dimnames(stringtable) <- NULL
  stacktable <- ldply(uniquerows, function(x) x)
  stringtable <- stringtable/sum(stringtable)*100
  stacktable <- data.frame(PctTime=stringtable[], stacktable)
  stacktable <- stacktable[order(stringtable, decreasing=TRUE),]
  rownames(stacktable) <- NULL
  stacktable <- head(stacktable, lines)
  na.cols <- which(sapply(stacktable, function(x) all(is.na(x))))
  stacktable <- stacktable[-na.cols]
  parent.cols <- which(sapply(stacktable, function(x) length(unique(x)))==1)
  parent.call <- paste0(paste(stacktable[1,parent.cols], collapse = " > ")," >")
  stacktable <- stacktable[,-parent.cols]
  calls <- aaply(as.matrix(stacktable[2:ncol(stacktable)]), 1, function(x) {
                   paste(na.omit(x), collapse= " > ")
                     })
  stacktable <- data.frame(PctTime=stacktable$PctTime, Call=calls)
  frac <- sum(stacktable$PctTime)
  attr(stacktable, "total.time") <- total.time
  attr(stacktable, "parent.call") <- parent.call
  attr(stacktable, "files") <- files
  attr(stacktable, "total.pct.time") <- frac
  cat("\n")
  print(stacktable, row.names=FALSE, right=FALSE, digits=3)
  cat("\n")
  cat(paste(files, collapse="\n"))
  cat("\n")
  cat(paste("\nParent Call:", parent.call))
  cat(paste("\n\nTotal Time:", total.time, "seconds\n"))
  cat(paste0("Percent of run time represented: ", format(frac, digits=3)), "%")

  invisible(stacktable)
}

Running this on the Henrik's example file, I get this:

在Henrik的示例文件中运行这个，我得到这个:

> Rprof("profile1.out", line.profiling=TRUE)
> source("http://pastebin.com/download.php?i=KjdkSVZq")
> Rprof(NULL)
> proftable("profile1.out", lines=10)

 PctTime Call                                                      
 20.47   1#17 > [ > 1#17 > [.data.frame                            
  9.73   1#17 > [ > 1#17 > [.data.frame > [ > [.factor             
  8.72   1#17 > [ > 1#17 > [.data.frame > [ > [.factor > NextMethod
  8.39   == > Ops.factor                                           
  5.37   ==                                                        
  5.03   == > Ops.factor > noNA.levels > levels                    
  4.70   == > Ops.factor > NextMethod                              
  4.03   1#17 > [ > 1#17 > [.data.frame > [ > [.factor > levels    
  4.03   1#17 > [ > 1#17 > [.data.frame > dim                      
  3.36   1#17 > [ > 1#17 > [.data.frame > length                   

#File 1: http://pastebin.com/download.php?i=KjdkSVZq

Parent Call: source > withVisible > eval > eval >

Total Time: 5.96 seconds
Percent of run time represented: 73.8 %

Note that the "Parent Call" applies to all the stacks represented on the table. This makes is useful when your IDE or whatever calls your code wraps it in a bunch of functions.

注意，“父调用”适用于表中表示的所有堆栈。当您的IDE或任何调用您的代码时，这将非常有用。

#3

I currently have R uninstalled here, but in SPlus you can interrupt the execution with the Escape key, and then do traceback(), which will show you the call stack. That should enable you to use this handy method.

我现在已经在这里卸载了R，但是在SPlus中，您可以使用Escape键来中断执行，然后执行traceback()，它将显示调用堆栈。这将使您能够使用这个方便的方法。

Here are some reasons why tools built on the same concepts as gprof are not very good at locating performance problems.

下面是基于gprof概念构建的工具不擅长定位性能问题的一些原因。

#4

A different solution comes from a different question: how to effectively use library(profr) in R:

不同的解决方案来自不同的问题:如何有效地使用R中的库(profr):

For example:

例如:

install.packages("profr")
devtools::install_github("alexwhitworth/imputation")

x <- matrix(rnorm(1000), 100)
x[x>1] <- NA
library(imputation)
library(profr)
a <- profr(kNN_impute(x, k=5, q=2), interval= 0.005)

It doesn't seem (to me at least), like the plots are at all helpful here (eg plot(a)). But the data structure itself does seem to suggest a solution:

看来(至少对我来说)，这里的情节毫无用处(如情节(a))。但数据结构本身似乎确实提供了一种解决方案:

