Re: [R] sciplot question
On May 26, 2009, at 4:37 , Frank E Harrell Jr wrote: Manuel Morales wrote: On Mon, 2009-05-25 at 06:22 -0500, Frank E Harrell Jr wrote: Jarle Bjørgeengen wrote: On May 24, 2009, at 4:42 , Frank E Harrell Jr wrote: Jarle Bjørgeengen wrote: On May 24, 2009, at 3:34 , Frank E Harrell Jr wrote: Jarle Bjørgeengen wrote: Great, thanks Manuel. Just for curiosity, any particular reason you chose standard error , and not confidence interval as the default (the naming of the plotting functions associates closer to the confidence interval ) error indication . - Jarle Bjørgeengen On May 24, 2009, at 3:02 , Manuel Morales wrote: You define your own function for the confidence intervals. The function needs to return the two values representing the upper and lower CI values. So: qt.fun - function(x) qt(p=.975,df=length(x)-1)*sd(x)/ sqrt(length(x)) my.ci - function(x) c(mean(x)-qt.fun(x), mean(x)+qt.fun(x)) Minor improvement: mean(x) + qt.fun(x)*c(-1,1) but in general confidence limits should be asymmetric (a la bootstrap). Thanks, if the date is normally distributed , symmetric confidence interval should be ok , right ? Yes; I do see a normal distribution about once every 10 years. Is it not true that the students-T (qt(... and so on) confidence intervals is quite robust against non-normality too ? A teacher told me that, the students-T symmetric confidence intervals will give a adequate picture of the variability of the data in this particular case. Incorrect. Try running some simulations on highly skewed data. You will find situations where the confidence coverage is not very close of the stated level (e.g., 0.95) and more situations where the overall coverage is 0.95 because one tail area is near 0 and the other is near 0.05. The larger the sample size, the more skewness has to be present to cause this problem. OK - I'm convinced. It turns out that the first change I made to sciplot was to allow for asymmetric error bars. Is there an easy way (i.e., existing package) to bootstrap confidence intervals in R. If so, I'll try to incorporate this as an option in sciplot. library(Hmisc) ?smean.cl.boot H(arrel)misc :-) Thanks for valuable input Frank. This seems to work fine. (slightly more time consuming , but what do we have CPU power for ) library(Hmisc) library(sciplot) my.ci - function(x) c(smean.cl.boot(x)[2],smean.cl.boot(x)[3]) lineplot .CI (V1 ,V2 ,data = d ,col = c (4 ),err .col = c (1 ),err .width = 0.02 ,legend=FALSE,xlab=Timeofday,ylab=IOPS,ci.fun=my.ci,cex=0.5,lwd=0.7) Have I understood you correct in that this is a more accurate way of visualizing variability in any dataset , than the students T confidence intervals, because it does not assume normality ? Can you explain the meaning of B, and how to find a sensible value (if not the default is sufficient) ? Best regards Jarle Bjørgeengen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] moving from Windows to Linux - need help
Hi Robert,
I agree with Robert.
It seems the cleanest way of getting R on Linux is to use one of the
Linux distros R is packaged for.
It may work otherwise, but there _will_ be more fiddling around
resolving dependencies manually or semi manually.
Best regards.
- Jarle Bjørgeengen
On May 26, 2009, at 12:56 , Paul Hiemstra wrote:
Hi Robert,
I had the exact same problem on my eeepc 900. I replaced the xandros-
like linux in this way:
- Download an Ubuntu iso file (I use 8.04, Kubuntu)
- Put the .iso file on a usb stick (use unetbootin)
- Install the ubuntu version
- Install the eeepc specific stuff from http://array.org/ubuntu/
(this is a repository with an eeepc kernel available and other
stuff, the site provides a lot of info on how to install the eeepc
specific things)
Now you have a normal linux distro (ubuntu) and you can use the
normal cran repositories (debian) to install R.
This worked very well for me, it was quite easy to get ubuntu
running. I know that this isn't an exact answer to your question,
but I found that re installing linux was the best option.
cheers and hth,
Paul
Robert Kinley wrote:
hi
I've used R for many years on windows machines, but
have now acquired an Asus eee 1000 linux machine.
In order to get the best out of the machine, I used the
'pimpmyeee.sh' script, to get the full KDE desktop.
