Using break/continue in a for loop
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Sherwin
am 30 Apr. 2022
Kommentiert: Image Analyst
am 30 Apr. 2022
I have this loop and I want the second loop to have steps with different sizes and I need the inner loop to end after each iteration and go to the next e:
N = 4;
ODN = [ 4 5 4 4];
NP = 17;
ODD = [10000 10000, 5000 5000];
for e = 1:N
for f = 1:ODN(e,1):NP
L = tp(f:ODN(e,1),1);
[X,Y] = min(L,[],1);
xpC(Y,1) = ODD(f,1);
continue
end
end
if I use continue it will continue the inner loop ( which I don't want and if I use break instead, it will reset f to 1. What can I do to get what I want?
0 Kommentare
Akzeptierte Antwort
Image Analyst
am 30 Apr. 2022
Reverse the order of the e anf f loops:
N = 4;
ODN = [ 4 5 4 4];
NP = 17;
ODD = [10000 10000, 5000 5000];
for f = 1:ODN(e,1):NP
for e = 1:N
L = tp(f:ODN(e,1),1);
[X,Y] = min(L,[],1);
xpC(Y,1) = ODD(f,1);
end
end
Now after it does this
L = tp(f:ODN(e,1),1);
[X,Y] = min(L,[],1);
xpC(Y,1) = ODD(f,1);
once, it will move to the next e in the list.
6 Kommentare
Image Analyst
am 30 Apr. 2022
You just need one loop to have them both increment in synchrony:
N = 4;
ODN = [ 4 5 4 4];
NP = 17;
ODD = [10000 10000; 5000 5000];
e = [1 2 3 4]
f = [1 5 10 14]
for k = 1 : length(f)
thisf = f(k);
thise = e(k);
L = tp(thisf:ODN(thise,1),1);
[X,Y] = min(L,[],1);
xpC(Y,1) = ODD(thisf,1);
end
Weitere Antworten (0)
Siehe auch
Kategorien
Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!