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()

我从不保存R工作区。我使用导入脚本和数据脚本，并将我不想经常重新创建的任何特别大的数据对象输出到文件。这样，我总是从一个新的工作空间开始，不需要清理大的物体。这是一个很好的函数。

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。

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 )

这是个好把戏。

另一个建议是尽可能使用内存效率高的对象:例如，使用矩阵而不是data.frame。

这并没有真正解决内存管理问题，但是一个不为人所知的重要函数是memory.limit()。可以使用memory.limit(size=2500)命令增加默认值，这里的大小以MB为单位。正如Dirk提到的，为了真正利用这一点，您需要使用64位。

这并没有增加上面的内容，而是以我喜欢的简单和大量注释的风格编写的。它生成一个对象大小排序表，但没有上面例子中给出的一些细节:

#Find the objects       
MemoryObjects = ls()    
#Create an array
MemoryAssessmentTable=array(NA,dim=c(length(MemoryObjects),2))
#Name the columns
colnames(MemoryAssessmentTable)=c("object","bytes")
#Define the first column as the objects
MemoryAssessmentTable[,1]=MemoryObjects
#Define a function to determine size        
MemoryAssessmentFunction=function(x){object.size(get(x))}
#Apply the function to the objects
MemoryAssessmentTable[,2]=t(t(sapply(MemoryAssessmentTable[,1],MemoryAssessmentFunction)))
#Produce a table with the largest objects first
noquote(MemoryAssessmentTable[rev(order(as.numeric(MemoryAssessmentTable[,2]))),])