The version of Linux is Xandros, which I believe is
a close relative of Debian, but sadly I have only a nodding
acquaintance with Linux at present.
Naturally I want to have the current version of R on it,
and I understand (or possibly misunderstand) that the binary for
the Debian flavour of Linux should do the trick.
I have tried -
1. using synaptic to add the appropriate (I think) CRAN
repository ... but every combination I have tried
gives a 404 error
2. downloading from CRAN what I think is a zipped-up version of
r-base software, and thewn using the eee's file-manager
'install DEB package' option ... but this returns 'cannot load ... '.
I'm a bit stuck ... can anyone help please ?
thanks Bob Kinley
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Re: [R] moving from Windows to Linux - need help
On May 26, 2009, at 1:13 , Jarle Bjørgeengen wrote: Hi Robert, I agree with Robert. You meant you agree with Paul, right ? Of course :-) BR Jarle Bjørgeengen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sciplot question
On May 26, 2009, at 3:02 , Frank E Harrell Jr wrote: Jarle Bjørgeengen wrote: On May 26, 2009, at 4:37 , Frank E Harrell Jr wrote: Manuel Morales wrote: On Mon, 2009-05-25 at 06:22 -0500, Frank E Harrell Jr wrote: Jarle Bjørgeengen wrote: On May 24, 2009, at 4:42 , Frank E Harrell Jr wrote: Jarle Bjørgeengen wrote: On May 24, 2009, at 3:34 , Frank E Harrell Jr wrote: Jarle Bjørgeengen wrote: Great, thanks Manuel. Just for curiosity, any particular reason you chose standard error , and not confidence interval as the default (the naming of the plotting functions associates closer to the confidence interval ) error indication . - Jarle Bjørgeengen On May 24, 2009, at 3:02 , Manuel Morales wrote: You define your own function for the confidence intervals. The function needs to return the two values representing the upper and lower CI values. So: qt.fun - function(x) qt(p=.975,df=length(x)-1)*sd(x)/ sqrt(length(x)) my.ci - function(x) c(mean(x)-qt.fun(x), mean(x)+qt.fun(x)) Minor improvement: mean(x) + qt.fun(x)*c(-1,1) but in general confidence limits should be asymmetric (a la bootstrap). Thanks, if the date is normally distributed , symmetric confidence interval should be ok , right ? Yes; I do see a normal distribution about once every 10 years. Is it not true that the students-T (qt(... and so on) confidence intervals is quite robust against non-normality too ? A teacher told me that, the students-T symmetric confidence intervals will give a adequate picture of the variability of the data in this particular case. Incorrect. Try running some simulations on highly skewed data. You will find situations where the confidence coverage is not very close of the stated level (e.g., 0.95) and more situations where the overall coverage is 0.95 because one tail area is near 0 and the other is near 0.05. The larger the sample size, the more skewness has to be present to cause this problem. OK - I'm convinced. It turns out that the first change I made to sciplot was to allow for asymmetric error bars. Is there an easy way (i.e., existing package) to bootstrap confidence intervals in R. If so, I'll try to incorporate this as an option in sciplot. library(Hmisc) ?smean.cl.boot H(arrel)misc :-) Thanks for valuable input Frank. This seems to work fine. (slightly more time consuming , but what do we have CPU power for ) library(Hmisc) library(sciplot) my.ci - function(x) c(smean.cl.boot(x)[2],smean.cl.boot(x)[3]) Don't double the executing time by running it twice! And this way you might possibly get an upper confidence interval that is lower than the lower one. Do function(x) smean.cl.boot(x)[-1] D'oh lineplot .CI (V1 ,V2 ,data = d ,col = c (4 ),err .col = c (1 ),err .width = 0.02 ,legend =FALSE,xlab=Timeofday,ylab=IOPS,ci.fun=my.ci,cex=0.5,lwd=0.7) Have I understood you correct in that this is a more accurate way of visualizing variability in any dataset , than the students T confidence intervals, because it does not assume normality ? Yes but instead of saying variability (which quantiles are good at) we are talking about the precision of the mean. Right. Can you explain the meaning of B, and how to find a sensible value (if not the default is sufficient) ? For most purposes the default is sufficient. There are great books and papers on the bootstrap for more info, including improved variations on the simple bootstrap percentile confidence interval used here. Frank Once again, thanks. Best regards - Jarle Bjørgeengen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sciplot question
