Error for exceeding the loop

Hi,
I have d = {[5:10] [20:30]}; e = [1 2 7 8 9 11 12 15 16 22 25 30 35 40 45]; I want to extract values from e which falls into vectors of d.
Result = { [7 8 9] [22 25 30] } because 7, 8 and 9 are within 5 and 10; 22, 25 and 30 are within 20 and 30.
The code I tried gave what I want but when i == 3, the error happened. How can I stop j loop when all vectors in d have been checked. Thank you!
clc; clear; close all
d = {[5:10] [20:30]};
e = [1 2 7 8 9 11 12 15 16 22 25 30 35 40 45];
E = length(e);
D = length(d);
for i = 1:D
while j <= max(d{end})
for j = 1:E
if e(j) >= min(d{i})
a(i) = e(j);
for k = j+1:E
if e(k) > max(d{i})
b(i) = e(k-1);
want{i} = [e(j:k-1)];
if ~isempty(b(i))
j=k;
i=i+1;
break;
end
end
end
end
end
end
end

 采纳的回答

Why you have made code that complex? It is a simple task. You can use intersect.
d = {[5:10] [20:30]};
e = [1 2 7 8 9 11 12 15 16 22 25 30 35 40 45];
N = length(d) ;
result = cell(N,1) ;
for i = 1:N
iwant{i} = intersect(e,d{i}) ;
end

6 个评论

thank you!
HYZ
HYZ 2020-5-23
编辑:HYZ 2020-5-23
Sorry ... I didn't think far. my real data are something like
d = {[5:10] [20:30]}; e = [1 2 7.5 8.5 9.5 11 12 15 16 22.5 25.5 30 35 40 45];
the elements in e might not be exactly same as in d. they are juz within the range of each vector in d.
Could you advise? thanks.
What is expected output here?
Output will be { [7.5 8.5 9.5] [22.5 25.5 30]}. thanks
Have a look on knnsearch, ismembertol.
I used 'find' to do it. Anyway thanks. I learnt "intercept"
clc; clear; close all
d = {[5:10] [20:30]};
e = [1 2 7.5 8.5 9.5 11 12 15 16 22.5 25.5 30 35 40 45];
D = length(d);
iwant = cell (1,D);
for i = 1:D
loc {i} = find(e >= min(d{i}) & e <= max(d{i}));
iwant{i} = [e(loc{i})];
end

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