Skip to main content

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.



Comments

Post a Comment

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...
Post a Comment
Read more

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?
Post a Comment
Read more

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...
Post a Comment
Read more