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

 Akzeptierte Antwort

KSSV
KSSV am 23 Mai 2020

0 Stimmen

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 Kommentare

HYZ
HYZ am 23 Mai 2020
thank you!
HYZ
HYZ am 23 Mai 2020
Bearbeitet: HYZ am 23 Mai 2020
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.
KSSV
KSSV am 23 Mai 2020
What is expected output here?
HYZ
HYZ am 23 Mai 2020
Output will be { [7.5 8.5 9.5] [22.5 25.5 30]}. thanks
KSSV
KSSV am 23 Mai 2020
Have a look on knnsearch, ismembertol.
HYZ
HYZ am 23 Mai 2020
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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HYZ
am 23 Mai 2020

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HYZ
am 23 Mai 2020

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