I have a big data frame taking about 900MB ram. Then I tried to modify it like this:
dataframe[[17]][37544]=0
It seems that makes R using more than 3G ram and R complains “Error: cannot allocate vector of size 3.0 Mb”, ( I am on a 32bit machine.)
I found this way is better:
dataframe[37544, 17]=0
but R’s footprint still doubled and the command takes quite some time to run.
From a C/C++ background, I am really confused about this behavior. I thought something like dataframe[37544, 17]=0 should be completed in a blink without costing any extra memory (only one cell should be modified). What is R doing for those commands I posted? What is the right way to modify some elements in a data frame then without doubling the memory footprint?
Thanks so much for your help!
Tao
Look up ‘copy-on-write’ in the context of R discussions related to memory. As soon as one part of a (potentially really large) data structure changes, a copy is made.
A useful rule of thumb is that if your largest object is N mb/gb/… large, you need around 3*N of RAM. Such is life with an interpreted system.
Years ago when I had to handle large amounts of data on machines with (relative to the data volume) relatively low-ram 32-bit machines, I got good use out of early versions of the bigmemory package. It uses the ‘external pointer’ interface to keep large gobs of memory outside of R. That save you not only the ‘3x’ factor, but possibly more as you may get away with non-contiguous memory (which is the other thing R likes).