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This question is similar to this: Nested list to graph. How to "flatten" an arbitrary expression, eg

expr = a[b[c, d[e][f], g], h]

to a list of key-value pairs representing Graph edges of the expression tree:

enter image description here

These can be extracted by applying WReach's exception-based method --> Can TreeForm be displayed “sideways”?:

  t_TreePlot := Throw@Hold@t;
  ][[1, 1]]


{{"a", "0", "a[b[c, d[e][f], g], h]"} -> {"b", "1", 
   "b[c, d[e][f], g]"}, {"b", "1", "b[c, d[e][f], g]"} -> {"c", "2", 
   "c"}, {"b", "1", "b[c, d[e][f], g]"} -> {"d[e]", "3", 
   "d[e][f]"}, {"d[e]", "3", "d[e][f]"} -> {"f", "4", "f"}, {"b", "1",
    "b[c, d[e][f], g]"} -> {"g", "5", "g"}, {"a", "0", 
   "a[b[c, d[e][f], g], h]"} -> {"h", "6", "h"}}

(Why are they cast to String?) Note non-atomic sub-expressions replaced with their Head. Based on the above, and the rule: ({h_, _, _} -> {a_, _, _}) :> ToExpression@h -> ToExpression@ a gives:

{a -> b, b -> c, b -> d[e], d[e] -> f, b -> g, a -> h}

TreeForm is a wrapper around TreePlot. TreePlot[%, VertexLabeling -> True] gives:

enter image description here

Since the layout is different, TreePlot must be making use of the other components of the output of the Block above.


How do TreeForm and SparseAray`ExpressionToTree (see below) extract these pairs of vertices? "Proof of work" is to extract the position (in the expression) of each vertex along with the edges.

Previously, I asked how to extract these "edges" based on a more restricted example Alternatives ordering affects pattern matching in Cases?. Also tried ReplaceList but don't know how to map it consistently across all levels.

share|improve this question
Related: (11458) – Mr.Wizard Jun 3 at 3:50
up vote 5 down vote accepted

Although kguler posted an answer using a nice internal function that does this (almost) directly I find this kind of expression manipulation interesting in itself so I'm going to see what can be done without it.

expr = a[b[c, d[e][f], g], h];

edges = Reap[Replace[expr, h_[___, c_[___] | c_?AtomQ, ___] /; Sow[h -> c] :> 1, {0, -1}]][[2, 1]];

TreePlot[edges, VertexLabeling -> True]

enter image description here

Or for the different layout:

TreePlot[edges, Automatic, Head @ expr, VertexLabeling -> True]

enter image description here

share|improve this answer
Thanks, nice use of Sow/Reap, and inner Alternatives. One issue, how to specify Head[expr] as RootVertex? Maybe using a function of Position[expr, h] (in Sow)? – alancalvitti Jun 27 '14 at 21:09
@alancalvitti I'm glad this helps. Regarding "RootVertex" does this do what you want? TreePlot[%, Automatic, Head@expr, VertexLabeling -> True] – Mr.Wizard Jun 28 '14 at 1:11

An alternative method to WReach's method is to use SparseArray`ExpressionToTree which produces the same output without string wrappers:

expr = a[b[c, d[e][f], g], h];
ett = SparseArray`ExpressionToTree[expr]
(* {{a,0,a[b[c,d[e][f],g],h]}->{b,1,b[c,d[e][f],g]},
    {a,0,a[b[c,d[e][f],g],h]}->{h,6,h}} *)

edges = ett[[All,All,1]] (* thanks: @Mr.Wizard *)
(* or edges = ett /. Rule[a_, b_] :> Rule[First[a], First[b]];*)
(* {a->b,b->c,b->d[e],d[e]->f,b->g,a->h} *)

Graph[edges, VertexLabels -> Placed["Name", {Center, Center}],
             VertexSize -> .3, VertexLabelStyle -> Directive[Red, Italic, 20],
             ImagePadding -> 20, ImageSize -> 400, 
             GraphLayout -> {"LayeredEmbedding", "RootVertex" -> edges[[1,1]]}]

enter image description here

share|improve this answer
More directly: SparseArray`ExpressionToTree[expr][[All, All, 1]]. Also, thanks for showing me ExpressionToTree! (Again?) – Mr.Wizard Jun 27 '14 at 1:24
Thank you indeed @Mr.Wizard. – kglr Jun 27 '14 at 1:35
@kguler, +1 - btw ETT --> ReleaseHold. But I'm asking how TreeForm (and ETT) convert the expr. I edited my Q to reflect this. – alancalvitti Jun 27 '14 at 18:40
@alancalvitti Is my answer of no interest to you? Are you only interested in the details of the internal implementation and not an equivalent method in top-level code? – Mr.Wizard Jun 27 '14 at 19:37
Yes that works, thank you, I had to test on some expressions of interest. – alancalvitti Jun 27 '14 at 21:06

Is this what you are seeking?

expr = a[b[c, d[e][f], g], h]
boxes = ToBoxes@TreeForm[expr]
positions = Cases[boxes, LineBox[{x__}] -> x, Infinity]
nodes =
    StyleBox[x_, __] :> ToExpression@x, Infinity] /. 
      {t_Times :> First@t, Verbatim[HoldForm][x_] -> x}
Rule @@@ Extract[nodes, List /@ positions]

{a -> b, a -> h, b -> c, b -> d, b -> g, d -> f}

share|improve this answer
No, for example that doesn't match d[e]. But also, How to extract it directly without using TreeForm? – alancalvitti Jun 26 '14 at 23:15

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