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I would like to create a simple decision tree for a 3-course meal that looks something like the following:

enter image description here

The tuples can be created using the code:

Tuples[{{""}, {"Soup", "Salad"}, {"Chicken", "Patty", "Liver", "Beef"}, {"C", "IC"}}]

to get

{{"", "Soup", "Chicken", "C"}, {"", "Soup", "Chicken", "IC"}, {"", "Soup", "Patty", "C"}, {"", "Soup", "Patty", "IC"}, {"", "Soup", "Liver", "C"}, {"", "Soup", "Liver", "IC"}, {"", "Soup", "Beef", "C"}, {"", "Soup", "Beef", "IC"}, {"", "Salad", "Chicken", "C"}, {"", "Salad", "Chicken", "IC"}, {"", "Salad", "Patty", "C"}, {"", "Salad", "Patty", "IC"}, {"", "Salad", "Liver", "C"}, {"", "Salad", "Liver", "IC"}, {"", "Salad", "Beef", "C"}, {"", "Salad", "Beef", "IC"}}

To get the decision tree, I've tried using the TreeGraph[] function, but it's too time-consuming writing out each paired rule and then I run into issues of repeats which I can't sort out. In any case, I would like to use the results from the code above, to generate a simple decisions tree. Any ideas?

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2 Answers 2

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layers = { {""}, {"Soup", "Salad"}, {"Chicken", "Patty", "Liver", "Beef"}, {"C", "IC"}}; eg = ExpressionGraph[ConstantArray[1, Rest[Length /@ layers]], PerformanceGoal -> "Quality", VertexStyle -> Black, ImageSize -> Large, AspectRatio -> 1, GraphLayout -> {"LayeredEmbedding", "Orientation" -> Left}]

enter image description here

Almost there ... except the labeling of vertices. To this end, we (1) use BreadthFirstScan to get a list of rules (vLabeling) to make the vertices in each layer indexed consecutively, (2) make appropriate number of copies of the elements of layers to get a list of labels (vlabels).

vLabeling = Thread[First @ Last @ Reap @ BreadthFirstScan[#, 1, {"PrevisitVertex" -> Sow}] -> VertexList[#]] &; vlabels = Join @@ MapThread[PadRight[##, "Periodic"]&, {layers, FoldList[Times, Length /@ layers]}]; SetProperty[eg, {VertexShapeFunction -> (Text[Style[vlabels[[#2 /. vLabeling[eg]]], 16], #] &)}]

enter image description here

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It sounds like you're interested in the visualization (i.e. TreeGraph), so what follows targets that. If you're actually interested in a data structure that could be used to implement a decision tree, then there is a better approach.

Here's your data (I've replaced the empty string with "root" to keep things a bit clearer).

nodes = {{"root"}, {"Soup", "Salad"}, {"Chicken", "Patty", "Liver", "Beef"}, {"C", "IC"}}

We will attempt a graph, but it won't be a tree because nodes can have mutiple predecessors. We'll attempt a tree later.

edges = Flatten@Map[Apply[DirectedEdge], Tuples /@ Partition[nodes, 2, 1], {-2}]; Graph[edges, VertexLabels -> "Name"]

In order to get a tree, we'll need to give the nodes unique names. We can do this by first getting all of the paths/decisions and then prepending ancestor node names to each node (this is like a trie).

paths = Tuples[nodes]; treePaths = FoldList[StringJoin, #] & /@ paths; treeEdges = Union@Flatten@Map[Apply[DirectedEdge], Partition[#, 2, 1] & /@ treePaths, {-2}]; (*At this point we have unique edges of the correct form for a tree, so we can use TreeGraph...*) TreeGraph[treeEdges, VertexLabels -> "Name"] (*You'll probably need to do some adjustments so that it looks nice, like setting the size or other display attributes.*)
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  • $\begingroup$ Wow! Thank you. This helps quite a bit. I still run into the issue of unique names. I would like to keep the non unique names but can’t figure out how to do this using TreeGraph $\endgroup$ Commented Mar 14, 2022 at 1:35
  • $\begingroup$ Maybe you can use special characters or invisible characters to preserve uniqueness. I don't think TreeGraph will work otherwise, because you will end up with cycles/diamonds, and TreeGraph explicitly disallows that. $\endgroup$ Commented Mar 14, 2022 at 3:45

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