jitter-bug? problematic behaviour of the jitter function

Dear Duncan, Dear Rui,

thanks for the responses and for pointing out that it is the ?fuzz? part
that is causing the problem. I agree that this is not a bug, but could be
undesirable/surprising behaviour, since it causes a large ?discontinuity?
in the jitter functions output depending on the input data.

I was (ab?)using the jitter function to break ties, where the desired
behaviour would be to add noise just small enough to make all values
unique. (Such a function can easily be hand coded of course.)

best regards,
Martin

Am 23.09.2020 um 22:25 schrieb Duncan Murdoch <murdoch.duncan at gmail.com
<mailto:murdoch.duncan at gmail.com>>:

On 23/09/2020 4:03 p.m., Rui Barradas wrote:
Hello,
I believe that though Duncan's explanation is right it is also not
explaining the value of the digits argument. round makes the first 2
numbers 0 but why?

If there had been rounding in their computation, you might see a
difference like 1e-15.  You wouldn't want to use that for the scale of
jittering, so some rounding is needed.

I think the documentation for the function is poor, but the intention was
probably to use the function in graphics (as the references did), and in
that case, any values too close together should be treated as equal and
jittering should separate them.  The particular computation used says that
if the range is in [1, 10), values equal to 3 decimal places will be too
close and need separation.

So I don't think this is a bug, but it might be a valid wishlist item:
document what "apart from fuzz" means, and perhaps allow it to be
controlled by the user.

Duncan Murdoch



The function below prints the digits argument and
then outputs d. The code is taken from jitter.
f <- function(x){
   z <- diff(r <- range(x[is.finite(x)]))
   cat("digits:", 3 - floor(log10(z)), "\n")
   diff(xx <- unique(sort.int(round(x, 3 - floor(log10(z))))))
}
Now see what cat outputs for 'digits'.
f(c(1,2,10^4))  # desired behaviour
#digits: 0
#[1]    1 9998
f(c(0,1,10^4))  # bad behaviour
#digits: -1
#[1] 10000
f(c(-1,0,10^4))  # bad behaviour
#digits: -1
#[1] 10000
f(c(1,2,10^5))  # bad behaviour
#digits: -1
#[1] 1e+05
And according to the documentation of ?round, negative digits are allowed:
Rounding to a negative number of digits means rounding to a power of
ten, so for example round(x, digits = -2) rounds to the nearest hundred.
But in this case two of the numbers are closer to 0 than they are of 10.
And unique keeps only 0 and the largest, then diff is big.
round(c(1,2,10^4),0)  # desired behaviour
#[1]     1     2 10000
round(c(0,1,10^4),-1)  # bad behaviour
#[1]     0     0 10000
round(c(-1,0,10^4),-1)  # bad behaviour
#[1]     0     0 10000
round(c(1,2,10^5),-1)  # bad behaviour
#[1] 0e+00 0e+00 1e+05
Isn't it still a bug?
Rui Barradas
?s 15:57 de 23/09/20, Duncan Murdoch escreveu:
On 23/09/2020 6:32 a.m., Martin Keller-Ressel wrote:
Dear all,

i have noticed some strange behaviour in the ?jitter? function in R.
On the help page for jitter it is stated that

"The result, say r, is r <- x + runif(n, -a, a) where n <- length(x)
and a is the amount argument (if specified).?

and

"If amount is NULL (default), we set a <- factor * d/5 where d is the
smallest difference between adjacent unique (apart from fuzz) x values.?

This works fine as long as there is no (very) large outlier

jitter(c(1,2,10^4))  # desired behaviour
[1]    1.083243    1.851571 9999.942716

But for very large outliers the added noise suddenly ?jumps? to a much
larger scale:

jitter(c(1,2,10^5)) # bad behaviour
[1] -19535.649   9578.702 115693.854
# Noise should be of order (2-1)/5  = 0.2 but is of much larger order.

This probably does not matter much when jitter is used for plotting,
but it can cause problems when jitter is used to break ties.

I think this is kind of documented:  "apart from fuzz" is what counts.
If you look at the code for jitter, you'll see this important line:

  d <- diff(xx <- unique(sort.int(round(x, 3 - floor(log10(z))))))

By the time you get here, z is the length of the rante of the data, so
it's 99999 in your example.  The rounding changes your values to
0,0,1e5, so the smallest difference is 1e5.

Duncan Murdoch