Clear Filters
Clear Filters

compare between vector and cell

1 view (last 30 days)
skysky2000
skysky2000 on 30 Dec 2016
Commented: Stephen23 on 31 Dec 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

Accepted Answer

KSSV
KSSV on 30 Dec 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 Comment
skysky2000
skysky2000 on 30 Dec 2016
That perfect KSSV, I appreciate it .
Thankssss alot

Sign in to comment.

More Answers (2)

Stephen23
Stephen23 on 30 Dec 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 Comment
skysky2000
skysky2000 on 30 Dec 2016
Thanks Stephen, What about the second question?

Sign in to comment.


skysky2000
skysky2000 on 30 Dec 2016
Edited: Stephen23 on 30 Dec 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 Comment
Stephen23
Stephen23 on 30 Dec 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.

Sign in to comment.

Categories

Find more on Creating and Concatenating Matrices in Help Center and File Exchange

Community Treasure Hunt

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

Start Hunting!