       Re: Efficient function to accumulate a list of {value,coord} into array

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• Subject: [mg132507] Re: Efficient function to accumulate a list of {value,coord} into array
• From: "Dr. Wolfgang Hintze" <weh at snafu.de>
• Date: Fri, 4 Apr 2014 03:55:38 -0400 (EDT)
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```schrieb im Newsbeitrag news:lhiudg\$dij\$1 at smc.vnet.net...

Dear all,

I am struggling with how to convert a list of (value, coords) tuples into an
array such that the an element in the array should represent the sum of all
value elements in that list with matching coords, and zero if no matches,
e.g.

{ {3.6, 7,4}, {3,4, 8,6}, {2.1, 7,4} }

if it were a 10X10 element array would become all zero's, except element
(7,4)->(3.6+2.1) and (8,6)->3.4

I've got some code which does what I want:

ParticleToDensity[particles_] := (
arr = ConstantArray[0, {144, 300}];
For[l = 1, l < NoParticles, l++,
(tp = particles[[l]];
If[tp[] != 0,
arr[[Round[tp[]], Round[tp[]]]] += tp[]]
)];
arr)

So, the argument to the function is a list of "particles", and the first
element of this particle is the value in the array I'd like to accumulate,
and the next two elements in that particle say where in the array they
should be accumulated.

It seems to work, but it is slow. I can see that this isn't really the
Mathematica way of doing things, but I'm struggling to think of a better
way.

For reference, NoParticles is typically set to something like 10,000, and it
takes about 0.1 secs to process on my machine. However this needs to be done
frequently, hence I am hoping for a better way.

Any help greatly appreciated.

Thanks,
Julian.

Julian,
try this function (I don't know how the speed compares to your code)

acum = {#[], Plus @@ Drop[#, 1]} &[
Sort[Union[Flatten[#, 1]], OrderedQ[{#2, #1}] &]] & /@
Split[Sort[{{#[], #[]}, #[]} & /@ #],
First[#1] == First[#2] &] &;

Example

x = {{3.6, 7, 4}, {3.4, 8, 6}, {2.1, 7, 4}};

acum[x]
{{{7, 4}, 5.7}, {{8, 6}, 3.4}}

The format is a bit different from the flat elements, but in this way it can
be treated easier by Mathematica functions.
You could apply

In:= Flatten[{#[], #[]}] & /@ %
Out= {{5.7, 7, 4}, {3.4, 8, 6}}

Best regards,
Wolfgang

```

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