[R] ... Error in names-.default(`*tmp*`, value = nmstrata)
Hi, OK, I will describe it in more detail.In the meantime I installed the newest debian package of r-base etc. and the error message changed from:Error in names-.default(`*tmp*`, value = nmstrata) : names attribute must be the same length as the vectorto:Error model is singular in: aov(rati ~ cs * spk + Error(sub/(cs * spk)), data = rating.df) The model is a repeated measures anova that's why I use the Error() term. I found the hint for this usage in Notes on the use of R for psychology experiments and questionaires (Jonathan Baron, 2003).cs and speaker are within-subjects factors. rati is the dependent variable: ratings done by subjects. sub are the subject-codes (between-subjects factor). SpaeVP is sorted because in the preprocessing I needed to recoded data of each subject and did: testfac - as.factor(spaeVP)levels(testfac) - vpnso I could do for ( i in vpn).after that data (rati) was sorted so I sorted spaeVp too, before putting it into the dataframe. (I found no better solution (newbie)) Now follows the syntax I already sent last time. After that is more data of my dataframe. with best regards, Wolfgang Pauli I use:spk =factor(rep(c(spk1,spk2,spk3,spk4,spk5,spk6,spk7,spk8), c(2,2,2,2,2,2,2,2)))cs = factor(rep(c(yes,no),c(8,8)))sub = factor(sort(spaeVP))rati = stack(data.frame(arousal))rati = rati[,1]rating.df - data.frame(sub, cs, spk, rati)summary(aov(rati ~ cs*spk + Error(sub/(cs*spk)), data=rating.df)) arousal (30 subjects, and 16 stimuli) before using stack() and putting it into rati [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [1,] -1 -2 -1 -1 -2 0 -1 0 -2 -1 0 0 -2 [2,] 0 0 0 0 -1 1 0 -1 -2 0 1 -1 -1 [3,] 1 -1 -1 -2 -1 0 0 0 -3 0 -1 0 2 [4,] 0 0 2 0 -1 1 0 1 0 2 -2 1 -2 [5,] 0 0 -1 -1 -1 0 0 0 0 0 1 0 1 [6,] -1 0 -2 0 -2 0 0 0 0 0 -1 1 0 [7,] -1 0 0 0 -1 0 0 -1 0 1 -1 1 -1 [8,] 1 0 0 0 -1 0 0 1 0 -1 -3 0 1 [9,] 0 -1 0 0 -1 0 0 1 0 -1 0 1 3[10,] 0 -2 -1 -1 0 2 0 -2 -2 1 -1 2 1[11,] 0 -1 0 0 0 -1 -1 0 -1 -1 1 1 0[12,] -1 -3 0 -2 0 0 0 0 -1 0 1 2 1[13,] 0 -1 -1 -1 0 -1 0 -1 -1 -1 -1 1 0[14,] 1 2 2 -1 0 0 0 0 1 -1 0 1 1[15,] 0 0 0 -1 0 1 -1 -1 -1 0 -1 -2 1[16,] 1 0 -2 -1 0 -1 0 0 -1 1 1 1 0 [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21] [,22] [,23] [,24] [,25] [1,] 0 0 0 0 0 0 1 1 0 0 1 1 [2,] -2 0 -1 -1 1 -3 3 0 0 -1 0 0 [3,] 1 0 0 0 1 -2 0 -1 -1 0 0 -1 [4,] 2 0 1 0 0 0 3 -1 1 0 -1 0 [5,] -2 1 -1 1 0 -2 1 0 0 0 -1 -1 [6,] 0 1 -1 0 -1 -2 3 3 2 0 -1 0 [7,] 0 1 1 0 0 -1 0 1 -1 1 0 0 [8,] 1 0 0 -2 0 0 -2 1 1 0 0 -1 [9,] 1 0 -2 0 0 -2 2 2 0 1 0 0[10,] 1 0 0 0 0 -1 1 1 0 1 1 2[11,] 0 -1 1 1 0 0 0 1 -2 1 0 1[12,] 1 1 0 1 -1 -1 1 2 0 1 0 0[13,] 2 0 -1 1 -1 -3 4 1 0 1 -1 -2[14,] -1 0 -1 2 -1 -1 1 0 -1 1 2 0[15,] 2 0 0 -2 -2 0 0 2 2 2 1 -1[16,] 0 0 0 1 0 -1 1 0 1 1 1 -1 [,26] [,27] [,28] [,29] [,30] [,31] [,32] [1,] 0 -1 -1 -1 -2 0 1 [2,] -2 -1 -1 1 0 1 2 [3,] 0 0 -1 -2 2 -2 1 [4,] 1 2 0 1 -1 0 1 [5,] 0 0 1 2 -4 -2 0 [6,] 0 1 -1 0 -1 2 0 [7,] -1 -2 -1 1 0 -2 1 [8,] 1 -2 0 0 -2 1 0 [9,] -2 0 0 0 -3 1 0[10,] 2 -2 -1 -1 1 2 0[11,] -1 -1 -1 -1 1 1 2[12,] -1 0 0 0 -1 0 1[13,] -2 1 -1 -1 -2 2 0[14,] 0 -1 -1 0 0 2 2[15,] 1 -1 0 0 0 3 1[16,] 0 0 2 1 0 2 1
[R] Error in names-.default(`*tmp*`, value = nmstrata) :
