Skip to main content

functions - How to Reap data from a ScheduledTask?


Here is an amusing little function mapped over a range just to iterate it. It ignores its argument and works entirely by side-effecting a global variable:


n$ = 1;

Map[(n$ = (n$++) + n$) &, Range[5]]


{3, 7, 15, 31, 63}

Here is a version of the same thing, just using Reap and Sow:


n$ = 1;
Reap[Map[Sow[n$ = (n$++) + n$] &, Range[5]]][[1]]



{3, 7, 15, 31, 63}

Here is a version that evaluates the function in an asynchronous task, capturing the result in a Dynamic (you'll have to key it in to a notebook, or evaluate the SEUploader (courtesy of @halirutan) notebook I linked at the end of this post, to see the variable n$ updating dynamically):


n$ = 1;
Dynamic[n$]
RunScheduledTask[n$ = (n$++) + n$, {0.25, 5}];


63


Clean up the task:


RemoveScheduledTask[ScheduledTasks[]]

Now, here is an attempt to also capture the results in a Reap (notice no semicolon at the end; I want to see the results of the Reap as well as to see the Dynamic):


n$ = 1;
Dynamic[n$]
Reap[RunScheduledTask[Sow[n$ = (n$++) + n$], {0.25, 5}]]


63


{ScheduledTaskObject[7, HoldForm[Sow[n$ = Increment[n$] + n$]], {0.25, 5},
Automatic, True, "AutoRemove" -> False, "EpilogFunction" :> Null,
"TaskGroup" -> "Global`"], {}}

and its cleanup


RemoveScheduledTask[ScheduledTasks[]]

Evidently, the Reap is evaluated too early and the sown results are lost. How can I get the intended results?


Here's the entire notebook, SE-Uploaded (just evaluate the following expression in a fresh notebook):



Import["http://goo.gl/NaH6rM"]["http://i.stack.imgur.com/x1zYS.png"]

Answer



You can't do this that way,


when you evaluate RunScheduledTask you are only sending a held procedure for scheduled evaluation to Kernel. But Reap[expr]:



gives the value of expr together with all expressions to which Sow has been applied during its evaluation.



RunScheduledTask is of course HoldFirst so Sow is not applied at this time.


You can put Reap inside but the you will need also some temporary variable to be able to retrive reaped value from scheduled evaluation.


Comments

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

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}]