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

如果您正在Linux上工作,希望使用多个进程,并且只需要对一个或多个大对象执行读取操作,请使用makeForkCluster而不是makePSOCKcluster。这也节省了将大对象发送给其他进程的时间。

其他回答

确保在可重复的脚本中记录您的工作。不时地重新打开R,然后source()您的脚本。您将清除不再使用的任何东西,作为一个额外的好处,您将测试您的代码。

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 )

基于@德克和@托尼的回答,我做了一个小小的更新。结果是在漂亮的大小值之前输出[1],所以我取出了捕获。解决问题的输出:

.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.prettysize <- napply(names, function(x) {
    format(utils::object.size(x),  units = "auto") })
obj.size <- napply(names, utils::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.prettysize, obj.dim)
names(out) <- c("Type", "Size", "PrettySize", "Rows", "Columns")
if (!missing(order.by))
    out <- out[order(out[[order.by]], decreasing=decreasing), ]
if (head)
    out <- head(out, n)

return(out)
}

# shorthand
lsos <- function(..., n=10) {
    .ls.objects(..., order.by="Size", decreasing=TRUE, head=TRUE, n=n)
}

lsos()

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。

使用环境而不是列表来处理占用大量工作内存的对象集合。

原因是:每当列表结构的一个元素被修改时,整个列表都会被临时复制。如果列表的存储需求大约是可用工作内存的一半,这就会成为一个问题,因为这时必须将数据交换到慢速硬盘上。另一方面,环境不受这种行为的影响,它们可以类似于列表。

这里有一个例子:

get.data <- function(x)
{
  # get some data based on x
  return(paste("data from",x))
}

collect.data <- function(i,x,env)
{
  # get some data
  data <- get.data(x[[i]])
  # store data into environment
  element.name <- paste("V",i,sep="")
  env[[element.name]] <- data
  return(NULL)  
}

better.list <- new.env()
filenames <- c("file1","file2","file3")
lapply(seq_along(filenames),collect.data,x=filenames,env=better.list)

# read/write access
print(better.list[["V1"]])
better.list[["V2"]] <- "testdata"
# number of list elements
length(ls(better.list))

结合结构,如大。矩阵或数据。表允许修改其内容的地方,非常有效的内存使用可以实现。