如何在R中高效使用Rprof？

发布于 2024-09-17 20:26:13 字数 3259 浏览 3 评论 0原文

我想知道是否可以通过类似于 matlab 的 Profiler 的方式从 R-Code 获取配置文件。也就是说，要知道哪些行号是特别慢的。

到目前为止，我所取得的成绩并不令人满意。我使用 Rprof 制作了一个配置文件。使用 summaryRprof 我得到类似以下内容：

<前><代码>$by.self self.time self.pct 总计.time 总计.pct [.data.frame 0.72 10.1 1.84 25.8 继承 0.50 7.0 1.10 15.4 数据帧 0.48 6.7 4.86 68.3 唯一默认 0.44 6.2 0.48 6.7 稀疏 0.36 5.1 1.18 16.6 结合 0.30 4.2 2.22 31.2 匹配 0.28 3.9 1.38 19.4 [<-.系数 0.28 3.9 0.56 7.9 水平 0.26 3.7 0.34 4.8 下一步方法 0.22 3.1 0.82 11.5 ...

和

$by.total
                      总.时间 总.pct 自我.时间 自我.pct
数据帧 4.86 68.3 0.48 6.7
绑定 2.22 31.2 0.30 4.2
调用 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
匹配 1.38 19.4 0.28 3.9
%% 1.26 17.7 0.14 2.0
系数 1.20 16.9 0.10 1.4
稀疏 1.18 16.6 0.36 5.1
...

老实说，从这个输出中我不知道我的瓶颈在哪里，因为（a）我经常使用 data.frame 并且（b）我从不使用例如 deparse。此外，什么是[？

所以我尝试了 Hadley Wickham 的 profr，但考虑到下图，它并没有任何用处： alt text

是否有更方便的方法来查看哪些行号和特定函数调用速度慢？
或者，是否有一些我应该查阅的文献？

任何提示表示赞赏。

编辑1：
根据哈德利的评论，我将粘贴下面的脚本代码和绘图的基本图形版本。但请注意，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脚本。 我正在寻找一种通用方法来查找瓶颈并加速 R 代码。

数据 (x) 如下所示：

键入单词响应 N 分类 classN
摘要愤怒苦涩 1 3a 3a
摘要愤怒控制 1 1a 1a
摘要愤怒父亲 1 3a 3a
摘要愤怒脸红了 1 3a 3a
抽象愤怒愤怒 1 1c 1c
抽象愤怒帽子 1 3a 3a
摘要愤怒帮助 1 3a 3a
抽象愤怒疯狂 13 3a 3a
摘要 愤怒管理 2 1a 1a
...直到第 1700 行

脚本（带有简短的解释）是这样的：

Rprof("profile1.out")

# 生成一个新的数据集，其中 x 的每一行包含 x$N 次 
y <- 向量('列表',长度(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"]))
}
所有 <- do.call('rbind',y)
列名（全部）<- 列名（x）

# 从单词 x 类表中创建一个数据框
table_all <- 表(全部$word,全部$classN)
dataf.all <- as.data.frame(table_all[,1:length(table_all[1,])])
dataf.all$words <- as.factor(rownames(dataf.all))
dataf.all$type <-“否”
# 获取单词的类型。
单词 <- 级别(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)

# 聚合响应类别
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（空）

图书馆（教授）
ggplot.profr(parse_rprof("profile1.out"))

最终数据如下所示：

1a 1b 1c 1d 1e 1f 2a 2b 2c 3a 3b pa 字类型 typeN c1 c2 c3 pa
3 0 8 0 0 0 0 0 0 24 0 0 愤怒 摘要 1 11 0 24 0
6 0 4 0 1 0 0 11 0 13 0 0 焦虑 摘要 1 11 11 13 0
2 11 1 0 0 0 0 4 0 17 0 0 态度 摘要 1 14 4 17 0
9 18 0 0 0 0 0 0 0 0 8 0 桶 混凝土 2 27 0 8 0
0 1 18 0 0 0 0 4 0 12 0 0 信念 摘要 1 19 4 12 0

基本图： alt text

今天运行脚本还对 ggplot2 图表进行了一些更改（基本上只是标签），请参见此处。

原文

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.