R> head(a, 10)
   level g_id t_id                f start   end n  leaf  time     source
9      1    1    1       kNN_impute 0.005 0.190 1 FALSE 0.185 imputation
10     2    1    1        var_tests 0.005 0.010 1 FALSE 0.005       <NA>
11     2    2    1            apply 0.010 0.190 1 FALSE 0.180       base
12     3    1    1         var.test 0.005 0.010 1 FALSE 0.005      stats
13     3    2    1              FUN 0.010 0.110 1 FALSE 0.100       <NA>
14     3    2    2              FUN 0.115 0.190 1 FALSE 0.075       <NA>
15     4    1    1 var.test.default 0.005 0.010 1 FALSE 0.005       <NA>
16     4    2    1           sapply 0.010 0.040 1 FALSE 0.030       base
17     4    3    1    dist_q.matrix 0.040 0.045 1 FALSE 0.005 imputation
18     4    4    1           sapply 0.045 0.075 1 FALSE 0.030       base

Single iteration solution:

That is the data structure suggests the use of tapply to summarize the data. This can be done quite simply for a single run of profr::profr

即数据结构建议使用tapply对数据进行汇总。这可以很简单地完成一次教授::教授

t <- tapply(a$time, paste(a$source, a$f, sep= "::"), sum)
t[order(t)] # time / function
R> round(t[order(t)] / sum(t), 4) # percentage of total time / function

base::!                    base::%in%                       base::|           base::anyDuplicated 
                       0.0015                        0.0015                        0.0015                        0.0015 
                      base::c                 base::deparse                     base::get                   base::match 
                       0.0015                        0.0015                        0.0015                        0.0015 
                   base::mget                     base::min                       base::t                   methods::el 
                       0.0015                        0.0015                        0.0015                        0.0015 
          methods::getGeneric        NA::.findMethodInTable               NA::.getGeneric      NA::.getGenericFromCache 
                       0.0015                        0.0015                        0.0015                        0.0015 
NA::.getGenericFromCacheTable                   NA::.identC             NA::.newSignature        NA::.quickCoerceSelect 
                       0.0015                        0.0015                        0.0015                        0.0015 
                NA::.sigLabel          NA::var.test.default                 NA::var_tests               stats::var.test 
                       0.0015                        0.0015                        0.0015                        0.0015 
                  base::paste                 methods::as<-     NA::.findInheritedMethods        NA::.getClassFromCache 
                       0.0030                        0.0030                        0.0030                        0.0030 
               NA::doTryCatch              NA::tryCatchList               NA::tryCatchOne               base::crossprod 
                       0.0030                        0.0030                        0.0030                        0.0045 
                    base::try                base::tryCatch          methods::getClassDef      methods::possibleExtends 
                       0.0045                        0.0045                        0.0045                        0.0045 
          methods::loadMethod                   methods::is     imputation::dist_q.matrix          methods::validObject 
                       0.0075                        0.0090                        0.0120                        0.0136 
       NA::.findNextFromTable        methods::addNextMethod               NA::.nextMethod                  base::lapply 
                       0.0166                        0.0346                        0.0361                        0.0392 
                 base::sapply     imputation::impute_fn_knn                  methods::new        imputation::kNN_impute 
                       0.0392                        0.0392                        0.0437                        0.0557 
      methods::callNextMethod      kernlab::as.kernelMatrix                   base::apply         kernlab::kernelMatrix 
                       0.0572                        0.0633                        0.0663                        0.0753 
          methods::initialize                       NA::FUN         base::standardGeneric 
                       0.0798                        0.0994                        0.1325 

From this, I can see that the biggest time users are kernlab::kernelMatrix and the overhead from R for S4 classes and generics.

从这里，我可以看到最大的时间用户是kernlab::kernelMatrix和S4类和泛型的R的开销。

Preferred:

I note that, given the stochastic nature of the sampling process, I prefer to use averages to get a more robust picture of the time profile:

我注意到，考虑到抽样过程的随机性，我更喜欢用平均值来获得更可靠的时间轮廓图:

prof_list <- replicate(100, profr(kNN_impute(x, k=5, q=2), 
    interval= 0.005), simplify = FALSE)

fun_timing <- vector("list", length= 100)
for (i in 1:100) {
  fun_timing[[i]] <- tapply(prof_list[[i]]$time, paste(prof_list[[i]]$source, prof_list[[i]]$f, sep= "::"), sum)
}

