Skip to main content

function construction - Another difference between Set and Setdelayed. Evaluation shortcut?



A little while ago I wondered why


f[x_] = f[x]

gives an infinite iteration. I ended up discovering the following difference in evaluation between Set and SetDelayed with Evaluate.


count = 0;
ClearAll@a
a /; (count++ < 20) = {a}
a // OwnValues
count = 0;
a


Output


{{{{{{{{{{{{{{{{{{{{{a}}}}}}}}}}}}}}}}}}}}}
{HoldPattern[a /; count++ < 20] :> {a}}
{a}

and


count = 0;
ClearAll@b
b /; (count++ < 20) := Evaluate@{b}

b // OwnValues
count = 0;
b

Output


{HoldPattern[b /; count++ < 20] :> {b}}
{{{{{{{{{{{{{{{{{{{{b}}}}}}}}}}}}}}}}}}}}

Can somebody explain the difference? Can we say that there is an evaluation shortcut at work here?


Related



This is a follow up question: Strange results of definitions using OwnValues


Why x = x doesn't cause an infinite loop, but f[x_] := f[x] does?


Does Set vs. SetDelayed have any effect after the definition was done?



Answer



I thought to give a bit more insight into why Update is needed, as pointed out in the other answers. Its documentation says Update may be needed when a change in 1 symbol changes another via a condition test.


In Jacob's example, setting count = 0 changes the condition test outcome, and thus a or b on the LHS. Consequently, a or b on the RHS is supposed to change. However, RHS a equals the old LHS a, which was undefined because count>=20, and needs Update to be changed. RHS b behaves the same, but was not evaluated in SetDelayed because Evaluate occurs before SetDelayed, so count is unchanged, and RHS b evaluates to LHS b with count<20. If we now reset count=0, evaluating b will return {b}.


To illustrate, I modify the example to separate LHS and RHS. MMA is clever enough to automatically update LHS declared as a variable, so I have to make a function:


count=0;
ClearAll[LHS,RHS];
LHS[]/;(count++<20)={RHS};

RHS=Unevaluated@LHS[];

count=0;
RHS (* Equals LHS[] with count >= 20 *)

(* Tell Wolfram Language about changes affecting RHS which depends on LHS *)
Update@Unevaluated@LHS;
RHS



LHS[]


{{{{{{{{{{{{{{{{{{{{LHS[]}}}}}}}}}}}}}}}}}}}}



Comments

Popular posts from this blog

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

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

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

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.