Re: [R] help change auto.key colors in xyplot
On Fri, Feb 13, 2009 at 11:22 AM, Dieter Menne
dieter.me...@menne-biomed.de wrote:
Dimitri Liakhovitski ld7631 at gmail.com writes:
the code below works just fine to produce a dotplot. However, I am not
successful changing the color of the lines in the legend (auto.key).
If I add col=..., it only changes the color of the letters in the
legend, but not the color of the lines.
I prefer to set pch and friends outside the panel function to
avoid clutter, and to set the parameters globally, thus forcing
my plots to be similar. This is a matter of taste, though.
On Windows, the key looks a bit ugly now.
Group 1 o
Group
2 o-
[That's because the code got wrapped around, breaking Group 2.]
I am not very happy that the lwd is not honored by the key.
Lines in lwd 2 (plot) and in lwd 1 (default key) do have a quite
different subjective color hue. Any way around this, Deepayan?
For the typical situation where you want the same colors for lines and
points, I find the
par.settings = simpleTheme(col=..., lwd=...)
construct very useful. Also note the slightly different key:
dotplot(c(d[[1]],d[[3]])~rep(d[[2]],2),
groups=rep(c(Group 1,Group 2), each=nrow(d)),
main=list(Chart Title,cex=1),
type=b,
par.settings = simpleTheme(col=c(blue,red),
pch=20, cex=1.3, lwd=2),
auto.key = list(space = top, points = FALSE,
lines = TRUE, type = b),
xlab=list(Title for X,cex=.9,font=2),
ylab=list(Title for Y,cex=.9,font=2),
panel = function(y,x,...) {
panel.grid(h = -1, v = -1,
col = gray, lty =dotted)
panel.superpose(x,y,... )
ltext(x, y, labels=round(y,3),cex=.8,col=black,font=2,
adj=c(-0.2,1))
})
For different colors for lines and points, you do need to be more
verbose, and change the line and symbol settings separately:
trellis.par.set(superpose.line = list(col=c(blue,red), lwd = 2),
superpose.symbol = list(cex = 1.3, pch = 20),
reference.line = list(col = gray, lty =dotted))
dotplot(c(d[[1]],d[[3]])~rep(d[[2]],2),
groups=rep(c(Group 1,Group 2), each=nrow(d)),
main=list(Chart Title,cex=1),
type=b,
auto.key = list(space = top, points = TRUE, lines = TRUE),
xlab=list(Title for X,cex=.9,font=2),
ylab=list(Title for Y,cex=.9,font=2),
panel = function(y,x,...) {
panel.grid(h = -1, v = -1)
panel.xyplot(x, y, ...)
ltext(x, y, labels=round(y,3),
cex=.8,col=black,font=2,
adj=c(-0.2,1))
})
-Deepayan
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Re: [R] help change auto.key colors in xyplot
On Fri, Feb 13, 2009 at 1:22 PM, Dieter Menne dieter.me...@menne-biomed.de wrote: Dimitri Liakhovitski ld7631 at gmail.com writes: the code below works just fine to produce a dotplot. However, I am not successful changing the color of the lines in the legend (auto.key). If I add col=..., it only changes the color of the letters in the legend, but not the color of the lines. I prefer to set pch and friends outside the panel function to avoid clutter, and to set the parameters globally, thus forcing my plots to be similar. This is a matter of taste, though. On Windows, the key looks a bit ugly now. Group 1 o Group 2 o- I am not very happy that the lwd is not honored by the key. Lines in lwd 2 (plot) and in lwd 1 (default key) do have a quite different subjective color hue. Any way around this, Deepayan? (Besides using ggplot2, as Hadley would argue ??) And here is a ggplot2 solution ;) df - data.frame( x = rep(d[[2]], 2), y = c(d[[1]], d[[3]]), group = rep(c(Group 1, Group 2), each = 3) ) ggplot(df, aes(x, y, group = group)) + geom_point(aes(colour = group), size = 3) + geom_line(aes(colour = group), size = 1.5) + geom_text(aes(label = y, y = y + 0.02), vjust = 0, size = 4) The legends are pretty good at figuring out exactly what should be displayed, based on what you've used in the graphic. Hadley -- http://had.co.nz/ __ 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] Optimizing Multiple Models...any suggestions?
