Displaying 20 results from an estimated 245 matches for "tensions".
Did you mean:
tension
2018 Jan 07
2
SpreadLevelPlot for more than one factor
Dear Ashim,
Try spreadLevelPlot(breaks ~ interaction(tension, wool), data=warpbreaks) .
I hope this helps,
John
-----------------------------
John Fox, Professor Emeritus
McMaster University
Hamilton, Ontario, Canada
Web: socialsciences.mcmaster.ca/jfox/
> -----Original Message-----
> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Ashim
> Kapoor
> Sent:
2008 Dec 12
2
Extracting the name of an object into a character string and vice versa
I am still struggling to map a character string to an object name and
vice versa in R.
I thought the as.name() function might work, but observe the following
behaviour ...
> attach(warpbreaks)
> levels(tension)
[1] "L" "M" "H"
> levels(as.name("tension"))
NULL
> objectname<-as.name("tension")
> objectname
tension
>
2018 Jan 09
0
SpreadLevelPlot for more than one factor
Dear Sir,
Many thanks for your reply.
I have a query.
I have a whole set of distributions which should be made normal /
homoscedastic. Take for instance the warpbreaks data set.
We have the following boxplots for the warpbreaks dataset:
a. boxplot(breaks ~ wool)
b. boxplot(breaks ~ tension)
c. boxplot(breaks ~ interaction(wool,tension))
d. boxplot(breaks ~ wool @ each level of tension)
e.
2018 Jan 14
1
SpreadLevelPlot for more than one factor
Dear Ashim,
I?ll address your questions briefly but they?re really not appropriate for
this list, which is for questions about using R, not general statistical
questions.
(1) The relevant distribution is within cells of the wool x tension
cross-classification because it?s the deviations from the cell means that
are supposed to be normally distributed with equal variance. In the
warpbreaks data
2012 Nov 29
2
Deleting certain observations (and their imprint?)
I'm manipulating a large dataset and need to eliminate some observations based on specific identifiers. This isn't a problem in and of itself (using which.. or subset..) but an imprint of the deleted observations seem to remain, even though they have 0 observations. This is causing me problems later on. I'll use the dataset warpbreaks to illustrate, I apologize if this isn't in
2018 Jan 07
2
SpreadLevelPlot for more than one factor
Dear All,
I want a transformation which will make the spread of the response at all
combinations
of 2 factors the same.
See for example :
boxplot(breaks ~ tension * wool, warpbreaks)
The closest I can do is :
spreadLevelPlot(breaks ~tension , warpbreaks)
spreadLevelPlot(breaks ~ wool , warpbreaks)
I want to do :
spreadLevelPlot(breaks ~tension * wool, warpbreaks)
But I get :
>
2009 Jan 14
3
Casting lists to data.frames, analog to SAS
I have a specific question and a general question.
Specific Question: I want to do an analysis on a data frame by 2 or more
class variables (i.e., use 2 or more columns in a dataframe to do
statistical classing). Coming from SAS, I'm used to being able to take a
data set and have the output of the analysis in a dataset for further
manipulation. I have a data set with vote totals, with one
2005 May 15
3
adjusted p-values with TukeyHSD?
hi list,
i have to ask you again, having tried and searched for several days...
i want to do a TukeyHSD after an Anova, and want to get the adjusted
p-values after the Tukey Correction.
i found the p.adjust function, but it can only correct for "holm",
"hochberg", bonferroni", but not "Tukey".
Is it not possbile to get adjusted p-values after
2007 Aug 14
4
Problem with "by": does not work with ttest (but with lme)
Hello,
I would like to do a large number of e.g. 1000 paired ttest using the by-function. But instead of using only the data within the 1000 groups, R caclulates 1000 times the ttest for the full data set(The same happens with Wilcoxon test). However, the by-function works fine with the lme function.
