Using Accumarray with @maxk instead of @max?

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Ayman Al-Sukhon
Ayman Al-Sukhon am 16 Dez. 2019
Kommentiert: Ayman Al-Sukhon am 17 Dez. 2019
Hi,
Say you have three vectors:
a = [1;2;3;4;5;1;3;1;4];
b = [100;200;300;400;500;400;300;200;100];
c = [123;456;221;111;800;1000;10;25;150];
And you use accumarray so that:
maxval = sparse(accumarray(a(:,1),max(b,c),[],@max))
You get:
maxval =
(1,1) 1000
(2,1) 456
(3,1) 300
(4,1) 400
(5,1) 800
Now lets, say I have a whole lot of variables for each subs (something like 2000 each), and I want the average of the top 3 values using the same method. How can I accomplish this? For example, in the same problem, I want my output to be:
The top three values with subs 1 are: 100,400 and 200, so their average is 233.33 and the first row in my sparse matrix is:
maxval =
(1,1) 1000
and so on.
Is it maybe possible to use maxk as a function handle?

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 16 Dez. 2019
accumarray(a(:,1), max(b,c), [], @(v) mean(maxk(v,3)), 0, true) %final parameter is sparse flag
  3 Kommentare
dpb
dpb am 17 Dez. 2019
@() is the preamble to define an anonymous function. The v is the dummy argument variable name Walter chose; you'll see it reflected in the argument to maxk().
All the details about anonymous functions is in the documentation under the general subject of functions.
Ayman Al-Sukhon
Ayman Al-Sukhon am 17 Dez. 2019
I see, thank you very much dpb!

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Weitere Antworten (1)

dpb
dpb am 16 Dez. 2019
Bearbeitet: dpb am 16 Dez. 2019
Don't see anyway around with accumarray because the VAL parameter must be 1:1 with rows of SUBS; use findgroups/splitapply or varfun (altho latter must be table or timetable).
g=findgroups(a);
mnk=splitapply(@(x) mean(maxk(x,3)),b,g);
yields
>> mnk =
233.3333
200.0000
300.0000
250.0000
500.0000
>>
  7 Kommentare
dpb
dpb am 16 Dez. 2019
What it looked like to me, too, Walter.
Your's works for the specific instance (or anywhere sum is the needed intemediary); however in the followup here I was thinking of the more general cases where the function needs the elements rather than a single result.
Ayman Al-Sukhon
Ayman Al-Sukhon am 17 Dez. 2019
Just wanted to thank everyone else for the help. All the suggestions were good, however Walter's got right to the point for me.

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