replacement - Is Replace doing double work?

In an answer to this question, rm -rf used the function Replace with level spefication to do a series of replacements inside out. As far as I can see, the order of substitutions when more than one level is involved is not documented, so I studied this a little bit deeper and constructed the following example for flattening a list, with an administration of the sublists that are replaced.

expr = {1, {2, {4, {5, 6}, {7, 8}}}};
usedsublists = {};
Replace[expr, x : {__} :> (AppendTo[usedsublists, x ]; Sequence @@ x), Infinity] 

usedsublists

(*  {1,2,4,5,6,7,8}  *)

(*  {{5,6},{7,8},{4,5,6,7,8},{5,6},{7,8},{2,4,5,6,7,8},{5,6},{7,8},{4,5,6,7,8},{5,6},{7,8}}  *)

Indeed, Replace works inside out. But it seems that Replace repeats substitutions it has already done for some reason. Do examples exist where this duplication is necessary, or is it a slight inefficiency in the implementation?

Answer

To understand how you end up with a length of 11 for your usedsubsets you must consider not only the traversal order^(1)(2)(3) in which the expression is scanned for pattern matching but the evaluation order as well. Much like Map as I summarized in Scan vs. Map vs. Apply Replace does not perform a sequential evaluation* but performs all replacements and then evaluates the entire expression.

If we change the head of your expr to Hold we not only witness the behavior described above but also clear illustration of the source of 11:

expr = Hold[1, {2, {4, {5, 6}, {7, 8}}}];

usedsubsets = {};
Replace[expr, x : {__} :> (AppendTo[usedsubsets, x]; Sequence @@ x), Infinity]

Hold[1, AppendTo[
  usedsubsets, {2, 
   AppendTo[usedsubsets, {4, AppendTo[usedsubsets, {5, 6}]; Sequence @@ {5, 6}, 
     AppendTo[usedsubsets, {7, 8}]; Sequence @@ {7, 8}}]; 

   Sequence @@ {4, AppendTo[usedsubsets, {5, 6}]; Sequence @@ {5, 6}, 
     AppendTo[usedsubsets, {7, 8}]; Sequence @@ {7, 8}}}]; 
 Sequence @@ {2, 
   AppendTo[usedsubsets, {4, AppendTo[usedsubsets, {5, 6}]; Sequence @@ {5, 6}, 
     AppendTo[usedsubsets, {7, 8}]; Sequence @@ {7, 8}}]; 
   Sequence @@ {4, AppendTo[usedsubsets, {5, 6}]; Sequence @@ {5, 6}, 
     AppendTo[usedsubsets, {7, 8}]; Sequence @@ {7, 8}}}]

Lest you think this has anything to do with AppendTo, Sequence, or even CompoundExpression or Hold, here is another more generic example:

Replace[List @@ expr, x : {__} :> f[g[x], h[x]], Infinity]

{1, f[g[{2, 
    f[g[{4, f[g[{5, 6}], h[{5, 6}]], f[g[{7, 8}], h[{7, 8}]]}], 
     h[{4, f[g[{5, 6}], h[{5, 6}]], f[g[{7, 8}], h[{7, 8}]]}]]}], 
  h[{2, f[g[{4, f[g[{5, 6}], h[{5, 6}]], f[g[{7, 8}], h[{7, 8}]]}], 
     h[{4, f[g[{5, 6}], h[{5, 6}]], f[g[{7, 8}], h[{7, 8}]]}]]}]]}

This is merely the result of the "inside-out" replacement you referenced in your question.

*The exception to this statement is the use of either RuleCondition or its more verbose cousin, the Trott-Strzebonski in-place evaluation technique. These cause early evaluation of the RHS of the rules, after pattern substitution but before replacement in the main expression.

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

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plotting - Mathematica: 3D plot based on combined 2D graphs

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