R inequality operator error? [duplicate] - r

Question

I'm trying to compare two numbers in R as a part of a if-statement condition:
(a-b) >= 0.5
In this particular instance, a = 0.58 and b = 0.08... and yet (a-b) >= 0.5 is false. I'm aware of the dangers of using == for exact number comparisons, and this seems related:
(a - b) == 0.5) is false, while
all.equal((a - b), 0.5) is true.
The only solution I can think of is to have two conditions: (a-b) > 0.5 | all.equal((a-b), 0.5). This works, but is that really the only solution? Should I just swear off of the = family of comparison operators forever?
Edit for clarity: I know that this is a floating point problem. More fundamentally, what I'm asking is: what should I do about it? What's a sensible way to deal with greater-than-or-equal-to comparisons in R, since the >= can't really be trusted?

Answer 1

I've never been a fan of all.equal for such things. It seems to me the tolerance works in mysterious ways sometimes. Why not just check for something greater than a tolerance less than 0.05
tol = 1e-5
(a-b) >= (0.05-tol)
In general, without rounding and with just conventional logic I find straight logic better than all.equal
If x == y then x-y == 0. Perhaps x-y is not exactly 0 so for such cases I use
abs(x-y) <= tol
You have to set tolerance anyway for all.equal and this is more compact and straightforward than all.equal.

Answer 2

You could create this as a separate operator or overwrite the original >= function (probably not a good idea) if you want to use this approach frequently:
# using a tolerance
epsilon <- 1e-10 # set this as a global setting
`%>=%` <- function(x, y) (x + epsilon > y)
# as a new operator with the original approach
`%>=%` <- function(x, y) (all.equal(x, y)==TRUE | (x > y))
# overwriting R's version (not advised)
`>=` <- function(x, y) (isTRUE(all.equal(x, y)) | (x > y))
> (a-b) >= 0.5
[1] TRUE
> c(1,3,5) >= 2:4
[1] FALSE FALSE TRUE

Answer 3

For completeness' sake, I'll point out that, in certain situations, you could simply round to a few decimal places (and this is kind of a lame solution by comparison to the better solution previously posted.)
round(0.58 - 0.08, 2) == 0.5

Answer 4

One more comment. The all.equal is a generic. For numeric values, it uses all.equal.numeric. An inspection of this function shows that it used .Machine$double.eps^0.5, where .Machine$double.eps is defined as
double.eps: the smallest positive floating-point number ‘x’ such that
‘1 + x != 1’. It equals ‘double.base ^ ulp.digits’ if either
‘double.base’ is 2 or ‘double.rounding’ is 0; otherwise, it
is ‘(double.base ^ double.ulp.digits) / 2’. Normally
‘2.220446e-16’.
(.Machine manual page).
In other words, that would be an acceptable choice for your tolerance:
myeq <- function(a, b, tol=.Machine$double.eps^0.5)
abs(a - b) <= tol

Answer 5

Choose some tolerance level:
epsilon <- 1e-10
Then use
(a-b+epsilon) >= 0.5

Answer 6

But, if your using tolerances anyway, why do you care that a-b == .5 (in fact) doesn't get evaluated? If you are using tolerances anyway you are saying I don't care about the end points exactly.
Here is what is true
if( (a-b) >= .5)
if( (a-b) < .5)
one of those should always evaluate true on every pair of doubles. Any code that uses one implicitly defines a no operation on the other one, at least. If your using tolerances to get actual .5 included in the first but your problem is defined on a continuous domain you arn't accomplishing much. In most problems involving continuous values in the underlying problem there will be very little point to that, since values arbitrarily over .5 will always evaluate as they should. Values arbitrarily close to .5 will go to the "wrong" flow control, but in continuous problems where you are using appropriate precision that doesn't matter.
The only time that tolerances make sense is when you are dealing with problems of the type
if( (a-b) == c)
if( (a-b) != c)
Here no amount of "appropriate precision" can help you. The reason is that you have to be prepared that the second will always evaluate to true unless you set the bits of a-b at a very low level by hand, when in fact you probably want the first to sometimes be true.

Answer 7

<= and >= comparisons are not language specific when numerical difficulty is raised in floating point numbers.
IsSmallerOrEqual <- function(a,b) { # To check a <= b
# Check whether "Mean relative difference..." exist in all.equal's result;
# If exists, it results in character, not logical
if ( class(all.equal(a, b)) == "logical" && (a<b | all.equal(a, b))) { return(TRUE)
} else if (a < b) { return(TRUE)
} else { return(FALSE) }
}
IsSmallerOrEqual(abs(-2-(-2.2)), 0.2) # TRUE; To check |-2-(-2.2)| <= 0.2
IsSmallerOrEqual(abs(-2-(-2.2)), 0.3) # TRUE
IsSmallerOrEqual(abs(-2-(-2.2)), 0.1) # FALSE
IsBiggerOrEqual <- function(a,b) { # To check a >= b
# Check whether "Mean relative difference..." exist in all.equal's result;
# If exists, it results in character, not logical
if ( class(all.equal(a, b)) == "logical" && (a>b | all.equal(a, b))) { return(TRUE)
} else if (a > b) { return(TRUE)
} else { return(FALSE) }
}
IsBiggerOrEqual(3,3) # TRUE
IsBiggerOrEqual(4,3) # TRUE
IsBiggerOrEqual(3,4) # FALSE
IsBiggerOrEqual(0.58 - 0.08,0.5) # TRUE
If all.equal is not processed, we may encounter errors.
The following is not necessary but useful:
abs(-2-(-2.2)) # 0.2
sprintf("%.54f",abs(-2-(-2.2))) # "0.200000000000000177635683940025046467781066894531250000"
sprintf("%.54f",0.2) # "0.200000000000000011102230246251565404236316680908203125"
all.equal(abs(-2-(-2.2)), 0.2) # TRUE; check nearly equivalence of floating point numbers
identical(abs(-2-(-2.2)), 0.2) # FALSE; check exact equivalence of floating point numbers