How do I identify unique rows based on multiple columns and calculate the average of the rest of the columns?

51 vues (au cours des 30 derniers jours)
I have a matrix as below:
A = [1 4 3 8; 4 5 6 9; 1 6 3 6; 2 6 9 3; 1 5 3 7];
My goal is to identify rows with both identical Column 1 and identical Column 3 values, and then calculate the average for the rest of the columns, i.e., Column 2 and Column 4, within these duplicate rows. In this example, the duplicate rows would be Rows # 1, 3, and 5. My ending matrix would be:
B = [1 5 3 7; 4 5 6 9; 2 6 9 3];
This is a much simplified example. In reality, I have 35 columns that need to be averaged, and millions of rows. What is the most efficient way of handling this? Do I have to write a loop and process each of the unique rows individually?
Many thanks!

Réponse acceptée

Kevin Holly
Kevin Holly le 19 Oct 2021
Modifié(e) : Kevin Holly le 19 Oct 2021
A = [1 4 3 8; 4 5 6 9; 1 6 3 6; 2 6 9 3; 1 5 3 7]
A = 5×4
1 4 3 8 4 5 6 9 1 6 3 6 2 6 9 3 1 5 3 7
I am going to assume that any row that columns 1 and 3 are identical, irregardless of what pair, you want to ignore those rows in the averaging of other columns.
Here is my approach:
t = table(A(:,1),A(:,3))
t = 5×2 table
Var1 Var2 ____ ____ 1 3 4 6 1 3 2 9 1 3
[C, ia, ic] = unique(t,'rows')
C = 3×2 table
Var1 Var2 ____ ____ 1 3 2 9 4 6
ia = 3×1
1 4 2
ic = 5×1
1 3 1 2 1
ic==1
ans = 5×1 logical array
1 0 1 0 1
A(ic==1,:)
ans = 3×4
1 4 3 8 1 6 3 6 1 5 3 7
mean(A(ic==1,:))
ans = 1×4
1 5 3 7
B = [mean(A(ic==1,:));A(ic~=1,:)]
B = 3×4
1 5 3 7 4 5 6 9 2 6 9 3
Your answer:
B = [1 5 3 7; 4 5 6 9; 2 6 9 3]
B = 3×4
1 5 3 7 4 5 6 9 2 6 9 3
Without showing work:
t = table(A(:,1),A(:,3));
[~,~,ic] = unique(t,'rows');
B = [mean(A(ic==1,:));A(ic~=1,:)]
B = 3×4
1 5 3 7 4 5 6 9 2 6 9 3

Plus de réponses (0)

Catégories

En savoir plus sur Matrix Indexing dans Help Center et File Exchange

Produits


Version

R2021a

Community Treasure Hunt

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

Start Hunting!

Translated by