# How to write a code for 3 Variables

2 vues (au cours des 30 derniers jours)
Gokhan Kayan le 25 Mar 2018
Commenté : Gokhan Kayan le 27 Mar 2018
I want to write a code to obtain a new value array that is dependent to 3 variables. My 3 variables are like the table shown below:
A B C
1 400 200
1 500 100
2 455 300
2 200 400
3 100 500
3 500 600
2 600 100
3 700 900
1 800 150
2 900 150
I want to calculate new 'D' array from A,B,C. My code should be like this if A=1 then calculate the sum of B and divide it to sum of C.
For example we have 3 row that A=1 and we have B=400,500,900 C=200,100,150 for A=1. So it shoulde be 400 +500+900 /200+100+150. The result (D) is 4 for A=1. I have so many A values and ı don't know how to calculate all of them. If you help me, I will be very happy. Thank you.
##### 0 commentairesAfficher -2 commentaires plus anciensMasquer -2 commentaires plus anciens

Connectez-vous pour commenter.

### Réponse acceptée

John D'Errico le 25 Mar 2018
Modifié(e) : John D'Errico le 25 Mar 2018
Easy peasy. What, 2 lines?
abc = [1 400 200
1 500 100
2 455 300
2 200 400
3 100 500
3 500 600
2 600 100
3 700 900
1 800 150
2 900 150];
[ai,bcsum] = consolidator(abc(:,1),abc(:,2:3),@sum)
ai =
1
2
3
bcsum =
1700 450
2155 950
1300 2000
bcratio = bcsum(:,1)./bcsum(:,2)
bcratio =
3.7778
2.2684
0.65
There is no need for the A values to be integers. As long as they are distinct will suffice.
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
Gokhan Kayan le 27 Mar 2018
Thank you very much John this is what I exactly want :)

Connectez-vous pour commenter.

### Plus de réponses (2)

Walter Roberson le 25 Mar 2018
sumB = accumarray(A, B);
sumC = accumarray(A, C);
D = sumB ./ sumC;
D(sumB == 0 & sumC == 0) = 0;
This applies directly only if the A values are positive integers, preferably small and consecutive. If they are not positive integers then there is an adjustment that can be made using unique()
The final setting to 0 is for the case where the sum of B and sum of C are both 0, replacing the NaN that would result with 0. The sums could be 0 if the entries can be positive and negative; the sums can also be 0 if there is a gap in the values of A, such as if A might be [1, 2, 4] with no 3 entry.
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
Gokhan Kayan le 27 Mar 2018
Thanks Walter :)

Connectez-vous pour commenter.

Geoff Hayes le 25 Mar 2018
Gokhan - you can do
A==1
to return an array of logical values, zeros and ones, that will tell you which element of A is a one (indicated by a one) and which element of A is not a one (indicated by a zero). For example,
A = [1 2 3 4 5 1 1]
then
A==1
returns
1 0 0 0 0 1 1
And so you could then do
sum(B(A==1))
to sum those elements of B that correspond to ones within A. See logical indexing for more details.
##### 1 commentaireAfficher -1 commentaires plus anciensMasquer -1 commentaires plus anciens
Gokhan Kayan le 27 Mar 2018
Thanks for your respond Geoff :)

Connectez-vous pour commenter.

### Catégories

En savoir plus sur Logical dans Help Center et File Exchange

### Community Treasure Hunt

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

Start Hunting!

Translated by