Filter löschen
Filter löschen

Count the number of same elements in an array

367 Ansichten (letzte 30 Tage)
luca
luca am 10 Sep. 2019
Bearbeitet: Vitek Stepien am 14 Aug. 2021
Hi given a vector
V = [ 1 2 4 3 4 2 3 5 6 4 5 6 8 4 2 3 5 7 8 5 3 1 3 5 7 8 9 5 3 2 4 6 7 8]
I would like to count how many times the value 1,2,3,4,5,6,7,8,9 are repeated inside V, and obtain a vector that report this values:
C = [2 4 6 5 6 3 3 4 1]
where 1 is repeated 2 times, 2 is repetead 4 times, 3 is repeated 6 times and so on..

Akzeptierte Antwort

madhan ravi
madhan ravi am 10 Sep. 2019
Bearbeitet: madhan ravi am 10 Sep. 2019
[~,~,ix] = unique(V);
C = accumarray(ix,1).'
  5 Kommentare
Stephen23
Stephen23 am 10 Sep. 2019
Bearbeitet: Stephen23 am 10 Sep. 2019
@madhan ravi : this is really quite neat:
accumarray(V(:),1)
Simple idea which works well:
>> V = [2 2 3 4 5 6 7 7 8 8 9 9]
>> accumarray(V(:),1)
ans =
0
2
1
1
1
1
2
2
2
madhan ravi
madhan ravi am 10 Sep. 2019
Thank you Stephen :) !

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (3)

Stephen23
Stephen23 am 10 Sep. 2019
Bearbeitet: Stephen23 am 10 Sep. 2019
Your 1st example:
>> V = [ 1 2 4 3 4 2 3 5 6 4 5 6 8 4 2 3 5 7 8 5 3 1 3 5 7 8 9 5 3 2 4 6 7 8];
>> C = hist(V,1:max(V))
C =
2 4 6 5 6 3 3 4 1
Your 2nd example:
>> V = [2 2 3 4 5 6 7 7 8 8 9 9]
>> C = hist(V,1:max(V))
C =
0 2 1 1 1 1 2 2 2

Vitek Stepien
Vitek Stepien am 14 Aug. 2021
Bearbeitet: Vitek Stepien am 14 Aug. 2021
I found this function extremely useful, and doing exactly what you need:
V = [ 1 2 4 3 4 2 3 5 6 4 5 6 8 4 2 3 5 7 8 5 3 1 3 5 7 8 9 5 3 2 4 6 7 8];
[gc,grps] = groupcounts(V'); % <- need column vector here
grps'
ans = 1×9
1 2 3 4 5 6 7 8 9
gc'
ans = 1×9
2 4 6 5 6 3 3 4 1
Where grps lists the unique values in order, and gc provides the count of each unique values found in v.
This is very similar to madhan ravi's accumarray, but even simpler.
P.S. I turned gc and grps into row vectors only for compactness of the post, it's purely aesthetical. However groupcounts requires a column vector, not a row.

Hugo Diaz
Hugo Diaz am 28 Nov. 2020
I use sparse(V(:),V(:), 1) for large arrays with missing indices.

Kategorien

Mehr zu Creating and Concatenating Matrices finden Sie in Help Center und File Exchange

Produkte


Version

R2019a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by