Hi Max, Thanks for the suggestion, that's exactly what I was looking for. Thanks again. Paul -- View this message in context: http://www.nabble.com/Optimizing-Multiple-Models...any-suggestions--tp21979556p22006494.html Sent from the R help mailing list archive at Nabble.com. __ 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] second axis title orientation
Hi, I have added a second y-axis via axis() to a plot. Then I tried to add an y axis title. But the axis() function seems to have no argument like ylab. and if I add a title with title() the text is centered at the first axis, not the second defined one. Is there a way to add an axis-title that orientates it's position at the last defined axis? __ 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] second axis title orientation
There is an example in: library(zoo) example(plot.zoo) See ?plot.zoo On Fri, Feb 13, 2009 at 6:19 PM, Jörg Groß jo...@licht-malerei.de wrote: Hi, I have added a second y-axis via axis() to a plot. Then I tried to add an y axis title. But the axis() function seems to have no argument like ylab. and if I add a title with title() the text is centered at the first axis, not the second defined one. Is there a way to add an axis-title that orientates it's position at the last defined axis? __ 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-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] PCA functions
Hi All, would appreciate an answer on this if you have a moment; Is there a function (before I try and write it !) that allows the input of a covariance or correlation matrix to calculate PCA, rather than the actual data as in princomp() Regards Glenn [[alternative HTML version deleted]] __ 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] question about tool collisions
Recently I got introduced to two packages: {seewave} and {audio} .
Turns out they both have a tool to call a system audio tool, and in both
cases the name of the tool is play(). Naturally these two tools do
slightly different things with different arguments.
So, what should a user do, and more to the point, what should R do? If
I've learned the right things about environments, I could include some
require() statements inside scripts to guarantee the right things are
loaded (and only loaded for that script).
But in general when there are two commands with the same name, it means
that if I distribute code which uses the command, a random user
elsewhere might have the other package loaded and bad things would happen.
In a perfect world, there would be some R control database that people
would check packages into. The database (or some poor volunteer) would
verify there were no naming conflicts with other posted packages, and so
on. Failing that, what are good rules for fail-safing code, or
figuring out how to check that some package I've loaded didn't blow away
some other tools I previously loaded?
Carl
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Re: [R] second axis title orientation
?mtext On Sat, 2009-02-14 at 00:19 +0100, Jörg Groß wrote: Hi, I have added a second y-axis via axis() to a plot. Then I tried to add an y axis title. But the axis() function seems to have no argument like ylab. and if I add a title with title() the text is centered at the first axis, not the second defined one. Is there a way to add an axis-title that orientates it's position at the last defined axis? __ 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. -- Alexandre Swarowsky Soils and Biogeochemistry Graduate Group University of California at Davis One Shields Avenue Davis CA 95618 Office: (530)752-4131 cell: (530)574-3028 __ 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] question about tool collisions
See the warn.conflicts arguments of require() and library(). People
thought about this a long time ago.
-- Bert Gunter
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Carl Witthoft
Sent: Friday, February 13, 2009 4:05 PM
To: r-help@r-project.org
Subject: [R] question about tool collisions
Recently I got introduced to two packages: {seewave} and {audio} .
Turns out they both have a tool to call a system audio tool, and in both
cases the name of the tool is play(). Naturally these two tools do
slightly different things with different arguments.
So, what should a user do, and more to the point, what should R do? If
I've learned the right things about environments, I could include some
require() statements inside scripts to guarantee the right things are
loaded (and only loaded for that script).
But in general when there are two commands with the same name, it means
that if I distribute code which uses the command, a random user
elsewhere might have the other package loaded and bad things would happen.
In a perfect world, there would be some R control database that people
would check packages into. The database (or some poor volunteer) would
verify there were no naming conflicts with other posted packages, and so
on. Failing that, what are good rules for fail-safing code, or
figuring out how to check that some package I've loaded didn't blow away
some other tools I previously loaded?
Carl
__
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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.
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Re: [R] Bootstrap or Wilcoxons' test?
