R：使用带有两个以上对象的all.equal

Question

When I compare more than two objects with identical the results make sense, but I do not understand what all.equal is doing. What is the logic behind the results that all.equal is outputting?

identical(25 + 2, 27)  #TRUE -- makes sense
identical(25, 2, 27)   #FALSE -- makes sense
identical(27, 22, 27)  #FALSE -- makes sense

all.equal(25 + 2, 27)  #TRUE -- makes sense
all.equal(25, 2, 27)   #TRUE -- why?
all.equal(27, 22, 27)  #TRUE -- why?

If either one of the last two all.equal() tests were false I would understand what R is doing -- either testing whether any sum of objects is equal (25+2==27) or testing whether any two objects are equal (27==27).

What is all.equal testing such that both of the last two equations are true?

Answer 1

The comments above appear to be correct. The arguments are taken in order, and the third argument is designated as the tolerance. However, the following results appear even stranger.

all.equal(target=25, current=2, 5) 
[1] TRUE

and

all.equal(target=25, current=2, tolerance=5) 
[1] TRUE

Here, the tolerance is clearly smaller than the difference, but TRUE is still returned. If we reduce tolerance to be even smaller, we get

all.equal(target=25, current=2, tolerance=.05) 
[1] "Mean relative difference: 0.92"

What is going on is that relative differences are calculated based on the mean value of the target argument. From the help file,

Numerical comparisons for scale = NULL (the default) are typically on relative difference scale unless the target values are close to zero: First, the mean absolute difference of the two numerical vectors is computed. If this is smaller than tolerance or not finite, absolute differences are used, otherwise relative differences scaled by the mean absolute target value . (Note that these comparisons are computed only for those vector elements where target is not NA and differs from current.)

So, reversing target and current arguments, we get

all.equal(target=2, current=25, tolerance=5) 
[1] "Mean absolute difference: 23"

which is what we might have expected.

Note that we could also get this by setting the scale argument to 1:

all.equal(target=25, current=2, tolerance=5, scale=1) 
[1] "Mean absolute difference: 23"

or only naming scale,

all.equal(25, 2, 5, scale=1) 
[1] "Mean absolute difference: 23"

which re-affirms the comments in the OP.