How to map a function to a specfic level spec

Question

I'm rewriting a function into a manipulatable graphic and while I'm happily surprised with how concise functional programming can be, I'm having trouble mapping the anonymous function Norm[#[[2]]-#[[1]]] & on to the pairs of points represented below:

Generate the data

timeSeries = 
  Table[MapThread[List,{list,#}] &/@  NestList[
   MapAt[RandomChoice[{
    {#[[1]],#[[2]]+RandomVariate[NormalDistribution[0,sigma]]},
    {#[[1]]+RandomVariate[NormalDistribution[0,sigma]],#[[2]]}
    }] &,#,{#} &/@ RandomInteger[{1,Length@list},Ceiling[Length@list * (1-precision)]]] &,
list,20],{sigma,{5,10,15}}];

---

Output:

{Length@timeSeries,Length /@ timeSeries,Length /@ timeSeries[[1]], Length /@timeSeries[[1,1]],timeSeries[[1,1,1]]}

{3,{21,21,21},{30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30,30},{2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2},{{114686.,132379.},{114686.,132379.}}}


Norm[#[[2]]-#[[1]]] & /@ timeSeries[[1,9]]

{2.43774,0.,8.73786,0.,0.,0.,0.,0.,0.446609,3.23359,3.73703,0.,8.29204,5.33826,0.,0.,0.,6.65264,3.52166,3.80975,8.55396,1.92847,0.,0.,0.,0.,0.,0.,0.,0.}

I had assumed this was a relatively common type of preprocessing for data sets but couldn't find a solution online. I've tried several configurations of Map, MapThread, etc. on different levelspec but haven't stumbled upon a combination that works yet. Moving past trial-and-error does anyone have any suggestions for how to properly map the above stated function on to all the pairs of 2D points?

Update: Added more info. list is the initial list of 2D points.

Answer 1

It's helpful to include a computable version of your question. I will fill in the missing items as follows:

list = RandomReal[10, {30, 2}];
precision = .1;

SeedRandom[1]
timeSeries = Table[
    MapThread[List, {list,#}]& /@ NestList[
        MapAt[
            RandomChoice[{
                {#[[1]],#[[2]]+RandomVariate[NormalDistribution[0,sigma]]},
                {#[[1]]+RandomVariate[NormalDistribution[0,sigma]],#[[2]]}
            }]&,
            #,
            RandomInteger[{1, Length@list},{Ceiling[Length@list * (1-precision)], 1}]
        ]&,
        list,
        20
    ],
    {sigma, {5,10,15}}
];

(I also used RandomInteger[range, {len, 1}] instead of {#}& /@ RandomInteger[range, len]).

Now, you can get the norms you want with:

r1 = Apply[Norm @* Subtract, timeSeries, {3}];

or the equivalent:

r2 = Apply[EuclideanDistance, timeSeries, {3}];

r1 === r2

True

Compare this with your short example:

Norm[#[[2]]-#[[1]]]& /@ timeSeries[[1, 9]]
r1[[1, 9]]

{26.0051, 11.037, 8.0916, 8.23104, 7.98833, 12.2301, 11.7063, 9.0764, 5.33768, 4.82262, 12.9593, 21.5366, 4.0383, 14.8998, 4.38089, 14.8319, 12.7268, 9.46539, 21.2087, 12.4623, 10.3275, 13.3517, 8.42573, 1.181, 11.5442, 10.7561, 5.60134, 8.46001, 21.9047, 15.4427}

{26.0051, 11.037, 8.0916, 8.23104, 7.98833, 12.2301, 11.7063, 9.0764, 5.33768, 4.82262, 12.9593, 21.5366, 4.0383, 14.8998, 4.38089, 14.8319, 12.7268, 9.46539, 21.2087, 12.4623, 10.3275, 13.3517, 8.42573, 1.181, 11.5442, 10.7561, 5.60134, 8.46001, 21.9047, 15.4427}