Charlotta, I'm not sure what you mean when you say simple linear regression. From your description you have two groups of people, for which you recorded contaminant concentration. Thus, I would think you would do something like a t-test to compare the mean concentration level. Where does the regression part come in? What are you regressing? As for the Wilcoxnin test, it is often thought of as a nonparametric t-test equivalent. This is only true if the observations were drawn, from a population with the same probability distribution. The null hypothesis of the Wilcoxin test is actually the observations were drawn, from the same probability distribution. Thus if your two samples had say different variances, there means could be the same, but since the variances are different, the Wilcoxin could give you a significant result. Don't know if this all makes sense, but if you have more questions, please e-mail your data and a more detailed description of what analysis you used and I'd be happy to try and help out. Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richs...@earthlink.net - Original Message - From: Charlotta Rylander z...@nilu.no To: r-help@r-project.org Sent: Friday, February 13, 2009 3:24 AM Subject: [R] Bootstrap or Wilcoxons' test? Hi! I'm comparing the differences in contaminant concentration between 2 different groups of people ( N=36, N=37). When using a simple linear regression model I found no differences between groups, but when evaluating the diagnostic plots of the residuals I found my independent variable to have deviations from normality (even after log transformation). Therefore I have used bootstrap on the regression parameters ( R= 1000 R=1) and this confirms my results , i.e., no differences between groups ( and the distribution is log-normal). However, when using wilcoxons' rank sum test on the same data set I find differences between groups. Should I trust the results from bootstrapping or from wilcoxons' test? Thanks! Regards Lotta Rylander [[alternative HTML version deleted]] __ 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-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] question about tool collisions
on 02/13/2009 06:05 PM Carl Witthoft wrote:
Recently I got introduced to two packages: {seewave} and {audio} .
Turns out they both have a tool to call a system audio tool, and in both
cases the name of the tool is play(). Naturally these two tools do
slightly different things with different arguments.
So, what should a user do, and more to the point, what should R do? If
I've learned the right things about environments, I could include some
require() statements inside scripts to guarantee the right things are
loaded (and only loaded for that script).
But in general when there are two commands with the same name, it means
that if I distribute code which uses the command, a random user
elsewhere might have the other package loaded and bad things would happen.
In a perfect world, there would be some R control database that people
would check packages into. The database (or some poor volunteer) would
verify there were no naming conflicts with other posted packages, and so
on. Failing that, what are good rules for fail-safing code, or
figuring out how to check that some package I've loaded didn't blow away
some other tools I previously loaded?
In addition to Bert's excellent recommendation (as well as ?conflicts),
this is one of the key reasons for using namespaces.
It would appear, from a review of the tarballs for both packages, that
audio is using a namespace, but seewave does not. You might make a
recommendation to the latter's author to implement a namespace for his
package. That would help avoid the collisions in third party code.
audio does export the play() function, so you can use:
audio::play(...)
to differentiate that function from:
seewave::play(...)
None of this is a guarantee of course, if other play() functions should
be loaded or come into being, which is why namespaces are a safer approach.
See also:
http://cran.r-project.org/doc/manuals/R-intro.html#Namespaces
HTH,
Marc Schwartz
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[R] Outlier Detection for timeseries
Hello R users, Can someone tell if there is a package in R that can do outlier detection that give outputs simiilar to what I got from SAS below. Many thanks in advance for any help! Outlier Details Approx Chi- Prob ObsTime ID Type Estimate Square ChiSq 12 12.00Additive 2792544.6 186.13 .0001 13 13.00Additive 954302.1 21.23 .0001 15 15.00Shift63539.3 9.060.0026 -- View this message in context: http://www.nabble.com/Outlier-Detection-for-timeseries-tp22008448p22008448.html Sent from the R help mailing list archive at Nabble.com. __ 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] Bootstrap or Wilcoxons' test?