Hi, I have a problem with aov(). I used it many times, but now I have new data, tried to use it in the same way. I get I strange Error message that i can't understand (see below). I guess it is caused by incompatible data types. The dataframe also looks ok to me. I use: spk =factor(rep(c(spk1,spk2,spk3,spk4,spk5,spk6,spk7,spk8), c(2,2,2,2,2,2,2,2))) cs = factor(rep(c(yes,no),c(8,8))) sub = factor(sort(spaeVP)) rati = stack(data.frame(arousal)) rati = rati[,1] rating.df - data.frame(sub, cs, spk, rati) summary(aov(rati ~ cs*spk + Error(sub/(cs*spk)), data=rating.df)) I get this Error (after about 5sec): Error in names-.default(`*tmp*`, value = nmstrata) : names attribute must be the same length as the vector rating.df sub cs spk rati 1 5 yes spk1 -1 2 5 yes spk10 3 5 yes spk21 4 5 yes spk20 5 5 yes spk30 6 5 yes spk3 -1 7 5 yes spk4 -1 8 5 yes spk41 9 5 no spk50 ... Hope somebody can help me. with best regards, Wolfgang __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] newbie question on contrasts and aov
In the meantime I figured out that the Difference-contrast is not quite what I was looking for. But I still have two questions 1) Why do I get different results for Helmert contrasts in SPSS and R. I guess the contrast matrixes of Helmert are about the same in SPSS and R. I probably make a mistake as i am a newbie to R. I thought that it might be, because I have a repaeted measures design. That's why I put the Error(sub) in the formula of aov. 2) I tried to make my own contrast matrix, to compute comparisons between adjacent factor levels, i.e. 1-2, 2-3 and 3-4. My matrix looks like this: [,1] [,2] [,3] [1,]100 [2,] -110 [3,]0 -11 [4,]00 -1 But then I get the same result as with contr.helmert(4). What is wrong I really don't get it! Thank you, Wolfgang Pauli On Sunday January 11 2004 18:07, you wrote: Notice `SPKType III Sum of Squares'. I don't believe your contrasts are orthogonal, and R's are sequential sum of squares. Also, are you sure these are the same contrasts? I presume this is contr.sdif from MASS (in which case it is churlish not to credit it), and SPSS's contrasts look more like Helmert contrasts from their labelling. Since it appears all your treatments are within subjects you do seem to be making life difficult for yourself. Although I would have done a simple fixed-effects analysis, applying summary.lm to the bottom stratum would give you simple t-tests for each contrast, including actual estimates of the magnitudes. On Sun, 11 Jan 2004, Wolfgang Pauli wrote: I try to move from SPSS to R/S and am trying to reproduce the results of SPSS in R. I calculated a one-way anova with spk as experimental factor and erp as depended variable. The result of the Anova are the same concearning the mean square, F and p values. But I also wanted to caculate the contr.sdif(4) contrast on spk. The results are completely different now. I hope anybody can help me. Thanks, Wolfgang This is what I get in SPSS: Tests of Within-Subjects Contrasts Measure: MEASURE_1 Source SPKType III Sum of Squares df Mean Square F Sig. SPK Level 2 vs. Level 1 3,493 1 3,493 2,026 ,178 Level 3 vs. Previous20,358 1 20,358 10,168 ,007 Level 4 vs. Previous18,808 1 18,808 15,368 ,002 Error(SPK) Level 2 vs. Level 1 22,414 13 1,724 Level 3 vs. Previous26,030 13 2,002 Level 4 vs. Previous15,911 13 1,224 This is the result in R: Error: sub Df Sum Sq Mean Sq F value Pr(F) Residuals 13 205.79 15.83 Error: Within Df Sum Sq Mean Sq F valuePr(F) spk3 29.425 9.808 9.4467 8.055e-05 *** spk: p 1 1.747 1.747 1.6821 0.2022649 spk: q 1 13.572 13.572 13.0719 0.0008479 *** spk: r 1 14.106 14.106 13.5861 0.0006915 *** Residuals 39 40.493 1.038 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 Spk.df - data.frame(sub,spk,erp) subset(Spk.df, subset=(sub!