take part of cell depend on vector
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skysky2000
am 31 Dez. 2016
Kommentiert: skysky2000
am 2 Jan. 2017
Dear all, I have cell (a) and vector (b), I wanna check each number in b that repeat it in a and take each cell from (a) make it as vector by loop.
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];
results should be as follow:
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};
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Akzeptierte Antwort
Stephen23
am 31 Dez. 2016
Bearbeitet: Stephen23
am 31 Dez. 2016
As you were already told in your last question, this 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:
>> 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];
>> idc = cellfun(@(x)any(bsxfun(@eq,x,b(:)),2),a,'Uni',0);
>> out = cellfun(@(x)[a{x}],num2cell([idc{:}],2),'Uni',0);
which gives exactly the output vectors that you specified, inside the cell array out:
>> out{1}
ans =
1 9 79 3 5 73 79 3
>> out{2}
ans =
Columns 1 through 15
1 9 79 3 2 29 16 7 3 3 4 74 3 5 73
Columns 16 through 22
79 3 6 56 3 7 3
>> out{3}
ans =
4 74 3
>> out{4}
ans =
[]
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