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。

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()

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。

我在推特上看到了这个，觉得德克的功能太棒了!根据JD Long的回答，为了方便用户阅读，我会这样做:

# 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.prettysize <- napply(names, function(x) {
                           format(utils::object.size(x), units = "auto") })
    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.prettysize, obj.dim)
    names(out) <- c("Type", "Size", "PrettySize", "Length/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)
}

lsos()

结果如下:

                      Type   Size PrettySize Length/Rows Columns
pca.res                 PCA 790128   771.6 Kb          7      NA
DF               data.frame 271040   264.7 Kb        669      50
factor.AgeGender   factanal  12888    12.6 Kb         12      NA
dates            data.frame   9016     8.8 Kb        669       2
sd.                 numeric   3808     3.7 Kb         51      NA
napply             function   2256     2.2 Kb         NA      NA
lsos               function   1944     1.9 Kb         NA      NA
load               loadings   1768     1.7 Kb         12       2
ind.sup             integer    448  448 bytes        102      NA
x                 character     96   96 bytes          1      NA

注:我补充的主要部分是(再次改编自JD的回答):

obj.prettysize <- napply(names, function(x) {
                           print(object.size(x), units = "auto") })

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]))),])

Unfortunately I did not have time to test it extensively but here is a memory tip that I have not seen before. For me the required memory was reduced with more than 50%. When you read stuff into R with for example read.csv they require a certain amount of memory. After this you can save them with save("Destinationfile",list=ls()) The next time you open R you can use load("Destinationfile") Now the memory usage might have decreased. It would be nice if anyone could confirm whether this produces similar results with a different dataset.