Skip to main content

list manipulation - How to make MapAt work with Span?


Span (;;) is very useful, but doesn't work with a lot of functions. Given the following input


list = {{"a", "b", "c"}, {"d", "e", "f", 

   "g"}, {"h", {{"i", "j"}, {"k", "l"}, {"m", "n"}, {"o", "pp"}}}}

We would like


MapAt[Framed, list, 1 ;; 2]
MapAt[Framed, list, {{1, 1}, {2, 2 ;; 3}, {3, 2, 1 ;; 3, 1}}]

to work as expected


enter image description here


Here is my first go at it:


SpanToRange[Span[x_:1,y_:1,z_:1]] := Module[{zNew = z},

    If[x>y && z==1, zNew = -1];
        Range[x, y, zNew]
    ] /; And[VectorQ[{z,y,z}, IntegerQ],
    And @@ Thread[{z,y,z} != 0]]

helper = Function[list,
    Module[{li=list},
        If[FreeQ[li, Span], li,
        li = Replace[li,s_ /; Head[s] =!= Span :> {s}, {1}];
        li = li /. s:_Span :> SpanToRange[s];

        Sequence @@ Flatten[
            Outer[List, Sequence @@ li],
            Depth[Outer[List, Sequence @@ li]]-3]]
    ]
];

protected = Unprotect[Span, MapAt];
Span /: MapAt[func_, list_, s:Span[x_:1,y_:1,z_:1]]:= MapAt[func, 
 list, Thread[{SpanToRange[s]}]];
MapAt[func_, list_, partspec_] /; !FreeQ[partspec, Span] := Module[{f,p = partspec},

    MapAt[func, list, Join[helper /@ p]]
];
Protect[Evaluate[protected]];

But this is far from finished, and the extended down values should support all valid uses of Span such as


MapAt[Framed, list, 3 ;;]
MapAt[Framed, list, ;; ;; 2]
MapAt[Framed, list, ;; 10 ;; 2]

Answer



There is a hidden update in V9: MapAt works with Span.



I've checked it does not work on V8 and V7.


enter image description here


I just started to do this once in the past and it worked. I was newbie in Mathematica when there was V8 or V7 so I have not realised it is new till Mr. Wizard poited out in comments that I'm smoking crack :).


I do not remember other case but it is the second, which I can recall, where there is no mark about this in documentation. I do not mean examples, I mean there is no "Last modyfied in 9" for MapAt only "New in 1.".


Couple of examples where I've used it:




I strongly recommend this, it is so handy, and, as Mr. Wizard noticed, fast!


big = Range[1*^5];
First@Timing@MapAt[#^2 &, big, List /@ Range[30000, 40000]] 

First@Timing@MapAt[#^2 &, big, 30000 ;; 40000]


10.202465
0.015600

Extended comparision inspired by RunnyKine:


test = {};
Do[ big = Range[10^i];
    AppendTo[test,

             {i,
              Mean@Last@Last@Reap@Do[
                  Sow@First@Timing@MapAt[#^2 &, big, List /@ Range[3000, 4000]], {10}],
              Mean@Last@Last@Reap@Do[
                  Sow@First@Timing@MapAt[#^2 &, big, 3000 ;; 4000], {10}]
             }]
  , {i, 5, 6.4, .2}]

ListLogPlot[Transpose[test][[2 ;;]], Joined -> True, DataRange -> {5, 6.4}]


enter image description here


Comments

Post a Comment

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...
Post a Comment
Read more

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...
Post a Comment
Read more

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more