Determining memory usage of objects?

Question

some time ago I stole this little nugget from here:

sort( sapply(ls(),function(x){object.size(get(x))})) 

it has served me well

1. by object size

to get memory allocation on an object-by-object basis, call object.size() and pass in the object of interest:

object.size(My_Data_Frame)

(unless the argument passed in is a variable, it must be quoted, or else wrapped in a get call.)variable name, then omit the quotes,

you can loop through your namespace and get the size of all of the objects in it, like so:

for (itm in ls()) { 
    print(formatC(c(itm, object.size(get(itm))), 
        format="d", 
        big.mark=",", 
        width=30), 
        quote=F)
}

2. by object type

to get memory usage for your namespace, by object type, use memory.profile()

memory.profile()

   NULL      symbol    pairlist     closure environment     promise    language 
      1        9434      183964        4125        1359        6963       49425 
special     builtin        char     logical     integer      double     complex 
    173        1562       20652        7383       13212        4137           1 

(There's another function, memory.size() but i have heard and read that it only seems to work on Windows. It just returns a value in MB; so to get max memory used at any time in the session, use memory.size(max=T)).

You could try the lsos() function from this question:

R> a <- rnorm(100)
R> b <- LETTERS
R> lsos()
       Type Size Rows Columns
b character 1496   26      NA
a   numeric  840  100      NA
R> 

This question was posted and got legitimate answers so much ago, but I want to let you know another useful tips to get the size of an object using a library called gdata and its ll() function.

library(gdata)
ll() # return a dataframe that consists of a variable name as rownames, and class and size (in KB) as columns
subset(ll(), KB > 1000) # list of object that have over 1000 KB
ll()[order(ll()$KB),] # sort by the size (ascending)

another (slightly prettier) option using dplyr

    data.frame('object' = ls()) %>% 
      dplyr::mutate(size_unit = object %>%sapply(. %>% get() %>% object.size %>% format(., unit = 'auto')),
                    size = as.numeric(sapply(strsplit(size_unit, split = ' '), FUN = function(x) x[1])),
                    unit = factor(sapply(strsplit(size_unit, split = ' '), FUN = function(x) x[2]), levels = c('Gb', 'Mb', 'Kb', 'bytes'))) %>% 
      dplyr::arrange(unit, dplyr::desc(size)) %>% 
      dplyr::select(-size_unit)

A data.table function that separates memory and unit for easier sorting:

ls.obj <- {as.data.table(sapply(ls(), function(x){format(object.size(get(x)), nsmall=3,digits=3,unit="Mb")}),keep.rownames=TRUE)[, c("mem","unit") := tstrsplit(V2, " ", fixed=TRUE)][, setnames(.SD,"V1","obj")][,.(obj,mem=as.numeric(mem),unit)][order(-mem)]}

ls.obj

                    obj     mem unit
 1:                obj1 848.283   Mb
 2:                obj2  37.705   Mb

...