The behavior of match function

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