CODEWITHSUNDEEP
rprecision
Michael
Michael

Reputation: 5898

Round to machine tolerance

I'd like to take a set of predictions and round them to machine tolerance, then calculate a hash from those numbers. The goal is to determine whether a complicated R process which generates those numbers has been altered. Each substantive change to the codebase to the process will generate these numbers, round them them, then compute the hash so identicalness of the process can be easily determined and tracked in a database.

I'm familiar with all.equal(), but using all.equal() would require every set of numbers to be compared with every other set of numbers directly in R--whereas what I'm envisioning the comparison of identicalness to a large set of process versions can be done just based on hashes in a table.

So my question is how would I round a numeric vector in R to a precision exactly comparable to the precision used in all.equal?

i.e., let's name the rounding function that accomplishes what I want machine.round: identical(machine.round(x),machine.round(y)) is TRUE if and only if all.equal(x,y) is TRUE.

what is the definition of machine.round?

#two vectors which are considered equal by all.equal but not by identical
x <- c(.3, .4 - .1, .5 - .2, .6 - .3, .7 - .4)
y <- c(.3, .3, .3, .3, .3)
all.equal(x,y)
identical(x,y)



#two vectors which are considered unequal by all.equal and identical:
x2 <- c(.30001, .4 - .1, .5 - .2, .6 - .3, .7 - .4)
y2 <- c(.3, .3, .3, .3, .3)
all.equal(x2,y2)
identical(x2,y2)

I want a function machine.round() that returns TRUE for identical(machine.round(x),machine.round(y)) but FALSE for identical(machine.round(x2),machine.round(y2))

Upvotes: 1

Views: 214

Answers (1)

James
James

Reputation: 66844

The default tolerance used by all.equal can be found by examinating its formal arguments:

formals(all.equal.numeric)$tolerance
sqrt(.Machine$double.eps)

Which evaluates to:

eval(formals(all.equal.numeric)$tolerance)
[1] 1.490116e-08

However, I don't think rounding then comparing is the correct approach. You should be checking closeness within this tolerance as all.equal does.

For example:

abs(x-y)<sqrt(.Machine$double.eps)
[1] TRUE TRUE TRUE TRUE TRUE
abs(x2-y2)<sqrt(.Machine$double.eps)
[1] FALSE  TRUE  TRUE  TRUE  TRUE

Upvotes: 1

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