How to code NESTED CYCLES
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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"
採用された回答
Star Strider
2023 年 9 月 9 日
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.
.
0 件のコメント
その他の回答 (2 件)
Bruno Luong
2023 年 9 月 9 日
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
2 件のコメント
Bruno Luong
2023 年 9 月 9 日
編集済み: Bruno Luong
2023 年 9 月 9 日
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
2023 年 9 月 9 日
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
2 件のコメント
Image Analyst
2023 年 9 月 10 日
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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