人们使用什么技巧来管理交互式R会话的可用内存?我使用下面的函数[基于Petr Pikal和David Hinds在2004年发布的r-help列表]来列出(和/或排序)最大的对象,并偶尔rm()其中一些对象。但到目前为止最有效的解决办法是……在64位Linux下运行,有充足的内存。

大家还有什么想分享的妙招吗?请每人寄一份。

# improved list of objects
.ls.objects <- function (pos = 1, pattern, order.by,
                        decreasing=FALSE, head=FALSE, n=5) {
    napply <- function(names, fn) sapply(names, function(x)
                                         fn(get(x, pos = pos)))
    names <- ls(pos = pos, pattern = pattern)
    obj.class <- napply(names, function(x) as.character(class(x))[1])
    obj.mode <- napply(names, mode)
    obj.type <- ifelse(is.na(obj.class), obj.mode, obj.class)
    obj.size <- napply(names, object.size)
    obj.dim <- t(napply(names, function(x)
                        as.numeric(dim(x))[1:2]))
    vec <- is.na(obj.dim)[, 1] & (obj.type != "function")
    obj.dim[vec, 1] <- napply(names, length)[vec]
    out <- data.frame(obj.type, obj.size, obj.dim)
    names(out) <- c("Type", "Size", "Rows", "Columns")
    if (!missing(order.by))
        out <- out[order(out[[order.by]], decreasing=decreasing), ]
    if (head)
        out <- head(out, n)
    out
}
# shorthand
lsos <- function(..., n=10) {
    .ls.objects(..., order.by="Size", decreasing=TRUE, head=TRUE, n=n)
}

当前回答

这是对这个优秀的老问题的一个新的回答。来自哈德利的高级R:

install.packages("pryr")

library(pryr)

object_size(1:10)
## 88 B

object_size(mean)
## 832 B

object_size(mtcars)
## 6.74 kB

(http://adv-r.had.co.nz/memory.html)

其他回答

For both speed and memory purposes, when building a large data frame via some complex series of steps, I'll periodically flush it (the in-progress data set being built) to disk, appending to anything that came before, and then restart it. This way the intermediate steps are only working on smallish data frames (which is good as, e.g., rbind slows down considerably with larger objects). The entire data set can be read back in at the end of the process, when all the intermediate objects have been removed.

dfinal <- NULL
first <- TRUE
tempfile <- "dfinal_temp.csv"
for( i in bigloop ) {
    if( !i %% 10000 ) { 
        print( i, "; flushing to disk..." )
        write.table( dfinal, file=tempfile, append=!first, col.names=first )
        first <- FALSE
        dfinal <- NULL   # nuke it
    }

    # ... complex operations here that add data to 'dfinal' data frame  
}
print( "Loop done; flushing to disk and re-reading entire data set..." )
write.table( dfinal, file=tempfile, append=TRUE, col.names=FALSE )
dfinal <- read.table( tempfile )

如果真的想避免泄漏,应该避免在全局环境中创建任何大对象。

我通常做的是有一个函数来完成这项工作并返回NULL -所有数据都在这个函数或它调用的其他函数中读取和操作。

我喜欢Dirk的.ls.objects()脚本,但我总是眯着眼睛数大小列中的字符。所以我做了一些丑陋的hack,使它呈现出漂亮的格式大小:

.ls.objects <- function (pos = 1, pattern, order.by,
                        decreasing=FALSE, head=FALSE, n=5) {
    napply <- function(names, fn) sapply(names, function(x)
                                         fn(get(x, pos = pos)))
    names <- ls(pos = pos, pattern = pattern)
    obj.class <- napply(names, function(x) as.character(class(x))[1])
    obj.mode <- napply(names, mode)
    obj.type <- ifelse(is.na(obj.class), obj.mode, obj.class)
    obj.size <- napply(names, object.size)
    obj.prettysize <- sapply(obj.size, function(r) prettyNum(r, big.mark = ",") )
    obj.dim <- t(napply(names, function(x)
                        as.numeric(dim(x))[1:2]))
    vec <- is.na(obj.dim)[, 1] & (obj.type != "function")
    obj.dim[vec, 1] <- napply(names, length)[vec]
    out <- data.frame(obj.type, obj.size,obj.prettysize, obj.dim)
    names(out) <- c("Type", "Size", "PrettySize", "Rows", "Columns")
    if (!missing(order.by))
        out <- out[order(out[[order.by]], decreasing=decreasing), ]
        out <- out[c("Type", "PrettySize", "Rows", "Columns")]
        names(out) <- c("Type", "Size", "Rows", "Columns")
    if (head)
        out <- head(out, n)
    out
}

除了以上回答中给出的更通用的内存管理技术外,我总是尽可能地减小对象的大小。例如,我处理非常大但非常稀疏的矩阵,换句话说,大多数值为零的矩阵。使用“矩阵”包(大写很重要),我能够将我的平均对象大小从~2GB减小到~200MB,简单如下:

my.matrix <- Matrix(my.matrix)

Matrix包包含的数据格式可以像常规矩阵一样使用(不需要更改其他代码),但能够更有效地存储稀疏数据,无论是加载到内存中还是保存到磁盘中。

此外,我收到的原始文件是“长”格式的,其中每个数据点都有变量x, y, z, I。将数据转换为只有变量I的x * y * z维度数组更有效。

了解你的数据并使用一些常识。

我非常喜欢Dirk开发的改进的对象函数。不过,大多数时候,一个包含对象名称和大小的更基本的输出对我来说就足够了。这是一个具有类似目标的简单函数。内存使用可以按字母顺序或大小排序,可以限制为一定数量的对象,并且可以按升序或降序排序。此外,我经常处理1GB以上的数据,因此该函数相应地改变单位。

showMemoryUse <- function(sort="size", decreasing=FALSE, limit) {

  objectList <- ls(parent.frame())

  oneKB <- 1024
  oneMB <- 1048576
  oneGB <- 1073741824

  memoryUse <- sapply(objectList, function(x) as.numeric(object.size(eval(parse(text=x)))))

  memListing <- sapply(memoryUse, function(size) {
        if (size >= oneGB) return(paste(round(size/oneGB,2), "GB"))
        else if (size >= oneMB) return(paste(round(size/oneMB,2), "MB"))
        else if (size >= oneKB) return(paste(round(size/oneKB,2), "kB"))
        else return(paste(size, "bytes"))
      })

  memListing <- data.frame(objectName=names(memListing),memorySize=memListing,row.names=NULL)

  if (sort=="alphabetical") memListing <- memListing[order(memListing$objectName,decreasing=decreasing),] 
  else memListing <- memListing[order(memoryUse,decreasing=decreasing),] #will run if sort not specified or "size"

  if(!missing(limit)) memListing <- memListing[1:limit,]

  print(memListing, row.names=FALSE)
  return(invisible(memListing))
}

下面是一些输出示例:

> showMemoryUse(decreasing=TRUE, limit=5)
      objectName memorySize
       coherData  713.75 MB
 spec.pgram_mine  149.63 kB
       stoch.reg  145.88 kB
      describeBy    82.5 kB
      lmBandpass   68.41 kB