=14oddball sub!=18odd sub!=19odd sub!=20oddball)) - Spk.selected.df contrasts(Spk.selected.df$spk) - contr.sdif(4) aov(erp ~ spk + Error(sub), data=Spk.selected.df) - Spk.aov summary(Spk.aov,data=Spk.selected.df,split=list(spk=list(p=1,q=2,r=3))) this is the the beginning of the dataframe, which I use: sub spkerp 1 10oddball spk1 2.587 2 11oddball spk1 -0.335 3 12oddball spk1 5.564 5 15oddball spk1 0.691 6 17oddball spk1 -1.846 10 21oddball spk1 1.825 11 22oddball spk1 0.370 12 2oddball spk1 3.234 13 3oddball spk1 1.462 14 5oddball spk1 2.535 15 6oddball spk1 9.373 16 7oddball spk1 2.132 17 8oddball spk1 -0.518 18 9oddball spk1 2.450 19 10oddball spk2 2.909 20 11oddball spk2 0.708 21 12oddball spk2 4.684 23 15oddball spk2 3.599 ... __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] newbie question on contrasts and aov
I try to move from SPSS to R/S and am trying to reproduce the results of SPSS in R. I calculated a one-way anova with spk as experimental factor and erp as depended variable. The result of the Anova are the same concearning the mean square, F and p values. But I also wanted to caculate the contr.sdif(4) contrast on spk. The results are completely different now. I hope anybody can help me. Thanks, Wolfgang This is what I get in SPSS: Tests of Within-Subjects Contrasts Measure: MEASURE_1 Source SPKType III Sum of Squares df Mean Square F Sig. SPK Level 2 vs. Level 1 3,493 1 3,493 2,026 ,178 Level 3 vs. Previous20,358 1 20,358 10,168 ,007 Level 4 vs. Previous18,808 1 18,808 15,368 ,002 Error(SPK) Level 2 vs. Level 1 22,414 13 1,724 Level 3 vs. Previous26,030 13 2,002 Level 4 vs. Previous15,911 13 1,224 This is the result in R: Error: sub Df Sum Sq Mean Sq F value Pr(F) Residuals 13 205.79 15.83 Error: Within Df Sum Sq Mean Sq F valuePr(F) spk3 29.425 9.808 9.4467 8.055e-05 *** spk: p 1 1.747 1.747 1.6821 0.2022649 spk: q 1 13.572 13.572 13.0719 0.0008479 *** spk: r 1 14.106 14.106 13.5861 0.0006915 *** Residuals 39 40.493 1.038 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 Spk.df - data.frame(sub,spk,erp) subset(Spk.df, subset=(sub!=14oddball sub!=18odd sub!=19odd sub!=20oddball)) - Spk.selected.df contrasts(Spk.selected.df$spk) - contr.sdif(4) aov(erp ~ spk + Error(sub), data=Spk.selected.df) - Spk.aov summary(Spk.aov,data=Spk.selected.df,split=list(spk=list(p=1,q=2,r=3))) this is the the beginning of the dataframe, which I use: sub spkerp 1 10oddball spk1 2.587 2 11oddball spk1 -0.335 3 12oddball spk1 5.564 5 15oddball spk1 0.691 6 17oddball spk1 -1.846 10 21oddball spk1 1.825 11 22oddball spk1 0.370 12 2oddball spk1 3.234 13 3oddball spk1 1.462 14 5oddball spk1 2.535 15 6oddball spk1 9.373 16 7oddball spk1 2.132 17 8oddball spk1 -0.518 18 9oddball spk1 2.450 19 10oddball spk2 2.909 20 11oddball spk2 0.708 21 12oddball spk2 4.684 23 15oddball spk2 3.599 ... __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
[R] Huynh-Feldt correction
Dear All,
I tried to help myself by searching in newsgroups etc. I found a test for
epsilon in a newsgroup (see attachement). I wrote a little skript to update
the summary.anova if epsilon 0.75.