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:

$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

$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 [?

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

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?

Any hints appreciated.

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.

The data (x) looks like this:

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:

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:

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:
alt text

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

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

盛夏尉蓝 2024-09-24 20:26:13

提醒读者昨天的突发新闻（R 3.0.0终于out）可能已经注意到一些与这个问题直接相关的有趣的事情：

通过 Rprof() 进行分析现在可以选择在语句级别记录信息，而不仅仅是函数级别。

事实上，这个新功能回答了我的问题，我将展示如何解决。

比方说，我们想要比较向量化和预分配在计算汇总统计（例如平均值）时是否真的比旧的 for 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)

要将此代码与 Rprof 一起使用，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，我将其上传到 pastebin，但它与本地文件的工作原理完全相同。

现在，我们

只需创建一个配置文件并表明我们要保存行号，
使用令人难以置信的组合 eval(parse(..., keep.source = TRUE)) （看似臭名昭著的 fortune(106) 在这里不适用，因为我还没有找到其他方法）
停止分析并指示我们想要基于行号的输出。

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")

给出：

$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

检查源代码告诉我们有问题的行 (#17) 确实是愚蠢的 if< /code>-for 循环中的语句。与使用矢量化代码（第 6 行）基本上没有时间计算相同的结果相比。

我还没有尝试过任何图形输出，但到目前为止我所得到的结果已经给我留下了深刻的印象。

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:

Profiling via Rprof() now optionally records information at the statement level, not just the function level.

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

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:

# 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.

Now, we

simply create a profile file and indicate that we want to save the line number,
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)
stop the profiling and indicate that we want the output based on the line numbers.

The code is:

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:

$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).

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

回复收藏 0 原文

黯淡〆 2024-09-24 20:26:13

更新：此函数已被重写以处理行号。它位于 github 此处。

我编写了这个函数来解析 Rprof 中的文件，并输出一个比 summaryRprof 更清晰的结果表。它显示完整的函数堆栈（如果 line.profiling=TRUE 则显示行号），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)
}

在 Henrik 的示例文件上运行此文件，我得到以下结果：

> 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 %

请注意，“ “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将其包装在一堆函数中时，这非常有用。

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

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:

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:

> 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.

回复收藏 0 原文

雪若未夕 2024-09-24 20:26:13

我目前在这里卸载了 R，但在 SPlus 中，您可以使用 Escape 键中断执行，然后执行 traceback()，这将显示调用堆栈。这应该使您能够使用这个方便的方法。

以下是为什么工具基于与gprof<相同的概念构建的一些原因/strong> 不太擅长定位性能问题。

回复收藏 0 原文

り繁华旳梦境 2024-09-24 20:26:13

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）：

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)

至少对我来说），这些图在这里根本没有帮助（例如 plot(a)）。但数据结构本身似乎确实提出了一个解决方案：

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

单次迭代解决方案：

即数据结构建议使用 tapply 来汇总数据。对于单次运行 profr::profr 来说，这可以非常简单地完成。

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

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

首选：

我注意到，考虑到采样过程的随机性，我更喜欢使用平均值来获得更可靠的时间概况：

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

删除不寻常的重复并转换为 data.frames：

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)
})

合并复制（几乎肯定可以更快）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

结果

从结果中，出现了与单个案例类似但更可靠的情况。也就是说，R 产生了大量开销，而且 library(kernlab) 也拖慢了我的速度。值得注意的是，由于 kernlab 是在 S4 中实现的，因此 R 中的开销是相关的，因为 S4 类比 S3 类慢得多。

我还要指出的是，我个人的观点是，它的清理版本可能是一个有用的拉取请求，作为

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

For example:

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:

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

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.

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:

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:

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:

# 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.

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!

回复收藏 0 原文

~没有更多了~