On Fri, 2005-10-21 at 11:19 +0800, ronggui wrote:> > x<-1:10
> > y<-x+1e-20
> > x
> [1] 1 2 3 4 5 6 7 8 9 10
> > y
> [1] 1 2 3 4 5 6 7 8 9 10
> > identical(x,y)
> [1] FALSE
> > match(x,y)
> [1] 1 2 3 4 5 6 7 8 9 10
>
> What's the principle the function use to determine if x match y?
>
> Thank you!
In this case, you are comparing x (an integer) with y (a numeric):
> x <- 1:10
> y <- x + 1e-20
> class(x)
[1] "integer"> class(y)
[1] "numeric"
Now:
> x == y
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
works element-wise, because the differences between the values (1e-20)
are less than:
> .Machine$double.eps
[1] 2.220446e-16
which is the smallest positive float such that 1 plus that value != 1.
See ?.Machine for more information on that.
For the same reason:
> match(x, y)
[1] 1 2 3 4 5 6 7 8 9 10
> x %in% y
[1] TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
both work element-wise.
However, if you used the following for 'y':
> y <- x + 1e-15
Note the results now:
> x == y
[1] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
because you are now have differences that are greater than .Machine
$double.eps.
In general however, when comparing floats, you will want to use
all.equal():
> all.equal(x, y)
[1] TRUE
which compares the values within a specified level of tolerance.
See ?all.equal for more information and importantly note the use of
isTRUE() as well:
> isTRUE(all.equal(x, y))
[1] TRUE
Using isTRUE() in this way will result in a single TRUE or FALSE result
depending upon the comparison. If the differences happen to be outside
the tolerance level, you get something like the following:
> y <- x + 1e-5
> all.equal(x, y)
[1] "Mean relative difference: 1.818182e-06"
which does not help if all you want is a single boolean result. Thus the
use of isTRUE() helps here:
> isTRUE(all.equal(x, y))
[1] FALSE
You should also read R FAQ 7.31 "Why doesn't R think these numbers are
equal?".
HTH,
Marc Schwartz