Did I just miss something or is it really not working? If not, is there any other possibility to
2001 Sep 05
3
Bug in ftable?? (Was: Two-way tables of data, etc)
Further to the discussion between Murray Jorgensen and Brian Ripley,
it seems to me better to choose tabulations that will not come and bite
you. Suppose your data are sligtly irregular, e.g. (for the sake of
the argument):
data( warpbreaks )
warpbreaks$variant <- rep( 1:5, len=54 )
attach( warpbreaks )
tb <- table( wool, tension, variant )
tb
# in this case you would like to see:
tp
2001 Sep 05
3
Bug in ftable?? (Was: Two-way tables of data, etc)
Further to the discussion between Murray Jorgensen and Brian Ripley,
it seems to me better to choose tabulations that will not come and bite
you. Suppose your data are sligtly irregular, e.g. (for the sake of
the argument):
data( warpbreaks )
warpbreaks$variant <- rep( 1:5, len=54 )
attach( warpbreaks )
tb <- table( wool, tension, variant )
tb
# in this case you would like to see:
tp
2010 May 11
1
Splines under tension
Does anyone know if R has a function for splines under tension. I know there
are numerous packages for spline interpolation within R i just can't find
one that lets you determine the tension factor.
Any help would be much appreciated!
Sam
--
View this message in context: http://r.789695.n4.nabble.com/Splines-under-tension-tp2173887p2173887.html
Sent from the R help mailing list archive at
2018 Jan 07
0
SpreadLevelPlot for more than one factor
Dear All,
we need to do :
library(car) for the spreadLevelPlot function
I forgot to say that.
Apologies,
Ashim
On Sun, Jan 7, 2018 at 10:37 AM, Ashim Kapoor <ashimkapoor at gmail.com> wrote:
> Dear All,
>
> I want a transformation which will make the spread of the response at all
> combinations
> of 2 factors the same.
>
> See for example :
>
>
2010 May 18
2
how to select rows per subset in a data frame that are max. w.r.t. a column
Hi,
I'd like to select one row in a data frame per subset which is maximal for a
particular value. I'm pretty close to the solution in the sense that I can
easily select the maximal values per subset using "aggregate", but I can't
really figure out how to select the rows in the original data frame that are
associated with these maximal values.
library(stats)
# this
2012 May 16
1
TukeyHSD plot error
Hi, I am seeking help with an error when running the example from R
Documentation for TukeyHSD. The error occurs with any example I run, from
any text book or website. thank you...
> plot(TukeyHSD(fm1, "tension")).
Error in plot(confint(as.glht(x)), ylim = c(0.5, n.contrasts + 0.5), ...) :
error in evaluating the argument 'x' in selecting a method for function
2012 Oct 23
1
How Rcmdr or na.exclude blocks TukeyHSD
Dear R-Helpers,
I was calling the TukeyHSD function and not getting confidence intervals or p-values. It turns out this was caused by missing data and the fact that I had previously turned on R Commander (Rcmdr). John Fox knew that Rcmdr sets na.action to na.exclude, which causes the problem. If you have this problem, you can either exit Rcmdr before calling TukeyHSD or you can set na.action to
2007 Sep 06
3
Warning message with aggregate function
Dear all,
When I use aggregate function as:
attach(warpbreaks)
aggregate(warpbreaks[, 1], list(wool = wool, tension = tension), sum)
The results are right but I get a warning message:
"number of items to replace is not a multiple of replacement length."
BTW: I use R version 2.4.1 in Ubuntu 7.04.
Your kind solutions will be great appreciated.
Best wishes
Yours, sincerely,
Xingwang
2006 Oct 27
0
glht for aov with Error() term
Dear all,
glht (from the multcomp package) needs a term and a model component
in it's fitted model.
In fitted models from e.g. repeated measurements ANOVAs I do not find
neither model nor term.
Is it possible to build together a model and term component myself,
so that glht will work for repeated measurements ANOVAs? If so, how
would I do that?
Best regards,
Michael Zehetleitner
2012 Jul 27
1
Understanding the intercept value in a multiple linear regression with categorical values
Hi!
I'm failing to understand the value of the intercept value in a
multiple linear regression with categorical values. Taking the
"warpbreaks" data set as an example, when I do:
> lm(breaks ~ wool, data=warpbreaks)
Call:
lm(formula = breaks ~ wool, data = warpbreaks)
Coefficients:
(Intercept) woolB
31.037 -5.778
I'm able to understand that the value of
2012 Jun 13
1
Tukey Kramer with ANOVA (glm)
Hello,
I am performing a BACI analysis with ANOVA using the following glm:
fit1<-glm(log(Cucs_m+1)~(BA*Otter)+BA+Otter+ID+Primary, data=b1)
The summary(aov(fit1)) shows significance in the interaction; however, now I
would like to determine what combinations of BA and Otter are significantly
different (each factor has two levels). ID and PRIMARY substrates are
categorical and included in