Filter löschen
Filter löschen

Find min and max in groups

6 Ansichten (letzte 30 Tage)
Sergio Rojas Blanco
Sergio Rojas Blanco am 14 Mai 2024
I have 2 vectors of equal dimension, A and B. To each value of A corresponds a value of B, but equal values of A can correspond to different values of B.
I need to obtain a cell array, "Min", whose cells are vectors with the indexes of the minimum values of B and another cell array, "Max", with the maximum values. The position in each cell array is indicated by the value of A.
For example, if
A = [12, 11, 12, 5, 3, 10, 5]; B = [0.1, 0.1, 0.1, 0.7, 0.4, 0.3, 0.2]
Then:
Min = {0 ,0 ,5 ,0 ,4 ,0 ,0 ,0 ,0 ,6 ,2 ,(1, 3)}; Max = {0 ,0 ,5 ,0 ,4 ,0 ,0 ,0 ,0 ,0 ,0 ,2 ,(1, 3)}
Because of:
a_1 = 12 ; b_1 = 0.1
a_3 = 12 ; b_3 = 0.1
min(0.1, 0,1) = 0.1 => C_12 = (1, 3); max(0.1, 0.1) = 0.1 => D_12 = (1, 3)
a_2 = 11 ; b_2 = 0.1
min(0.1) = 0.1 => C_11 = (2); max(0.1) = 0.1 => D_11 = 2
...
Intiutively, it seems that it could be done in batch using groups.
Would anyone know how to do it? Thanks in advance

Akzeptierte Antwort

Stephen23
Stephen23 am 14 Mai 2024
Bearbeitet: Stephen23 am 14 Mai 2024
A = [12, 11, 12, 5, 3, 10, 5];
B = [0.1, 0.1, 0.1, 0.7, 0.4, 0.3, 0.2];
X = 1:numel(A);
Fmax = @(x){x(max(B(x))==B(x))};
Fmin = @(x){x(min(B(x))==B(x))};
Cmax = accumarray(A(:),X(:),[],Fmax,{0});
Cmax{:}
ans = 0
ans = 0
ans = 5
ans = 0
ans = 4
ans = 0
ans = 0
ans = 0
ans = 0
ans = 6
ans = 2
ans = 2x1
1 3
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
Cmin = accumarray(A(:),X(:),[],Fmin,{0});
Cmin{:}
ans = 0
ans = 0
ans = 5
ans = 0
ans = 7
ans = 0
ans = 0
ans = 0
ans = 0
ans = 6
ans = 2
ans = 2x1
1 3
<mw-icon class=""></mw-icon>
<mw-icon class=""></mw-icon>
  1 Kommentar
Sergio Rojas Blanco
Sergio Rojas Blanco am 15 Mai 2024
Thank you very much. it works perfectly and elegantly!
You answer everything :D

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Shifting and Sorting Matrices finden Sie in Help Center und File Exchange

Tags

Produkte


Version

R2024a

Community Treasure Hunt

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

Start Hunting!

Translated by