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$begingroup$
Consider the following expression.
expr = {a, {b1, b2}, {c, {d1, d2}}};
One can get the levels in an expression as follows:
ClearAll[levels];
SetAttributes[levels, {HoldAllComplete}];
levels[expr_] :=
Column @ Table[Level[expr, {level}, Heads -> True], {level, 0, Depth[expr]-1}];
levels[expr]
But when I look at the TreeForm of it expr
TreeForm[expr]
I don't see what I expected: the leaf count for this expression should be 10.
LeafCount[expr]
One can try to get the true level tree as follows:
Graph[
{
Sequence @@ (expr[UndirectedEdge]#& /@ {List, a, {b1, b2}, {c,{d1, d2}}}),
Sequence @@ (expr[[2]][UndirectedEdge]#& /@ {List2, b1, b2}),
Sequence @@ (expr[[3]][UndirectedEdge]#& /@ {List3, c, {d1, d2}}),
Sequence @@ (expr[[3,2]][UndirectedEdge]#& /@ {List4, d1, d2})
}, VertexLabels -> "Name"]
Is there a way to produce this graph for arbitrary expression?
Also, multiple vertices with the same name List get joined so I have to rename them to List1, List2, ..., etc. Is there a way to fix this while keeping the layout of the graph?
`
asically, I want to display heads at the same level as their parts, which is their true position in the tree.
trees
edited 5 hours ago
m_goldberg
86.6k872196
asked 13 hours ago
user13892user13892
1,111514
$endgroup$
add a comment |
$begingroup$
Consider the following expression.
expr = {a, {b1, b2}, {c, {d1, d2}}};
One can get the levels in an expression as follows:
ClearAll[levels];
SetAttributes[levels, {HoldAllComplete}];
levels[expr_] :=
Column @ Table[Level[expr, {level}, Heads -> True], {level, 0, Depth[expr]-1}];
levels[expr]
But when I look at the TreeForm of it expr
TreeForm[expr]
I don't see what I expected: the leaf count for this expression should be 10.
LeafCount[expr]
One can try to get the true level tree as follows:
Graph[
{
Sequence @@ (expr[UndirectedEdge]#& /@ {List, a, {b1, b2}, {c,{d1, d2}}}),
Sequence @@ (expr[[2]][UndirectedEdge]#& /@ {List2, b1, b2}),
Sequence @@ (expr[[3]][UndirectedEdge]#& /@ {List3, c, {d1, d2}}),
Sequence @@ (expr[[3,2]][UndirectedEdge]#& /@ {List4, d1, d2})
}, VertexLabels -> "Name"]
Is there a way to produce this graph for arbitrary expression?
Also, multiple vertices with the same name List get joined so I have to rename them to List1, List2, ..., etc. Is there a way to fix this while keeping the layout of the graph?
`
asically, I want to display heads at the same level as their parts, which is their true position in the tree.
trees
edited 5 hours ago
m_goldberg
86.6k872196
asked 13 hours ago
user13892user13892
1,111514
$endgroup$
add a comment |
$begingroup$
Consider the following expression.
expr = {a, {b1, b2}, {c, {d1, d2}}};
One can get the levels in an expression as follows:
ClearAll[levels];
SetAttributes[levels, {HoldAllComplete}];
levels[expr_] :=
Column @ Table[Level[expr, {level}, Heads -> True], {level, 0, Depth[expr]-1}];
levels[expr]
But when I look at the TreeForm of it expr
TreeForm[expr]
I don't see what I expected: the leaf count for this expression should be 10.
LeafCount[expr]
One can try to get the true level tree as follows:
Graph[
{
Sequence @@ (expr[UndirectedEdge]#& /@ {List, a, {b1, b2}, {c,{d1, d2}}}),
Sequence @@ (expr[[2]][UndirectedEdge]#& /@ {List2, b1, b2}),
Sequence @@ (expr[[3]][UndirectedEdge]#& /@ {List3, c, {d1, d2}}),
Sequence @@ (expr[[3,2]][UndirectedEdge]#& /@ {List4, d1, d2})
}, VertexLabels -> "Name"]
Is there a way to produce this graph for arbitrary expression?
Also, multiple vertices with the same name List get joined so I have to rename them to List1, List2, ..., etc. Is there a way to fix this while keeping the layout of the graph?
