Why Matlab does not plot "for loop" generated vector?

Question

Anne 2020-2-16

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/505722-why-matlab-does-not-plot-for-loop-generated-vector

评论： Walter Roberson 2020-2-19

I am quite a beginner and after hours of googling I just could not figure out why Matlab does not plot what needed. I can see only a blank.

My script is:

c = 7;

n=12;

for i = 0:1:n

u=i;

p_u = (2.3/c)*((u/c)^1.3)*exp(-(u/c)^2.3);

hours =8760*p_u*(1-0.09);

hold on;

plot(u, hours)

end

Thank you in advance!

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Answer 1

David Hill 2020-2-16

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/505722-why-matlab-does-not-plot-for-loop-generated-vector#answer_415771

在 MATLAB Online 中打开

Best to use arrays/vectors.

c=7;
n=12;
u=0:.1:n;
hours = (2.3/c)*((u/c).^1.3).*exp(-(u/c).^2.3)*(1-0.09)*8760;
plot(u,hours)

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Rik 2020-2-18

You can define u however you like. That is the benefit of writing code like this.

Anne 2020-2-18

I found a mistake. I didn't see that there was a dot.

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Answer 2

Anne 2020-2-18

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/505722-why-matlab-does-not-plot-for-loop-generated-vector#answer_416249

It is a pity that when I need to use Matlab fast, then I will have errors all the time.

This code is giving 1x11176 vector for power and I don't know why. It should be 1x151 vector. because i=0:0.1:15 wll be vector with 151 elements. It worked well when I chose i=0:15

n=15;

uu=0:0.1:n;

u_cutin=1;

u_N=3;

u_cutout=7;

ii=1

for i=0:0.1:n

if i<u_cutin

power1(ii)=0

elseif i>=u_N && i<=u_cutout

power1(ii)=0.42*0.718*0.5*1.2*pi*((38/2)^2)*u_N^3

elseif i>u_cutout

power1(ii)=0

else

power1(ii)=0.42*0.718*0.5*1.2*pi*((38/2)^2)*i^3

end

ii=ii+i*10

end

plot(uu,power1)

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Walter Roberson 2020-2-19

You do not initialize power in this code.

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