# Here is where the stochastic nature of the profiler complicates things.
# Because of randomness, each replication may have slightly different 
# functions called during profiling
sapply(fun_timing, function(x) {length(names(x))})

# we can also see some clearly odd replications (at least in my attempt)
> sapply(fun_timing, sum)
[1]    2.820    5.605    2.325    2.895    3.195    2.695    2.495    2.315    2.005    2.475    4.110    2.705    2.180    2.760
 [15] 3130.240    3.435    7.675    7.155    5.205    3.760    7.335    7.545    8.155    8.175    6.965    5.820    8.760    7.345
 [29]    9.815    7.965    6.370    4.900    5.720    4.530    6.220    3.345    4.055    3.170    3.725    7.780    7.090    7.670
 [43]    5.400    7.635    7.125    6.905    6.545    6.855    7.185    7.610    2.965    3.865    3.875    3.480    7.770    7.055
 [57]    8.870    8.940   10.130    9.730    5.205    5.645    3.045    2.535    2.675    2.695    2.730    2.555    2.675    2.270
 [71]    9.515    4.700    7.270    2.950    6.630    8.370    9.070    7.950    3.250    4.405    3.475    6.420 2948.265    3.470
 [85]    3.320    3.640    2.855    3.315    2.560    2.355    2.300    2.685    2.855    2.540    2.480    2.570    3.345    2.145
 [99]    2.620    3.650

Removing the unusual replications and converting to data.frames:

删除不寻常的复制和转换到数据。帧:

fun_timing <- fun_timing[-c(15,83)]
fun_timing2 <- lapply(fun_timing, function(x) {
  ret <- data.frame(fun= names(x), time= x)
  dimnames(ret)[[1]] <- 1:nrow(ret)
  return(ret)
})

Merge replications (almost certainly could be faster) and examine results:

合并复制(几乎肯定会更快)并检查结果:

# function for merging DF's in a list
merge_recursive <- function(list, ...) {
  n <- length(list)
  df <- data.frame(list[[1]])
  for (i in 2:n) {
    df <- merge(df, list[[i]], ... = ...)
  }
  return(df)
}

# merge
fun_time <- merge_recursive(fun_timing2, by= "fun", all= FALSE)
# do some munging
fun_time2 <- data.frame(fun=fun_time[,1], avg_time=apply(fun_time[,-1], 1, mean, na.rm=T))
fun_time2$avg_pct <- fun_time2$avg_time / sum(fun_time2$avg_time)
fun_time2 <- fun_time2[order(fun_time2$avg_time, decreasing=TRUE),]
# examine results
R> head(fun_time2, 15)
                         fun  avg_time    avg_pct
4      base::standardGeneric 0.6760714 0.14745123
20                   NA::FUN 0.4666327 0.10177262
12       methods::initialize 0.4488776 0.09790023
9      kernlab::kernelMatrix 0.3522449 0.07682464
8   kernlab::as.kernelMatrix 0.3215816 0.07013698
11   methods::callNextMethod 0.2986224 0.06512958
1                base::apply 0.2893367 0.06310437
7     imputation::kNN_impute 0.2433163 0.05306731
14              methods::new 0.2309184 0.05036331
10    methods::addNextMethod 0.2012245 0.04388708
3               base::sapply 0.1875000 0.04089377
2               base::lapply 0.1865306 0.04068234
6  imputation::impute_fn_knn 0.1827551 0.03985890
19           NA::.nextMethod 0.1790816 0.03905772
18    NA::.findNextFromTable 0.1003571 0.02188790

Results

From the results, a similar but more robust picture emerges as with a single case. Namely, there is a lot of overhead from R and also that library(kernlab) is slowing me down. Of note, since kernlab is implemented in S4, the overhead in R is related since S4 classes are substantially slower than S3 classes.

从结果来看，一个类似但更健壮的画面出现在一个单一的案例中。也就是说，从R和库(kernlab)中有大量的开销使我放慢了脚步。值得注意的是，由于kernlab是在S4中实现的，所以R中的开销是相关的，因为S4类比S3类要慢得多。

I'd also note that my personal opinion is that a cleaned up version of this might be a useful pull request as a summary method for profr. Although I'd be interested to see others' suggestions!

我还想指出的是，我个人的观点是，对于profr来说，一个清理过的版本可能是一个有用的摘要方法。虽然我很想看看别人的建议!