`
asically, I want to display heads at the same level as their parts, which is their true position in the tree.
trees
edited 5 hours ago
m_goldberg
86.6k872196
asked 13 hours ago
user13892user13892
1,111514
$endgroup$
Consider the following expression.
expr = {a, {b1, b2}, {c, {d1, d2}}};
One can get the levels in an expression as follows:
ClearAll[levels];
SetAttributes[levels, {HoldAllComplete}];
levels[expr_] :=
Column @ Table[Level[expr, {level}, Heads -> True], {level, 0, Depth[expr]-1}];
levels[expr]
But when I look at the TreeForm of it expr
TreeForm[expr]
I don't see what I expected: the leaf count for this expression should be 10.
LeafCount[expr]
One can try to get the true level tree as follows:
Graph[
{
Sequence @@ (expr[UndirectedEdge]#& /@ {List, a, {b1, b2}, {c,{d1, d2}}}),
Sequence @@ (expr[[2]][UndirectedEdge]#& /@ {List2, b1, b2}),
Sequence @@ (expr[[3]][UndirectedEdge]#& /@ {List3, c, {d1, d2}}),
Sequence @@ (expr[[3,2]][UndirectedEdge]#& /@ {List4, d1, d2})
}, VertexLabels -> "Name"]
Is there a way to produce this graph for arbitrary expression?
Also, multiple vertices with the same name List get joined so I have to rename them to List1, List2, ..., etc. Is there a way to fix this while keeping the layout of the graph?
`
asically, I want to display heads at the same level as their parts, which is their true position in the tree.
trees
trees
edited 5 hours ago
m_goldberg
86.6k872196
asked 13 hours ago
user13892user13892
1,111514
edited 5 hours ago
m_goldberg
86.6k872196
asked 13 hours ago
user13892user13892
1,111514
edited 5 hours ago
m_goldberg
86.6k872196
edited 5 hours ago
m_goldberg
86.6k872196
edited 5 hours ago
m_goldberg
86.6k872196
86.6k872196
asked 13 hours ago
user13892user13892
1,111514
asked 13 hours ago
user13892user13892
1,111514
asked 13 hours ago
user13892user13892
1,111514
1,111514
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
GraphComputation`ExpressionGraph[expr /. List -> (List[List, ##] &)]
TreeForm[expr /. List -> (List[List, ##] &)]
rules = List @@@ SparseArray`ExpressionToTree[expr /. List -> (List[List, ##] &)];
edges = DirectedEdge @@@ (rules[[All, All, 2]] + 1);
vertices = Property[#2 + 1, {VertexLabels -> #3}] & @@@ DeleteDuplicates[Flatten[rules, 1]];
TreeGraph[vertices, edges, ImagePadding -> 40, ImageSize -> 600, VertexSize -> Medium]
Update: An alternative approach is to use the original expression with ExpressionToTree and add new edges:
g1 = Graph[SparseArray`ExpressionToTree[{a, {b1, b2}, {c, foo[d1, d2]}}],
VertexLabels -> "Name", VertexLabelStyle -> 14, ImageSize -> 600]
newedges = # [DirectedEdge]
{Symbol[ToString[Head[First@Last[#]]] <> ToString[#[[2]]]]} & /@
Select[VertexList[g1], Head[#[[1]]] === Symbol &];
VertexReplace[EdgeAdd[g1, newedges], v_ :> Last[v]]
answered 12 hours ago
kglrkglr
185k10202420
$endgroup$
add a comment |
$begingroup$
Try the code
levelTree[expr_] := Replace[expr, {h_[x___] -> {h, x}}, {0, Infinity}];
levelTree @ {a, {b1, b2}, {c, {d1, d2}}}
which returns
{List, a, {List, b1, b2}, {List, c, {List, d1, d2}}}
A simple exmaple
levelTree[a b + c d]
which returns
{Plus, {Times, a, b}, {Times, c, d}}
I like the lispy variation
levelTree[expr_] := Replace[expr, (h : Except[List])[x___] -> {h, x}, {0, Infinity}];
levelTree @ plus[car[{1, 2}], cdr[{3, 4}]]
which returns
{plus, {car, {1, 2}}, {cdr, {3, 4}}}
Given any of these results, you can now use TreeForm[] or ExpressionGraph[] or some other custom Graph display.
