similar to: Splines under tension

Displaying 20 results from an estimated 4000 matches similar to: "Splines under tension"

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
2017 Nov 01
3
Cox Regression : Spline Coefficient Interpretation?
Hi, I'm using a Cox-Regression to estimate hazard rates on prepayments. I'm using the "pspline" function to face non-linearity, but I have no clue how to interpret the result. Unfortunately I did not find enough information on the "pspline" function wether in the survival package nor using google.. I got following output: * library(survival)* > > > >
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.
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 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 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:
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 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
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 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
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 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
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
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
2001 Dec 05
4
Questions about piecewise spline fitting
Hi All, I want to fit a piecewise spline of degree 1, i.e. a spline consisting of a straight line over each piece. I downloaded the R package pspline, then I have following questions: 1. in the program, the degree of the spline is specified by 2*norder-1. Why do they adopt such scheme that we can only fit a spline with odd degree? 2. norder cannot be set to 1. Is there any specific reason
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 Jan 15
1
Multiple comparison and two-way ANOVA design
Dear useRs, I'm working on multiple comparison design on two factor (2 3 levels) ANOVA. Each of the tests I have tried (Tukey, multcomp package) seem to do only with one factor at a time. fm1 <- aov(breaks ~ wool * tension, data = warpbreaks) tHSD <- TukeyHSD(fm1, "tension", ordered = FALSE) $tension diff lwr upr p adj M-L -10.000000 -19.35342
2006 Aug 01
1
plot() with TukeyHSD
Hello, When plotting the results of a TukeyHSD multiple comparisons procedure with an ANOVA (lm) object, an extra line appears in the confidence intervals that contain 0. For example (this is straight from the TukeyHSD helpfile): > summary(fm1 <- aov(breaks ~ wool + tension, data = warpbreaks)) > TukeyHSD(fm1, "tension", ordered = TRUE) > plot(TukeyHSD(fm1,