[R] jitter-bug? problematic behaviour of the jitter function
Duncan Murdoch
murdoch@dunc@n @end|ng |rom gm@||@com
Thu Sep 24 19:24:20 CEST 2020
Those seem like useful properties if jitter() is used in plotting (as it
was originally intended), but that use isn't even mentioned in the help
page. Martin wanted to "add a small amount of noise to a numeric
vector" "in order to break ties" (quoting from that help page).
For Martin's use, it sounds as though quantreg::dither might be a better
solution (though I think it won't work when numerical error splits ties,
so some differences are extremely small, if the scale of the values
varies too much, but I'd guess that's a fairly rare circumstance).
Duncan Murdoch
On 24/09/2020 1:03 p.m., Bert Gunter wrote:
> Folks: Please note:
>
> There is *no* way to "jitter" the 3 values 1,2, and 1e5 so that:
>
> a) the jittered values differ from the original ones by a fraction of
> their original value;
> b) the plotting symbols for the jittered values will be distinguishable
> on a linear scale holding all 3 values.
>
> Cheers,
> Bert
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Thu, Sep 24, 2020 at 8:39 AM Martin Keller-Ressel
> <martin.keller-ressel using tu-dresden.de
> <mailto:martin.keller-ressel using tu-dresden.de>> wrote:
>
> 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 using gmail.com
> <mailto:murdoch.duncan using gmail.com><mailto:murdoch.duncan using gmail.com
> <mailto:murdoch.duncan using 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 <http://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
> Às 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 <http://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
>
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