compare between vector and cell

2 vues (au cours des 30 derniers jours)
skysky2000
skysky2000 le 30 Déc 2016
Commenté : Stephen23 le 31 Déc 2016
Dear all, I have one vector (b) and one cell (a) as shown below: How I know how many times each number in vector b repeat it in a.
a=[{[1,9,79,3] [2,29,16,7,3] 3 [4,74,3] [5,73,79,3] [6,56,3] [7,3]}]
b=[ 79 3 74 10];
the expect result should be;
result= [ 2 7 1 0];
Thanks alot

Réponse acceptée

KSSV
KSSV le 30 Déc 2016
a=[{[1,9,79,3] [2,29,16,7,3] 3 [4,74,3] [5,73,79,3] [6,56,3] [7,3]}] ;
b=[ 79 3 74 10];
result= [ 2 7 1 0];
a_mat = cell2mat(a) ;
for i = 1:length(b)
result(i) = length(find(a_mat==b(i)))
end
  1 commentaire
skysky2000
skysky2000 le 30 Déc 2016
That perfect KSSV, I appreciate it .
Thankssss alot

Connectez-vous pour commenter.

Plus de réponses (2)

Stephen23
Stephen23 le 30 Déc 2016
A much simpler solution:
>> a = {[1,9,79,3],[2,29,16,7,3],3,[4,74,3],[5,73,79,3],[6,56,3],[7,3]};
>> b = [79,3,74,10];
>> sum(bsxfun(@eq,[a{:}],b(:)),2)
ans =
2
7
1
0
  1 commentaire
skysky2000
skysky2000 le 30 Déc 2016
Thanks Stephen, What about the second question?

Connectez-vous pour commenter.


skysky2000
skysky2000 le 30 Déc 2016
Modifié(e) : Stephen23 le 30 Déc 2016
Dear KSSV, another question. Can I take the cell that number repeat it on it with same loop. like for example:
a=[{[1,9,79,3] [2,29,16,7,3] 3 [4,74,3] [5,73,79,3] [6,56,3] [7,3]}] ;
b=[ 79 3 74 10];
result= [ 2 7 1 0];
cell-part 79= [{[1,9,79,3] [5,73,79,3]}]
cell_part 3 = [{[1,9,79,3] [2,29,16,7,3] 3 [4,74,3] [5,73,79,3] [6,56,3] [7,3]}];
cell_part 74 = {[4,74,3]}
cell_part 10 = {0};
thankss
  1 commentaire
Stephen23
Stephen23 le 30 Déc 2016
@skysky2000: that is not a good idea: naming variables dynamically will make your code slow, buggy, and hard to follow. Read this to know why:
A much better solution is to learn to use indexing, which is fast and efficient.

Connectez-vous pour commenter.

Catégories

En savoir plus sur Matrix Indexing 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