I must disagree with both this general characterization of the Wilcoxon test and with the specific example offered. First, we ought to spell the author's correctly and then clarify that it is the Wilcoxon rank-sum test that is being considered. Next, the WRS test is a test for differences in the location parameter of independent samples conditional on the samples having been drawn from the same distribution. The WRS test would have no discriminatory power for samples drawn from the same distribution having equal location parameters but only different with respect to unequal dispersion. Look at the formula, for Pete's sake. It summarizes differences in ranking, so it is in fact designed NOT to be sensitive to the spread of the values in the sample. It would have no power, for instance, to test the variances of two samples, both with a mean of 0, and one having a variance of 1 with the other having a variance of 3. One can think of the WRS as a test for unequal medians. -- David Winsemius, MD. MPH Heritage Laboratories On Feb 13, 2009, at 7:48 PM, Murray Cooper wrote: Charlotta, I'm not sure what you mean when you say simple linear regression. From your description you have two groups of people, for which you recorded contaminant concentration. Thus, I would think you would do something like a t-test to compare the mean concentration level. Where does the regression part come in? What are you regressing? As for the Wilcoxnin test, it is often thought of as a nonparametric t-test equivalent. This is only true if the observations were drawn, from a population with the same probability distribution. The null hypothesis of the Wilcoxin test is actually the observations were drawn, from the same probability distribution. Thus if your two samples had say different variances, there means could be the same, but since the variances are different, the Wilcoxin could give you a significant result. Don't know if this all makes sense, but if you have more questions, please e-mail your data and a more detailed description of what analysis you used and I'd be happy to try and help out. Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richs...@earthlink.net - Original Message - From: Charlotta Rylander z...@nilu.no To: r-help@r-project.org Sent: Friday, February 13, 2009 3:24 AM Subject: [R] Bootstrap or Wilcoxons' test? Hi! I'm comparing the differences in contaminant concentration between 2 different groups of people ( N=36, N=37). When using a simple linear regression model I found no differences between groups, but when evaluating the diagnostic plots of the residuals I found my independent variable to have deviations from normality (even after log transformation). Therefore I have used bootstrap on the regression parameters ( R= 1000 R=1) and this confirms my results , i.e., no differences between groups ( and the distribution is log-normal). However, when using wilcoxons' rank sum test on the same data set I find differences between groups. Should I trust the results from bootstrapping or from wilcoxons' test? Thanks! Regards Lotta Rylander [[alternative HTML version deleted]] __ 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-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-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] Bootstrap or Wilcoxons' test?
First of all, sorry for my typing mistakes. Second, the WRS test is most certainly not a test for unequal medians. Although under specified models it would be. Just as under specified models it can be a test for other measures of location. Perhaps I did not word my explanation correctly, but I did not mean to imply that it would be a test of equality of variance. It is plain and simple a test for the equality of distributions. When the results of a properly applied parametric test do not agree with the WRS, it is usually do to a difference in the empirical density function of the two samples. Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richs...@earthlink.net - Original Message - From: David Winsemius dwinsem...@comcast.net To: Murray Cooper myrm...@earthlink.net Cc: Charlotta Rylander z...@nilu.no; r-help@r-project.org Sent: Friday, February 13, 2009 9:19 PM Subject: Re: [R] Bootstrap or Wilcoxons' test? I must disagree with both this general characterization of the Wilcoxon test and with the specific example offered. First, we ought to spell the author's correctly and then clarify that it is the Wilcoxon rank-sum test that is being considered. Next, the WRS test is a test for differences in the location parameter of independent samples conditional on the samples having been drawn from the same distribution. The WRS test would have no discriminatory power for samples drawn from the same distribution having equal location parameters but only different with respect to unequal dispersion. Look at the formula, for Pete's sake. It summarizes differences in ranking, so it is in fact designed NOT to be sensitive to the spread of the values in the sample. It would have no power, for instance, to test the variances of two samples, both with a mean of 0, and one having a variance of 1 with the other having a variance of 3. One can think of the WRS as a test for unequal medians. -- David Winsemius, MD. MPH Heritage Laboratories On Feb 13, 2009, at 7:48 PM, Murray Cooper wrote: Charlotta, I'm not sure what you mean when you say simple linear regression. From your description you have two groups of people, for which you recorded contaminant concentration. Thus, I would think you would do something like a t-test to compare the mean concentration level. Where does the regression part come in? What are you regressing? As for the Wilcoxnin test, it is often thought of as a nonparametric t-test equivalent. This is only true if the observations were drawn, from a population with the same probability distribution. The null hypothesis of the Wilcoxin test is actually the observations were drawn, from the same probability distribution. Thus if your two samples had say different variances, there means could be the same, but since the variances are different, the Wilcoxin could give you a significant result. Don't know if this all makes sense, but if you have more questions, please e-mail your data and a more detailed description of what analysis you used and I'd be happy to try and help out. Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richs...@earthlink.net - Original Message - From: Charlotta Rylander z...@nilu.no To: r-help@r-project.org Sent: Friday, February 13, 2009 3:24 AM Subject: [R] Bootstrap or Wilcoxons' test? Hi! I'm comparing the differences in contaminant concentration between 2 different groups of people ( N=36, N=37). When using a simple linear regression model I found no differences between groups, but when evaluating the diagnostic plots of the residuals I found my independent variable to have deviations from normality (even after log transformation). Therefore I have used bootstrap on the regression parameters ( R= 1000 R=1) and this confirms my results , i.e., no differences between groups ( and the distribution is log-normal). However, when using wilcoxons' rank sum test on the same data set I find differences between groups. Should I trust the results from bootstrapping or from wilcoxons' test? Thanks! Regards Lotta Rylander [[alternative HTML version deleted]] __ 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-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-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
[R] several ifelse problems...