answered 12 hours ago
SomosSomos
1,12819
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
GraphComputation`ExpressionGraph[expr /. List -> (List[List, ##] &)]
TreeForm[expr /. List -> (List[List, ##] &)]
rules = List @@@ SparseArray`ExpressionToTree[expr /. List -> (List[List, ##] &)];
edges = DirectedEdge @@@ (rules[[All, All, 2]] + 1);
vertices = Property[#2 + 1, {VertexLabels -> #3}] & @@@ DeleteDuplicates[Flatten[rules, 1]];
TreeGraph[vertices, edges, ImagePadding -> 40, ImageSize -> 600, VertexSize -> Medium]
Update: An alternative approach is to use the original expression with ExpressionToTree and add new edges:
g1 = Graph[SparseArray`ExpressionToTree[{a, {b1, b2}, {c, foo[d1, d2]}}],
VertexLabels -> "Name", VertexLabelStyle -> 14, ImageSize -> 600]
newedges = # [DirectedEdge]
{Symbol[ToString[Head[First@Last[#]]] <> ToString[#[[2]]]]} & /@
Select[VertexList[g1], Head[#[[1]]] === Symbol &];
VertexReplace[EdgeAdd[g1, newedges], v_ :> Last[v]]
answered 12 hours ago
kglrkglr
185k10202420
$endgroup$
add a comment |
$begingroup$
GraphComputation`ExpressionGraph[expr /. List -> (List[List, ##] &)]
TreeForm[expr /. List -> (List[List, ##] &)]
rules = List @@@ SparseArray`ExpressionToTree[expr /. List -> (List[List, ##] &)];
edges = DirectedEdge @@@ (rules[[All, All, 2]] + 1);
vertices = Property[#2 + 1, {VertexLabels -> #3}] & @@@ DeleteDuplicates[Flatten[rules, 1]];
TreeGraph[vertices, edges, ImagePadding -> 40, ImageSize -> 600, VertexSize -> Medium]
Update: An alternative approach is to use the original expression with ExpressionToTree and add new edges:
g1 = Graph[SparseArray`ExpressionToTree[{a, {b1, b2}, {c, foo[d1, d2]}}],
VertexLabels -> "Name", VertexLabelStyle -> 14, ImageSize -> 600]
newedges = # [DirectedEdge]
{Symbol[ToString[Head[First@Last[#]]] <> ToString[#[[2]]]]} & /@
Select[VertexList[g1], Head[#[[1]]] === Symbol &];
VertexReplace[EdgeAdd[g1, newedges], v_ :> Last[v]]
answered 12 hours ago
kglrkglr
185k10202420
$endgroup$
add a comment |
$begingroup$
GraphComputation`ExpressionGraph[expr /. List -> (List[List, ##] &)]
TreeForm[expr /. List -> (List[List, ##] &)]
rules = List @@@ SparseArray`ExpressionToTree[expr /. List -> (List[List, ##] &)];
edges = DirectedEdge @@@ (rules[[All, All, 2]] + 1);
vertices = Property[#2 + 1, {VertexLabels -> #3}] & @@@ DeleteDuplicates[Flatten[rules, 1]];
TreeGraph[vertices, edges, ImagePadding -> 40, ImageSize -> 600, VertexSize -> Medium]
Update: An alternative approach is to use the original expression with ExpressionToTree and add new edges:
g1 = Graph[SparseArray`ExpressionToTree[{a, {b1, b2}, {c, foo[d1, d2]}}],
VertexLabels -> "Name", VertexLabelStyle -> 14, ImageSize -> 600]
newedges = # [DirectedEdge]
{Symbol[ToString[Head[First@Last[#]]] <> ToString[#[[2]]]]} & /@
Select[VertexList[g1], Head[#[[1]]] === Symbol &];
VertexReplace[EdgeAdd[g1, newedges], v_ :> Last[v]]
answered 12 hours ago
kglrkglr
185k10202420
$endgroup$
GraphComputation`ExpressionGraph[expr /. List -> (List[List, ##] &)]
TreeForm[expr /. List -> (List[List, ##] &)]
rules = List @@@ SparseArray`ExpressionToTree[expr /. List -> (List[List, ##] &)];
edges = DirectedEdge @@@ (rules[[All, All, 2]] + 1);
vertices = Property[#2 + 1, {VertexLabels -> #3}] & @@@ DeleteDuplicates[Flatten[rules, 1]];
TreeGraph[vertices, edges, ImagePadding -> 40, ImageSize -> 600, VertexSize -> Medium]
Update: An alternative approach is to use the original expression with ExpressionToTree and add new edges:
