How to make the for loop the length as the array inside the for loop?

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Matthew 2023-10-10

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2031294-how-to-make-the-for-loop-the-length-as-the-array-inside-the-for-loop

评论： Star Strider 2023-10-10

I am making a for loop and I want to subtract each element by each other. I was able to do this but my array is now 1x200 and my original array was 1x201. How do I make it so the for loop array is 1x201?

x1 = [1e6:20000:5e6];
L = 4e6; % in m
D_c = 5000;% in m
D1 = [0:25:5000];
x_o = 1e6; % in m
x_2 = [x_o:20000:5e6];
Tw_o = 300; % in K
LR_w = 0.004; % in K/m
LR_c = -0.007; % in K/m
pz_0 = 1000; % in mb
D_1 = D_c.*(sin((pi/L).*(x1-x_o)));
T_c2 = Tw_o-(((LR_w-LR_c).*D_1)-((LR_c.*D1)));
g = 9.81;
Rd = 287;
p1 = pz_0*(((((T_c2)-(LR_c.*D1))./(T_c2))).^((g)/(Rd*LR_c)));
for i = 1:length(p1)-1
    deltap(i) = p1(i+1)-p1(i);
end

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Answer 1

Star Strider 2023-10-10

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You do not need the loop at all. Just use the diff function —

x1 = [1e6:20000:5e6];
L = 4e6; % in m
D_c = 5000;% in m
D1 = [0:25:5000];
x_o = 1e6; % in m
x_2 = [x_o:20000:5e6];
Tw_o = 300; % in K
LR_w = 0.004; % in K/m
LR_c = -0.007; % in K/m
pz_0 = 1000; % in mb
D_1 = D_c.*(sin((pi/L).*(x1-x_o)));
T_c2 = Tw_o-(((LR_w-LR_c).*D_1)-((LR_c.*D1)));
g = 9.81;
Rd = 287;
p1 = pz_0*(((((T_c2)-(LR_c.*D1))./(T_c2))).^((g)/(Rd*LR_c)));
deltap = diff(p1)
deltap = 1×200
   -2.8534   -2.8634   -2.8734   -2.8835   -2.8935   -2.9036   -2.9136   -2.9237   -2.9337   -2.9437   -2.9536   -2.9635   -2.9733   -2.9831   -2.9927   -3.0023   -3.0118   -3.0212   -3.0305   -3.0396   -3.0486   -3.0574   -3.0661   -3.0746   -3.0830   -3.0911   -3.0990   -3.1068   -3.1143   -3.1215
for i = 1:length(p1)-1
    deltap(i) = p1(i+1)-p1(i);
end
deltap
deltap = 1×200
   -2.8534   -2.8634   -2.8734   -2.8835   -2.8935   -2.9036   -2.9136   -2.9237   -2.9337   -2.9437   -2.9536   -2.9635   -2.9733   -2.9831   -2.9927   -3.0023   -3.0118   -3.0212   -3.0305   -3.0396   -3.0486   -3.0574   -3.0661   -3.0746   -3.0830   -3.0911   -3.0990   -3.1068   -3.1143   -3.1215

If you want to calculate the numerical derivative at eash point instead, use the gradient function —

deltap = gradient(p1)
deltap = 1×201
   -2.8534   -2.8584   -2.8684   -2.8784   -2.8885   -2.8985   -2.9086   -2.9186   -2.9287   -2.9387   -2.9486   -2.9585   -2.9684   -2.9782   -2.9879   -2.9975   -3.0071   -3.0165   -3.0258   -3.0350   -3.0441   -3.0530   -3.0618   -3.0704   -3.0788   -3.0870   -3.0951   -3.1029   -3.1105   -3.1179

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Matthew 2023-10-10

Is there a way to get these values positive instead of negative? Thanks.

Star Strider 2023-10-10

One way would be to take the absolute value (the abs function), the other, since they are uniformly negative, would be to multiply them by -1 or just use a unary negative:

deltap = -gradient(p1)

Use whatever works best in your application.

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