Dear R users, From the code below, I try to compute y value. (In fact, y looks like a trapezoid) -- x - seq(0,1,.01) y - ifelse(abs(x-.5)=0.3,0, ifelse(abs(w-.5)=0.4,-1, ifelse((0.1w w0.2),10*x-2,-10*x+8))) -- So, results are... -- x [1] 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12 0.13 0.14 [16] 0.15 0.16 0.17 0.18 0.19 0.20 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29 [31] 0.30 0.31 0.32 0.33 0.34 0.35 0.36 0.37 0.38 0.39 0.40 0.41 0.42 0.43 0.44 [46] 0.45 0.46 0.47 0.48 0.49 0.50 0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 [61] 0.60 0.61 0.62 0.63 0.64 0.65 0.66 0.67 0.68 0.69 0.70 0.71 0.72 0.73 0.74 [76] 0.75 0.76 0.77 0.78 0.79 0.80 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 [91] 0.90 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99 1.00 y [1] -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 8 8 8 8 8 8 8 8 8 0 0 0 0 0 [26] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [51] 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 [76] 0 0 0 0 0 8 8 8 8 8 8 8 8 8 8 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 [101] -1 -- However, even though the results show that y=8 for x=0.11, when x=0.11, actual y value is -0.9. And, y=-0.8 for x=0.88. I cannot understand the above results. Any comments will be greatly appreciated. Kathryn Lord -- View this message in context: http://www.nabble.com/several-%22ifelse%22-problems...-tp22009321p22009321.html Sent from the R help mailing list archive at Nabble.com. __ 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] Bootstrap or Wilcoxons' test?
Hi Charlotta, to be more constructive toward your goal. If you bootstrap the regression when the regression is ill-specified, the bootstrap may not help you. Further, a test as difficult as a regression does not seem to be necessary in your case. A t-test if your dependent variable is (approxiamately) normal for both groups and if variances are equal or a Wilcoxon test if your dependent variable is not normal should do. The bootstrap should be very powerful if you do NOT perform it on the regression (again, bootstrapping the regression may just mean to do the wrong thing over and over again, which is no improvement). Just bootstrap sample means for the two groups and compare them appropriately (see: http://www.stat.berkeley.edu/users/rodwong/Stat131a/boot_diff_twomeans.pdf ). Otherwise, rely on the result of the Wilcoxon test as it is likely more appropriate if your dependent variable is not normal in the two groups. Daniel - cuncta stricte discussurus - -Ursprüngliche Nachricht- Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im Auftrag von David Winsemius Gesendet: Friday, February 13, 2009 9:19 PM An: Murray Cooper Cc: r-help@r-project.org Betreff: Re: [R] Bootstrap or Wilcoxons' test? I must disagree with both this general characterization of the Wilcoxon test and with the specific example offered. First, we ought to spell the author's correctly and then clarify that it is the Wilcoxon rank-sum test that is being considered. Next, the WRS test is a test for differences in the location parameter of independent samples conditional on the samples having been drawn from the same distribution. The WRS test would have no discriminatory power for samples drawn from the same distribution having equal location parameters but only different with respect to unequal dispersion. Look at the formula, for Pete's sake. It summarizes differences in ranking, so it is in fact designed NOT to be sensitive to the spread of the values in the sample. It would have no power, for instance, to test the variances of two samples, both with a mean of 0, and one having a variance of 1 with the other having a variance of 3. One can think of the WRS as a test for unequal medians. -- David Winsemius, MD. MPH Heritage Laboratories On Feb 13, 2009, at 7:48 PM, Murray Cooper wrote: Charlotta, I'm not sure what you mean when you say simple linear regression. From your description you have two groups of people, for which you recorded contaminant concentration. Thus, I would think you would do something like a t-test to compare the mean concentration level. Where does the regression part come in? What are you regressing? As for the Wilcoxnin test, it is often thought of as a nonparametric t-test equivalent. This is only true if the observations were drawn, from a population with the same probability distribution. The null hypothesis of the Wilcoxin test is actually the observations were drawn, from the same probability distribution. Thus if your two samples had say different variances, there means could be the same, but since the variances are different, the Wilcoxin could give you a significant result. Don't know if this all makes sense, but if you have more questions, please e-mail your data and a more detailed description of what analysis you used and I'd be happy to try and help out. Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richs...@earthlink.net - Original Message - From: Charlotta Rylander z...@nilu.no To: r-help@r-project.org Sent: Friday, February 13, 2009 3:24 AM Subject: [R] Bootstrap or Wilcoxons' test? Hi! I'm comparing the differences in contaminant concentration between 2 different groups of people ( N=36, N=37). When using a simple linear regression model I found no differences between groups, but when evaluating the diagnostic plots of the residuals I found my independent variable to have deviations from normality (even after log transformation). Therefore I have used bootstrap on the regression parameters ( R= 1000 R=1) and this confirms my results , i.e., no differences between groups ( and the distribution is log-normal). However, when using wilcoxons' rank sum test on the same data set I find differences between groups. Should I trust the results from bootstrapping or from wilcoxons' test? Thanks! Regards Lotta Rylander [[alternative HTML version deleted]] __ 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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting
Re: [R] Outlier Detection for timeseries
what are you looking for? I am not familiar with SAS as I am a poor scientist. I am not promising anything, but if you were to tell me what you wanted - method etc. I may know of a package or something that would work. Stephen On Fri, Feb 13, 2009 at 9:16 PM, Pele drdi...@yahoo.com wrote: Hello R users, Can someone tell if there is a package in R that can do outlier detection that give outputs simiilar to what I got from SAS below. Many thanks in advance for any help! Outlier Details Approx Chi- Prob ObsTime ID Type Estimate Square ChiSq 12 12.00Additive 2792544.6 186.13 .0001 13 13.00Additive 954302.1 21.23 .0001 15 15.00Shift63539.3 9.060.0026 -- View this message in context: http://www.nabble.com/Outlier-Detection-for-timeseries-tp22008448p22008448.html Sent from the R help mailing list archive at Nabble.com. __ 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. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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] superscript
Dear R-users. I'm struggeling to fix the superscript of a label of a figure axis. For some reason R doesn't recognize the hat symbol. plot(1,1,xlab=ligth intensity (PAR),ylab=expression(mass Pteridium rhizomes (gr/0.25m^2))) A very similiar scriptline does not give any problem at all: plot(1,1,xlab=expression(balsa plot basal area (m^2/ha)),ylab=light intensity (PAR)) Someone? _ s. It's easy! aspxmkt=en-us __ 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] superscript
Try: plot(1,1,xlab=ligth intensity (PAR),ylab=expression(mass Pteridium rhizomes (gr/0.25*m^2))) Daniel Moreira, MD Research Associate Duke University Medical Center DUMC 2626, MSRB-I Room 455 571 Research Drive Durham, North Carolina 27710 Telephone: (919) 681-7132 Fax: (919) 668-7093 E-mail: daniel.more...@duke.edu David Douterlungne daviddou...@hotmail.com Sent by: r-help-boun...@r-project.org 02/14/2009 12:21 AM To r-help@r-project.org cc Subject [R] superscript Dear R-users. I'm struggeling to fix the superscript of a label of a figure axis. For some reason R doesn't recognize the hat symbol. plot(1,1,xlab=ligth intensity (PAR),ylab=expression(mass Pteridium rhizomes (gr/0.25m^2))) A very similiar scriptline does not give any problem at all: plot(1,1,xlab=expression(balsa plot basal area (m^2/ha)),ylab=light intensity (PAR)) Someone? _ s. It's easy! aspxmkt=en-us __ 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. [[alternative HTML version deleted]] __ 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] several ifelse problems...
Dear Kathie, On Fri, 13 Feb 2009 21:08:25 -0800 (PST) kathie kathryn.lord2...@gmail.com wrote: Dear R users, [snip] However, even though the results show that y=8 for x=0.11, when x=0.11, actual y value is -0.9. And, y=-0.8 for x=0.88. I cannot understand the above results. It may help you to understand those results when you look at the difference between 0.1 x x 0.2 and 0.1 x x 0.2 Your code used the latter but I strongly suspect you wanted the former. HTH. Best wishes, Berwin === Full address = Berwin A TurlachTel.: +65 6516 4416 (secr) Dept of Statistics and Applied Probability+65 6516 6650 (self) Faculty of Science FAX : +65 6872 3919 National University of Singapore 6 Science Drive 2, Blk S16, Level 7 e-mail: sta...@nus.edu.sg Singapore 117546http://www.stat.nus.edu.sg/~statba __ 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] Bootstrap or Wilcoxons' test?