Although I doublechecked my statistic book for the formula, the result is
always the half of the epsilon calculated by SPSS. It is really strange. If I
multiply espilon by two, I get the right results...
regards,
WP
#begin -
correctSummary - function(sum,eps) {
# eps is the epsilon-value returned by function hf
# sum is the summary.aov object in which the Huynh-Fldt correction has to be
down
len - length(sum)
newSum - sum
for (i in 2: len) {
newSum[[i]][[1]][4,] - newSum[[i]][[1]][2,]
newSum[[i]][[1]][3,] - newSum[[i]][[1]][2,]
newSum[[i]][[1]][2,] - newSum[[i]][[1]][1,]
rownames(newSum[[i]][[1]]) - c(rownames(newSum[[i]][[1]][1,]),
Huynh-Feldt, rownames(newSum[[i]][[1]][2,]), Huynh-Feldt)
newSum[[i]][[1]][2,][1] - newSum[[i]][[1]][2,][1]*epsilon*2
newSum[[i]][[1]][4,][1] - newSum[[i]][[1]][4,][1]*epsilon*2
newSum[[i]][[1]][2,][3] - newSum[[i]][[1]][2,][2]/newSum[[i]][[1]][2,][1]
newSum[[i]][[1]][4,][3] - newSum[[i]][[1]][4,][2]/newSum[[i]][[1]][4,][1]
newSum[[i]][[1]][2,][4] - newSum[[i]][[1]][2,][3]/newSum[[i]][[1]][4,][3]
newSum[[i]][[1]][2,][5] - pf(as.numeric(newSum[[i]][[1]][2,][4]),
as.numeric(newSum[[i]][[1]][2,][1]), as.numeric(newSum[[i]][[1]][4,][1]),
lower.tail = FALSE)
}
newSum
}
hf - function(m){
# m is a matrix with subjects as rows and conditions as columns
# note that checking for worst case scenarios F correction first
might # be a good idea using J/(J-1) as the df correction factor
n- length(m[,1])
J-length(m[1,])
X-cov(m)
r- length(X[,1])
D-0
for (i in 1: r) D- D+ X[i,i]
D-D/r
SPm- mean(X)
SPm2- sum(X^2)
SSrm-0
for (i in 1: r) SSrm- SSrm + mean(X[i,])^2
epsilon - (1/(J-1))*((J^2*(D-SPm)^2)/(SPm2 - 2*J*SSrm + J^2*SPm^2))
epsilon
}
#
On Thursday December 18 2003 12:00, Wolfgang Pauli wrote:
Dear R-helpers,
Does anybody know, whether there is an option to tell aov/anova, or do
something similar, to get a Huynh-Feldt correction of dfs in the aov/anova
function?
Thanks in advance!
__
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[R] Huynh-Feldt correction
Dear R-helpers, Does anybody know, whether there is an option to tell aov/anova, or do something similar, to get a Huynh-Feldt correction of dfs in the aov/anova function? Thanks in advance! __ [EMAIL PROTECTED] mailing list https://www.stat.math.ethz.ch/mailman/listinfo/r-help