g1 = Graph[SparseArray`ExpressionToTree[{a, {b1, b2}, {c, foo[d1, d2]}}],
VertexLabels -> "Name", VertexLabelStyle -> 14, ImageSize -> 600]
newedges = # [DirectedEdge]
{Symbol[ToString[Head[First@Last[#]]] <> ToString[#[[2]]]]} & /@
Select[VertexList[g1], Head[#[[1]]] === Symbol &];
VertexReplace[EdgeAdd[g1, newedges], v_ :> Last[v]]
answered 12 hours ago
kglrkglr
185k10202420
edited 8 hours ago
answered 12 hours ago
kglrkglr
185k10202420
answered 12 hours ago
kglrkglr
185k10202420
answered 12 hours ago
kglrkglr
185k10202420
185k10202420
add a comment |
add a comment |
$begingroup$
Try the code
levelTree[expr_] := Replace[expr, {h_[x___] -> {h, x}}, {0, Infinity}];
levelTree @ {a, {b1, b2}, {c, {d1, d2}}}
which returns
{List, a, {List, b1, b2}, {List, c, {List, d1, d2}}}
A simple exmaple
levelTree[a b + c d]
which returns
{Plus, {Times, a, b}, {Times, c, d}}
I like the lispy variation
levelTree[expr_] := Replace[expr, (h : Except[List])[x___] -> {h, x}, {0, Infinity}];
levelTree @ plus[car[{1, 2}], cdr[{3, 4}]]
which returns
{plus, {car, {1, 2}}, {cdr, {3, 4}}}
Given any of these results, you can now use TreeForm[] or ExpressionGraph[] or some other custom Graph display.
answered 12 hours ago
SomosSomos
1,12819
$endgroup$
add a comment |
$begingroup$
Try the code
levelTree[expr_] := Replace[expr, {h_[x___] -> {h, x}}, {0, Infinity}];
levelTree @ {a, {b1, b2}, {c, {d1, d2}}}
which returns
{List, a, {List, b1, b2}, {List, c, {List, d1, d2}}}
A simple exmaple
levelTree[a b + c d]
which returns
{Plus, {Times, a, b}, {Times, c, d}}
I like the lispy variation
levelTree[expr_] := Replace[expr, (h : Except[List])[x___] -> {h, x}, {0, Infinity}];
levelTree @ plus[car[{1, 2}], cdr[{3, 4}]]
which returns
{plus, {car, {1, 2}}, {cdr, {3, 4}}}
Given any of these results, you can now use TreeForm[] or ExpressionGraph[] or some other custom Graph display.
answered 12 hours ago
SomosSomos
1,12819
$endgroup$
add a comment |
$begingroup$
Try the code
levelTree[expr_] := Replace[expr, {h_[x___] -> {h, x}}, {0, Infinity}];
levelTree @ {a, {b1, b2}, {c, {d1, d2}}}
which returns
{List, a, {List, b1, b2}, {List, c, {List, d1, d2}}}
A simple exmaple
levelTree[a b + c d]
which returns
{Plus, {Times, a, b}, {Times, c, d}}
I like the lispy variation
levelTree[expr_] := Replace[expr, (h : Except[List])[x___] -> {h, x}, {0, Infinity}];
levelTree @ plus[car[{1, 2}], cdr[{3, 4}]]
which returns
{plus, {car, {1, 2}}, {cdr, {3, 4}}}
Given any of these results, you can now use TreeForm[] or ExpressionGraph[] or some other custom Graph display.
answered 12 hours ago
SomosSomos
1,12819
$endgroup$
Try the code
levelTree[expr_] := Replace[expr, {h_[x___] -> {h, x}}, {0, Infinity}];
levelTree @ {a, {b1, b2}, {c, {d1, d2}}}
which returns
{List, a, {List, b1, b2}, {List, c, {List, d1, d2}}}
A simple exmaple
levelTree[a b + c d]
which returns
{Plus, {Times, a, b}, {Times, c, d}}
I like the lispy variation
levelTree[expr_] := Replace[expr, (h : Except[List])[x___] -> {h, x}, {0, Infinity}];
levelTree @ plus[car[{1, 2}], cdr[{3, 4}]]
which returns
{plus, {car, {1, 2}}, {cdr, {3, 4}}}
Given any of these results, you can now use TreeForm[] or ExpressionGraph[] or some other custom Graph display.
answered 12 hours ago
SomosSomos
1,12819
edited 8 hours ago
answered 12 hours ago
SomosSomos
1,12819
answered 12 hours ago
SomosSomos
1,12819
answered 12 hours ago
SomosSomos
1,12819
1,12819
add a comment |
add a comment |
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