Info

この質問は閉じられています。 編集または回答するには再度開いてください。

I don't know what is wrong with my code when recreating a table

1 回表示 (過去 30 日間)
Sarah Kneer
Sarah Kneer 2020 年 11 月 12 日
閉鎖済み: MATLAB Answer Bot 2021 年 8 月 20 日
There's this table I need to create, in which, depending on the time(t), the value of Clo and Met changes. My code is:
t = 0:0.5:24;
for i = 0:0.5:24
if t(i)>=0 && t(i)<8
Clo = 3;
Met = 0.4;
end
if t(i)>=8 && t(i)<12
Clo = 1;
Met = 1;
elseif t(i)>=12 && t(i)<17
Clo = 0.5;
Met = 1.6;
elseif t(i)>=17 && t(i)<22
Clo = 1;
Met = 0.9;
elseif t(i)>=22 && t(i)<24
Clo = 2.5;
Met = 0.4;
else
plot(t,Clo,'g');
xlabel ('Time (hours)');
ylabel ('Clo');
end
end

回答 (1 件)

Subhadeep Koley
Subhadeep Koley 2020 年 11 月 12 日
loop indices must be positive integers or logical values. In your code, you're trying to use fraction values as index. Also, your plot command is inside an else branch. The below modifications might help.
t = 0:0.5:24;
Clo = zeros(size(t));
Met = zeros(size(t));
for idx = 1:numel(t)
if t(idx)>=0 && t(idx)<8
Clo(idx) = 3;
Met(idx) = 0.4;
elseif t(idx)>=8 && t(idx)<12
Clo(idx) = 1;
Met(idx) = 1;
elseif t(idx)>=12 && t(idx)<17
Clo(idx) = 0.5;
Met(idx) = 1.6;
elseif t(idx)>=17 && t(idx)<22
Clo(idx) = 1;
Met(idx) = 0.9;
elseif t(idx)>=22 && t(idx)<=24
Clo(idx) = 2.5;
Met(idx) = 0.4;
end
end
plot(t, Clo,'g')
xlabel ('Time (hours)')
ylabel ('Clo')
figure
plot(t, Met,'r')
xlabel ('Time (hours)')
ylabel ('Met')
  6 件のコメント
Sarah Kneer
Sarah Kneer 2020 年 11 月 12 日
I got it know, I missed the (i) after Clo and Met. Thanks.
Peter Perkins
Peter Perkins 2020 年 11 月 19 日
It's also worth noting that the loop in this code is unnecessary:
t = 0:0.5:24;
Clo = zeros(size(t));
Met = zeros(size(t));
i = t>=0 & t<8
Clo(i) = 3;
Met(i) = 0.4;
i = t>=8 && t<12
Clo(i) = 1;
Met(i) = 1;
i = t>=12 && t<17
Clo(i) = 0.5;
Met(i) = 1.6;
i = t>=17 && t<22
Clo(i) = 1;
Met(i) = 0.9;
i = t>=22 && t<=24
Clo(i) = 2.5;
Met(i) = 0.4;

タグ

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by