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