Skip to main content

save - Saving memoization to disk


I like using memoization (i.e. the construct myFunction[x_]:=myFunction[x]=...) when I have a heavy function that I need to re-evaluate on the same arguments. However, I find it frustrating that each time I quit the kernel(s), all the advantage goes lost.


Is there a way of saving the results? I can think of a very cumbersome way of doing it, such as this block upon defining our function


memo = If[FileExistsQ[FileNameJoin[{Directory[], "memo.mx"}]], Import["memo.mx"], {}];
myFunction[x_] := Module[{value = ...},
AppendTo[memo, "myFunction[" <> ToString[x] <> "]=" <> ToString[value]];
Export["memo.mx", Union@memo];

myFunction[x] = value]
Evaluate[ToExpression/@memo];

Is this okay, or is there a better (or even designated) way of doing this?



Answer



Not much different from your approach and maybe not the best/safest approach, but DumpSave helps a bit because at least you don't have to works with strings:


cacheFile = FileNameJoin[{$TemporaryDirectory, "fibonacciCache" <> ".mx"}];
If[FileExistsQ[cacheFile],
Get[cacheFile],
fibonacci[1] = 1;

fibonacci[2] = 1;
fibonacci[n_Integer] := Module[{},
fibonacci[n] = fibonacci[n - 1] + fibonacci[n - 2];
DumpSave[cacheFile, fibonacci];
fibonacci[n]
]
]

For example




In[3]:= fibonacci[500] // Timing


Out[3]= {0.562500, \ 1394232245616978801397243828704072839500702565876973072641089629483255\ 71622863290691557658876222521294125}



Now Quit[]-ing and reevaluating the previous cell:



In[3]:= fibonacci[500] // Timing


Out[3]= {0., \ 1394232245616978801397243828704072839500702565876973072641089629483255\ 71622863290691557658876222521294125}



Comments

Popular posts from this blog

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...

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...

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}]