Displaying 20 results from an estimated 2000 matches similar to: "SpreadLevelPlot for more than one factor"
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
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 :
>
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 :
>
>
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
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
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
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
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
2013 Feb 25
1
quesion about SS of ANOVA
Hi all:
I have a quesion about ANOVA: Is SS(Sum of Square) of a specific factor constant with the number of factors changing?
dat1 includes one factor g1,and g1's SS is called SS_g1_dat1.
dat2 includes two factors g1,g2,and g1's SS is called SS_g1_dat2.
My quesion is: Is SS_g1_dat1 equals to SS_g1_dat2?
I have both "yes" and "no" reasons for the quesion,but
2006 Apr 25
1
by() and CrossTable()
I am attempting to produce crosstabulations between two variables for
subgroups defined by a third factor variable. I'm using by() and
CrossTable() in package gmodels. I get the printing of the tables first
and then a printing of each level of the INDICES. For example:
library(gmodels)
by(warpbreaks, warpbreaks$tension, function(x){CrossTable(x$wool,
x$breaks > 30,
2001 Sep 05
3
Bug in ftable?? (Was: Two-way tables of data, etc)
Further to the discussion between Murray Jorgensen and Brian Ripley,
it seems to me better to choose tabulations that will not come and bite
you. Suppose your data are sligtly irregular, e.g. (for the sake of
the argument):
data( warpbreaks )
warpbreaks$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
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 09
1
how to convert by lists in data.frames
Dear R-helpers,
I’ve got a summary of results from a by() call that I am making with a list
of more than two of factors not very different from the example in the by()
help page
require(stats)
by(warpbreaks[, 1], warpbreaks[, -1], summary)
The result of the command gives a list of the form
wool: A
tension: L
Min. 1st Qu. Median Mean 3rd Qu. Max.
25.00 26.00
2013 Apr 10
2
Table with n(%) formatting: A better way to do this?
Hello All,
Am learning to create tables with n(%) formatting using R. Below is a working example. I think this is not bad but wondered if there are better ways of doing it. Although it can be quite humbling, seeing good code helps me assess my progress as an R programmer.
Ultimately want to have code that I can turn into a function. Will then use the output produced to make tables using
2009 Jul 02
1
From xtabs to matrix
Hi list,
is it possible convert the xtabs result
xtabs(breaks~tension+wool,data=warpbreaks)
wool
tension A B
L 401 254
M 216 259
H 221 169
to a simple matrix or data frame?
A B
L 401 254
M 216 259
H 221 169
Thanks a lot! Gianandrea
--
View this message in context: http://www.nabble.com/From-xtabs-to-matrix-tp24304588p24304588.html
2006 Dec 31
3
tabulate: switching columns and rows
Hi all,
Please, is there any way of controlling factors in row/columns when using ftable/xtabs? As far as I can see, the last cross-clasifing variable in the formula will appear in columns. The previous ones, in rows. For instance, is it possible to make tension and replicate appear in columns?
ftable(xtabs(breaks ~ wool + tension + replicate, data = warpbreaks))
After some years using SAS
2005 Jul 15
1
Adjusted p-values with TukeyHSD (patch)
Dear R-developeRs,
Attached follows a patch against svn 34959 that adds the
printing of p-values to the TukeyHSD.aov function in stats package. I
also updated the corresponding documentation file and added a 'see also'
reference to the simint function of the multcomp package.
As it was already brought up in a previous thread [1] in R-help,
one can obtain the adjusted
2008 Sep 07
1
an error to call 'gee' function in R
Dear List:
I found an error when I called the 'gee' function. I cannot solve and explain it. There are no errors when I used the 'geeglm' function. Both functions fit the gee model. The project supervisor recommends me to use the 'gee' function. But I cannot explain to him why this error happens. Would you help me solve this problem? I appreciate your help.
In