The Wilcoxon rank sum test is not plain and simple a test equality of distributions. If it were such, it would be able to test for differences in variance when locations were similar. For that purpose it would, in point of fact, be useless. Compare these simple situations w.r.t. the WRS: x - rnorm(100) # mean=0, sd=1 y - rnorm(100, mean=0, sd=4) wilcox.test(x,y) Wilcoxon rank sum test with continuity correction data: x and y W = 4518, p-value = 0.2394 alternative hypothesis: true location shift is not equal to 0 y - rnorm(100, mean=.2, sd=0) wilcox.test(x,y) Wilcoxon rank sum test with continuity correction data: x and y W = 3900, p-value = 0.004079 alternative hypothesis: true location shift is not equal to 0 It is a test of the equality of location (and the median is a readily understood non-parametric measure of location). The test is derived under the *assumption* that the samples are drawn from the *same* distribution differing only by a shift. If the distributions were not of the same family, the test would be invalidated. The wilcox.test help page is informative, saying the null hypothesis is that the distributions of xand y differ by a location shift of mu. The pseudomedian is optionally estimated when conf.int is set to TRUE. I also suggest looking at the formula for the statistic. It is available with getAnywhere(wilcox.test.default). If one wants a test for equality of distribution, one could turn to a more general test (with loss of power but with at least some potential for detecting differences in dispersion) such as the Kolmogorov-Smirnov or Kuiper tests. With x and y as above: ks.test(x,y) Two-sample Kolmogorov-Smirnov test data: x and y D = 0.61, p-value 2.2e-16 alternative hypothesis: two-sided Warning message: In ks.test(x, y) : cannot compute correct p-values with ties Returning to the OP's question, rather than worrying about normality in samples, the greater threat to validity in regression methods is unequal variances across groups or the range of continuous predictors. -- David Winsemius On Feb 13, 2009, at 11:12 PM, Murray Cooper wrote: First of all, sorry for my typing mistakes. Second, the WRS test is most certainly not a test for unequal medians. Although under specified models it would be. Just as under specified models it can be a test for other measures of location. Perhaps I did not word my explanation correctly, but I did not mean to imply that it would be a test of equality of variance. It is plain and simple a test for the equality of distributions. When the results of a properly applied parametric test do not agree with the WRS, it is usually do to a difference in the empirical density function of the two samples. Murray M Cooper, Ph.D. Richland Statistics 9800 N 24th St Richland, MI, USA 49083 Mail: richs...@earthlink.net - Original Message - From: David Winsemius dwinsem...@comcast.net To: Murray Cooper myrm...@earthlink.net Cc: Charlotta Rylander z...@nilu.no; r-help@r-project.org Sent: Friday, February 13, 2009 9:19 PM Subject: Re: [R] Bootstrap or Wilcoxons' test? I must disagree with both this general characterization of the Wilcoxon test and with the specific example offered. First, we ought to spell the author's correctly and then clarify that it is the Wilcoxon rank-sum test that is being considered. Next, the WRS test is a test for differences in the location parameter of independent samples conditional on the samples having been drawn from the same distribution. The WRS test would have no discriminatory power for samples drawn from the same distribution having equal location parameters but only different with respect to unequal dispersion. Look at the formula, for Pete's sake. It summarizes differences in ranking, so it is in fact designed NOT to be sensitive to the spread of the values in the sample. It would have no power, for instance, to test the variances of two samples, both with a mean of 0, and one having a variance of 1 with the other having a variance of 3. One can think of the WRS as a test for unequal medians. -- David Winsemius, MD. MPH Heritage Laboratories On Feb 13, 2009, at 7:48 PM, Murray Cooper wrote: Charlotta, I'm not sure what you mean when you say simple linear regression. From your description you have two groups of people, for which you recorded contaminant concentration. Thus, I would think you would do something like a t-test to compare the mean concentration level. Where does the regression part come in? What are you regressing? As for the Wilcoxnin test, it is often thought of as a nonparametric t-test equivalent. This is only true if the observations were drawn, from a population with the same probability distribution. The null hypothesis of the Wilcoxin test is actually the observations were drawn, from the same probability distribution. Thus if your two
