This is probably a FAQ, and I don't really have a question about it, but I
just ran across this in something I was working on:
as.integer(1000*1.003)
[1] 1002
I didn't expect it, but maybe I should have. I guess it's about the
machine precision added to the fact that as.integer always rounds down:
as.integer(1000*1.003 + 255 * .Machine$double.eps)
[1] 1002
as.integer(1000*1.003 + 256 * .Machine$double.eps)
[1] 1003
This does it right...
as.integer( round( 1000*1.003 ) )
[1] 1003
...but this seems to always give the same answer and it is a little faster
in my application:
as.integer( 1000*1.003 + .1 )
[1] 1003
FYI - I'm reading in a long vector of numbers from a text file with no
more than three digits to the right of the decimal. I'm converting them
to integers and saving them in binary format.
Best,
Mike
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