Log plus one transformation in R
On 12/12/2016 12:26 PM, John Sorkin wrote:
David, I did read the help page. All it says is log1p(x) computes log(1+x) accurately also for |x| << 1 (and less accurately when x is approximately -1). This gives me pause. Does it mean that log(x) does not give accurate results? If log1p gives more accurate values than log, why is the log function not written to use the more accurate computation performed by log1p.
I don't think anyone has directly answered these questions. The problem isn't with log(), it's with the representation of floating point numbers in R. The log of 1 + 10^(-100) is very close to 10^(-100), but since 1 + 1e-100 evaluates to 1 (we only keep 15 or 16 digits of precision), log(1 + 1e-100) will come out as zero. On the other hand, log1p(1e-100) evaluates correctly to 1e-100.
I don't believe I can look directly at the code for log and log1p,
R is open source, so you could, but those are likely coming from system libraries, so it isn't easy to see how the approximations are being done. Duncan Murdoch
so I need to rely on the kindness of others to explain the differences between the computations performed by the functions. I guess the test I ran, log1p(0.000001)/log(0.000001+1), did not have enough precision to demonstrate a difference between the two functions. John John David Sorkin M.D., Ph.D. Professor of Medicine Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing)
David Winsemius <dwinsemius at comcast.net> 12/12/16 12:05 PM >>>
On Dec 12, 2016, at 8:53 AM, John Sorkin <jsorkin at grecc.umaryland.edu>
wrote:
At the risk of being flamed . . . What is the difference between log1p(x) and log(x+1)? The two methods appear to give the same results:
log1p(0.000001)/log(0.000001+1)
[1] 1 John
Read the help page more carefully.