Alternatively, you could fit a "tree-based" mixed logit model based on continuation ratio logits. See a recent JSS paper by De Boeck and Partchev (2012, http://www.jstatsoft.org/v48/c01/). The idea is to convert the three-level response into a binary one using a decision tree. In one possible tree, the first node is indifferent w.r.t. size (response both) vs. picky (response small or big). The second node is small given picky vs. big given picky. You need to extend your data by a two-level node variable. The new response is then binary. The model can be fit using glmer(). Best, Florian --- Florian Wickelmaier Department of Psychology University of Tuebingen Schleichstr. 4, 72076 Tuebingen, Germany
Running multinomial models with random effects
3 messages · florian.wickelmaier at uni-tuebingen.de, MORAN LOPEZ, TERESA
18 days later
Hi Florian, thanks a lot for your help. I have been reading De Boeck and Partchev paper and I have some questions. First of all I am having some problems when applying dendrify function. My dataframe has the following structure: RISK Area CHOICE 1 MONTE Sopie 2 2 MONTE Sopie 2 3 MONTE Sopie 2 4 MONTE Anchurones 3 5 MONTE Anchurones 1 6 MONTE Anchurones 2 being 1 (both acorns choice, 2 chosing the big one and 3 chosing the small one) Following your instructions I mapped my tree as: mapping <- cbind(c(0, 1, 1), c(NA, 0, 1)) Then I remove the first two columns in order to implement dendrify function dendrify(jay[,-(1:2)],mapping) The following error arise: Error: is.matrix(mat) is not TRUE I have not been able to find the problem. I have followed tutorial package and paper instructions. mat=(jay[,-(1:2)], has only one column since I only have one item and each row corresponds to one choice event. Any suggestions? Quoting :
Alternatively, you could fit a "tree-based" mixed logit model based on continuation ratio logits. See a recent JSS paper by De Boeck and Partchev (2012, http://www.jstatsoft.org/v48/c01/). The idea is to convert the three-level response into a binary one using a decision tree. In one possible tree, the first node is indifferent w.r.t. size (response both) vs. picky (response small or big). The second node is small given picky vs. big given picky. You need to extend your data by a two-level node variable. The new response is then binary. The model can be fit using glmer(). Best, Florian --- Florian Wickelmaier Department of Psychology University of Tuebingen Schleichstr. 4, 72076 Tuebingen, Germany
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Dear Florian, I have just figured out what was going on. I haven?t converted my dataframe into matrix. Sorry for such an obvious question! Quoting "MORAN LOPEZ, TERESA":
Hi Florian, thanks a lot for your help. I have been reading De Boeck and Partchev paper and I have some questions. First of all I am having some problems when applying dendrify function. My dataframe has the following structure: RISK Area CHOICE 1 MONTE Sopie 2 2 MONTE Sopie 2 3 MONTE Sopie 2 4 MONTE Anchurones 3 5 MONTE Anchurones 1 6 MONTE Anchurones 2 being 1 (both acorns choice, 2 chosing the big one and 3 chosing the small one) Following your instructions I mapped my tree as: mapping <- cbind(c(0, 1, 1), c(NA, 0, 1)) Then I remove the first two columns in order to implement dendrify function dendrify(jay[,-(1:2)],mapping) The following error arise: Error: is.matrix(mat) is not TRUE I have not been able to find the problem. I have followed tutorial package and paper instructions. mat=(jay[,-(1:2)], has only one column since I only have one item and each row corresponds to one choice event. Any suggestions? Quoting :
Alternatively, you could fit a "tree-based" mixed logit model based on continuation ratio logits. See a recent JSS paper by De Boeck and Partchev (2012, http://www.jstatsoft.org/v48/c01/). The idea is to convert the three-level response into a binary one using a decision tree. In one possible tree, the first node is indifferent w.r.t. size (response both) vs. picky (response small or big). The second node is small given picky vs. big given picky. You need to extend your data by a two-level node variable. The new response is then binary. The model can be fit using glmer(). Best, Florian --- Florian Wickelmaier Department of Psychology University of Tuebingen Schleichstr. 4, 72076 Tuebingen, Germany
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