How to code NESTED CYCLES
조회 수: 1 (최근 30일)
이전 댓글 표시
f=[3 6 3 9]
b=[5 8 10 12]
a is a function with parameter f and b!
for l=1:numel(b)
for i=1:numel(f)
a(f(i),b(l))
end
end
but this is a problem:
if b is empty the function won't loop me with "f"
채택된 답변
Star Strider
2023년 9월 9일
I would set the empty array to 1 (so that it iterates one time only) and be done with it —
a = @(x,y) [x y];
f = [3 6 3 9]; % Neither Empty
b = [5 8 10 12];
if isempty(f)
f = NaN;
elseif isempty(b)
b = NaN;
end
for l=1:numel(b)
for i=1:numel(f)
q = a(f(i),b(l))
end
end
f = []; % 'f' Empty
b = [5 8 10 12];
if isempty(f)
f = NaN;
elseif isempty(b)
b = NaN;
end
for l=1:numel(b)
for i=1:numel(f)
q = a(f(i),b(l))
end
end
f = [3 6 3 9]; % 'b' Empty
b = [];
if isempty(f)
f = NaN;
elseif isempty(b)
b = NaN;
end
for l=1:numel(b)
for i=1:numel(f)
q = a(f(i),b(l))
end
end
The ‘default’ value (here ‘NaN’) can be any value that makes sense in the context of whatever ‘a’ does. If a given argument is added, then set it to the identity element for that operation, so if it is added set it to 0 if the associated vector is empty, if it is multiplied, set it to 1.
.
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추가 답변 (2개)
Bruno Luong
2023년 9월 9일
Reverse the 2 loops, then it loops on f
f=[3 6 3 9]
b=[]
for i=1:numel(f)
f(i)
for l=1:numel(b)
a(f(i),b(l))
end
end
댓글 수: 2
Bruno Luong
2023년 9월 9일
편집: Bruno Luong
2023년 9월 9일
Split in three loops if you have to
for i=1:numel(f)
for l=1:numel(b)
% ... do something with both f(i) and b(l)
a(f(i),b(l))
end
end
for i=1:numel(f)
% ... do something with f(i) ALONE
end
for l=1:numel(b)
% ... do something with b(l) ALONE
end
Image Analyst
2023년 9월 9일
Check them in advance:
if isempty(b)
% What to do
return;
end
if isempty(f)
% What to do
return;
end
for l=1:numel(b)
for i=1:numel(f)
something = a(f(i),b(l))
end
end
댓글 수: 2
Image Analyst
2023년 9월 10일
Correct. Why would you want to continue if one is empty? If you do you'll just get an error. If you can "fix" the situation, do so inside the if -- for example Star set them to nan instead. If you can't fix the situation, you should just exit the code after alerting the user.
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