jitter-bug? problematic behaviour of the jitter function

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.

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.

best regards,
Martin

--------------------------------
Martin Keller-Ressel
Professor f?r Stochastische Analysis und Finanzmathematik
Technische Universit?t Dresden
Institut f?r Mathematische Stochastik
Willersbau B 316, Zellescher Weg 12-14
01062 Dresden
--------------------------------


	[[alternative HTML version deleted]]

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

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?

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

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

On Sep 24, 2020, at 8:08 AM, Martin Keller-Ressel <martin.keller-ressel at 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 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

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

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 at tu-dresden.de 
<mailto:martin.keller-ressel at 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 at gmail.com
    <mailto:murdoch.duncan at gmail.com><mailto: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 <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
    ?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 <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