On May 24, 2009, at 4:42 , Frank E Harrell Jr wrote: Jarle Bjørgeengen wrote: On May 24, 2009, at 3:34 , Frank E Harrell Jr wrote: Jarle Bjørgeengen wrote: Great, thanks Manuel. Just for curiosity, any particular reason you chose standard error , and not confidence interval as the default (the naming of the plotting functions associates closer to the confidence interval ) error indication . - Jarle Bjørgeengen On May 24, 2009, at 3:02 , Manuel Morales wrote: You define your own function for the confidence intervals. The function needs to return the two values representing the upper and lower CI values. So: qt.fun - function(x) qt(p=.975,df=length(x)-1)*sd(x)/ sqrt(length(x)) my.ci - function(x) c(mean(x)-qt.fun(x), mean(x)+qt.fun(x)) Minor improvement: mean(x) + qt.fun(x)*c(-1,1) but in general confidence limits should be asymmetric (a la bootstrap). Thanks, if the date is normally distributed , symmetric confidence interval should be ok , right ? Yes; I do see a normal distribution about once every 10 years. Is it not true that the students-T (qt(... and so on) confidence intervals is quite robust against non-normality too ? A teacher told me that, the students-T symmetric confidence intervals will give a adequate picture of the variability of the data in this particular case. Best rgds Jarle Bjørgeengen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sciplot question
Great, thanks Manuel. Just for curiosity, any particular reason you chose standard error , and not confidence interval as the default (the naming of the plotting functions associates closer to the confidence interval ) error indication . - Jarle Bjørgeengen On May 24, 2009, at 3:02 , Manuel Morales wrote: You define your own function for the confidence intervals. The function needs to return the two values representing the upper and lower CI values. So: qt.fun - function(x) qt(p=.975,df=length(x)-1)*sd(x)/sqrt(length(x)) my.ci - function(x) c(mean(x)-qt.fun(x), mean(x)+qt.fun(x)) lineplot.CI(x.factor = dose, response = len, data = ToothGrowth, ci.fun=my.ci) Manuel On Fri, 2009-05-22 at 18:38 +0200, Jarle Bjørgeengen wrote: Hi, I would like to have lineplot.CI and barplot.CI to actually plot confidence intervals , instead of standard error. I understand I have to use the ci.fun option, but I'm not quite sure how. Like this : qt(0.975,df=n-1)*s/sqrt(n) but how can I apply it to visualize the length of the student's T confidence intervals rather than the stdandard error of the plotted means ? -- http://mutualism.williams.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sciplot question
On May 24, 2009, at 3:34 , Frank E Harrell Jr wrote: Jarle Bjørgeengen wrote: Great, thanks Manuel. Just for curiosity, any particular reason you chose standard error , and not confidence interval as the default (the naming of the plotting functions associates closer to the confidence interval ) error indication . - Jarle Bjørgeengen On May 24, 2009, at 3:02 , Manuel Morales wrote: You define your own function for the confidence intervals. The function needs to return the two values representing the upper and lower CI values. So: qt.fun - function(x) qt(p=.975,df=length(x)-1)*sd(x)/ sqrt(length(x)) my.ci - function(x) c(mean(x)-qt.fun(x), mean(x)+qt.fun(x)) Minor improvement: mean(x) + qt.fun(x)*c(-1,1) but in general confidence limits should be asymmetric (a la bootstrap). Thanks, if the date is normally distributed , symmetric confidence interval should be ok , right ? When plotting the individual sample , it looks normally distributed. Best regards. Jarle Bjørgeengen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] sciplot question
Hi,
I would like to have lineplot.CI and barplot.CI to actually plot
confidence intervals , instead of standard error.
I understand I have to use the ci.fun option, but I'm not quite sure
how.
Like this :
qt(0.975,df=n-1)*s/sqrt(n)
but how can I apply it to visualize the length of the student's T
confidence intervals rather than the stdandard error of the plotted
means ?
--
~~
Best regards .~.
Jarle Bjørgeengen /V\
Mob: +47 9155 7978// \\
http://www.uio.no/sok?person=jb /( )\
while(){if(s/^(.*\?)$/42 !/){print $1 $_}}^`~'^
~~
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
