Determining memory usage of objects?
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)