How does the tolerance parameter of all.equal() work?

r

Solution

If `target` is greater than `tolerance`, it seems to check for `relative error <= tolerance`. That is, `abs(current-target)/target <= tolerance` in:

all.equal(target, current, tolerance)

For ex:

all.equal(3, 6, tolerance = 1)
# TRUE --> abs(6-3)/3 <= 1

Instead, if `target` is smaller than `tolerance`, `all.equal` uses `mean absolute difference`.

all.equal(0.01, 4, tolerance = 0.01)
# [1] "Mean absolute difference: 3.99"

all.equal(0.01, 4, tolerance = 0.00999)
# [1] "Mean relative difference: 399"

all.equal(4, 0.01, tolerance = 0.01)
# [1] "Mean relative difference: 0.9975"

However, this is not what the documentation states. To look further as to why this is happening, let's look at the relevant snippet from `all.equal.numeric`:

# take the example: all.equal(target=0.01, current=4, tolerance=0.01)
cplx <- is.complex(target) # FALSE
out <- is.na(target) # FALSE
out <- out | target == current # FALSE

target <- target[!out] # = target (0.01)
current <- current[!out] # = current (4)

xy <- mean((if(cplx) Mod else abs)(target - current)) # else part is run = 3.99

# scale is by default NULL
what <- if (is.null(scale)) {
    xn <- mean(abs(target)) # 0.01
    if (is.finite(xn) && xn > tolerance) { # No, xn = tolerance
        xy <- xy/xn
        "relative"
    }
    else "absolute" # this is computed for this example
}
else {
    xy <- xy/scale
    "scaled"
}

All that is being checked in the code above (shown only the necessary parts for the example from OP) is: to remove any NA and equal values (of `target` and `current`) from `target` and `current`. Then compute `xy` as the mean absolute difference of `target` and `current`. But deciding if it is going to be `relative` or `absolute` depends on the part `what`. And here `xy` is not checked for any conditions. It depends only on `xn` which is `mean(abs(target))`.

So, in conclusion, the part pasted by the OP (pasted here for convenience):

If this (meaning, mean absolute difference) is smaller than tolerance or not finite, absolute differences are used, otherwise relative differences scaled by the mean absolute difference.

seems wrong/misleading.

Problem

Original source