Thanks, Peter. You are correct in that I was incorrectly interpreting
the purpose of stripplot and as a result, my expectations were
invalid. Thanks for getting me back on track and your promptness in
responding.
@Pascal Thanks for the feedback. However, FTR, specifying the package
by using lattice::stripplot produced the same result. This is not
surprising given Peter's response.
Finally, and also FTR, I have upgraded my R installation to 2.15.1
and lattice package to 0.18-3 with the same results. Again, not
surprising given that I was misinterpreting what to expect. Bottom
line is that in all cases, R 2.10.1, specifying the package, and R
2.15.1 the function is and was performing as it should have been.
Operator error.
Best,
[hidden email]
On Fri, Jun 29, 2012 at 7:47 AM, Peter Ehlers [via R]
<
[hidden email]> wrote:
> On 2012-06-28 21:06, meonline wrote:
>
>> For the following example,
>>
>>> library(lattice)
>>> df<-data.frame(i=1:100,p=runif(100),id=rep(c('a','b'),100))
>>> summary(df[,'p'])
>> Min. 1st Qu. Median Mean 3rd Qu. Max.
>> 0.01165 0.33580 0.57520 0.53290 0.74540 0.98610
>>> stripplot(p~i|id,df)
>>
>> The plot that is output is as expected with the exception that the values
>> are scaled by a factor of 100 in the plot (and the yaxis has too many
>> labels) -- see attached screenshot.
>>
>> I'm running R.2.10 under Ubuntu 10.04.
>>
>> Why does this happen and how do I get the stripplot to plot the values as
>> they are in the data (from 0 to 1)?
>
> You need to understand what a stripplot is for; as the
> help page says: "stripplot produces one-dimensional
> scatterplots".
>
> It's usually used to compare several such one-dim plots
> for different levels of a factor variable; see the
> example on the help page or the examples on the help
> page for stripchart which does the same thing using
> traditional graphics. Using your data, try:
>
> stripplot(p ~ id, data = df)
>
> With stripplot(p ~ i | id, df), lattice coerces 'p' to
> a factor, assigning levels '1', '2', ..., '100' since
> you have 100 different values. It then plots 100
> strips _horizontally_ (you can't tell since each
> plot has only a single value). You get 100 labels on
> the y-axis. No scaling has taken place.
>
> The measured variable is taken to be 'i'. Its values
> are also 1-100 and that's what you see on the x-axis.
>
> Try adding 'horizontal = FALSE' to the command:
>
> stripplot(p ~ i | id, data = df, horizontal = FALSE)
>
> This will give you 100 _vertical_ strips, using 'i' as
> the factor variable. The measured variable is then 'p'
> and you'll see the values you were expecting.
>
> But I suspect that you never wanted a stripplot at all.
>
> Peter Ehlers
>
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