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I'm confusing about width(height) argument in geom_jitter() #1775

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kimchpekr opened this issue Sep 24, 2016 · 6 comments
Closed

I'm confusing about width(height) argument in geom_jitter() #1775

kimchpekr opened this issue Sep 24, 2016 · 6 comments

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@kimchpekr
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@kimchpekr kimchpekr commented Sep 24, 2016

"width(height)" argument in geom_jitter() is described as belows.

width

Amount of vertical and horizontal jitter. The jitter is added in both positive and negative directions, so the total spread is twice the value specified here. If omitted, defaults to 40% of the resolution of the data: this means the jitter values will occupy 80% of the implied bins.

So, I expected geom_jitter() and geom_jitter(width = 0.4) will produce same plot and "width = 0.4" spreads points across a range "base ± resolution * 0.4". but the results are not.

> p <- ggplot(mtcars, aes(gear, mpg))
> p + geom_jitter()  ## jitter range is 3 ± 0.4

default

> p + geom_jitter(width = 0.4) ## jitter range is narrow than 3 ± 0.4

width 0 4

I'm confusing about width argument and how can i calculate a jitter range.

I expect like belows but the result doesn't

geom_jitter(width = 0.2) will spread points across a range [2.8 ~ 3.2] : 3 ± 0.2
geom_jitter(width = 0.4) will spread points across a range [2.6 ~ 3.4] : 3 ± 0.4
geom_jitter(width = 0.6) will spread points across a range [2.4 ~ 3.6] : 3 ± 0.6

but, geom_point(aes(x = jitter(gear, amount = 0.2)) produce a plot what i expected.
geom_point jitter

Please, give me some hints...

@has2k1
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@has2k1 has2k1 commented Sep 24, 2016

That is clearly a bug. I think the width is being treated as a substitute for the resolution, if that is the case then the effective amount is 0.4 * 0.4 = 0.16

@kimchpekr
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@kimchpekr kimchpekr commented Sep 25, 2016

@has2k1 Thanks for comment.

data resolution at "gear = 3" is 1, so amount should be calculated 1 * 0.4 = 0.4 instead of 0.4 * 0.4 = 0.16

> resolution(mtcars$gear[mtcars$gear == 3])
[1] 1

but you right.
geom_jitter(width = 0.4) spreads points across a range you calculated 3±0.16 like below.
width 0 16

I'm still confusing....

@has2k1
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@has2k1 has2k1 commented Sep 25, 2016

I've since looked at the code and I'll submit a PR that fixes it.

@kimchpekr
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@kimchpekr kimchpekr commented Sep 25, 2016

@has2k1
I'm looking forward to it.

has2k1 added a commit to has2k1/ggplot2 that referenced this issue Sep 25, 2016
The problem was the %||% operator accepts single tokens and so
care (bracketing as required) must be taken avoid miscalculations.

Some other places in the code-base had it right.

Fixes tidyverse#1775
@hadley hadley closed this in #1777 Sep 27, 2016
hadley added a commit that referenced this issue Sep 27, 2016
The problem was the %||% operator accepts single tokens and so
care (bracketing as required) must be taken avoid miscalculations.

Some other places in the code-base had it right.

Fixes #1775
@svenhalvorson
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@svenhalvorson svenhalvorson commented Apr 13, 2017

I'm having a hard time knowing what this means:
'If omitted, defaults to 40% of the resolution of the data: this means the jitter values will occupy 80% of the implied bins.'

You're saying that the range of values that point could be mapped to is 3 +/- (resolution) * 0.4 so how is resolution determined?

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3 participants