Matlab should have give array but it gives a number. How to fix?

Question

Ege Tunç 2019-3-16

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/450447-matlab-should-have-give-array-but-it-gives-a-number-how-to-fix

编辑： Jan 2019-3-18

I have a code to calculate some lambda values then uses these values with our r values to find d values. However, these values are 1x21 array but the given answer is just a number why? and how to fix this?

U=[10 9.7 9.4 9.1 8.8 8.5 8.2 7.9 7.6 7.3 7.0 6.7 6.4 6.1 5.8 5.5 5.2 4.9 4.6 4.3 4.0];
h=6.625.*10.^(-34);
m=9.109.*10.^(-31);
e=1.602.*10.^(-19);
lambda=h./sqrt(2.*m.*e.*U.*1000);
R=65.*10.^(-3);
r1i=[0.7 0.75 0.8 0.85 0.9 0.9 1.05 1.05 1.1 1.1 1.1 1.15 1.15 1.10 1.10 1.10 1.10 1.15 1.10 1.15 1.2];
r1o=[1.2 1.3 1.35 1.4 1.4 1.45 1.5 1.55 1.55 1.6 1.6 1.67 1.67 1.75 1.75 1.8 1.9 1.9 2 2.05 2.1];
r2i=[0.75 0.75 0.85 0.95 1 1 1.05 1.1 1.1 1.1 1.1 1.15 1.2 1.2 1.25 1.25 1.25 1.3 1.3 1.3 1.35];
r2o=[1.25 1.3 1.4 1.4 1.45 1.45 1.5 1.55 1.6 1.65 1.65 1.7 1.7 1.75 1.85 1.9 1.9 1.95 2 2.05 2.1];
r3i=[0.7 0.75 0.75 0.8 0.85 0.85 0.9 0.95 1 1 1 1 1.05 1.05 1.05 1.1 1.1 1.15 1.15 1.15 1.2];
r3o=[1.2 1.25 1.3 1.35 1.4 1.45 1.5 1.5 1.55 1.6 1.6 1.65 1.7 1.7 1.75 1.8 1.85 1.9 1.95 2 2.05];
d1i=(2*R.*lambda)/(10.^(-2).*r1i);
d1o=(2*R.*lambda)/(10.^(-2).*r1o);
d2i=(2*R.*lambda)/(10.^(-2).*r2i);
d2o=(2*R.*lambda)/(10.^(-2).*r2o);
d3i=(2*R.*lambda)/(10.^(-2).*r3i);
d3o=(2*R.*lambda)/(10.^(-2).*r3o);
dinner=(d1i+d2i+d3i)/3
douter=(d1o+d2o+d3o)/3

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Answer 1

Jan 2019-3-16

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/450447-matlab-should-have-give-array-but-it-gives-a-number-how-to-fix#answer_365711

编辑：Jan 2019-3-18

在 MATLAB Online 中打开

Use the elementwise division:

d1i=(2*R.*lambda) ./ (10.^(-2).*r1i);
%                 ^ 

You have used the elementwise operations exhaustively, even if they are not needed, because the operands are scalars only. But at the only location, where it is needed, the matrix division / is applied. I'd use:

d1i = 2 * R * lambda ./ (1e-2 * r1i);
% or even nicer:
d1i = 200 * R * lambda ./ r1i;

By the way, compare

h=6.625.*10.^(-34);

with

h = 6.625e-34;

The first is an expensive power operation, while the second is a cheap constant - and nicer. Then your code becomes:

U = [10 9.7 9.4 9.1 8.8 8.5 8.2 7.9 7.6 7.3 7.0 6.7 6.4 6.1 5.8 5.5 5.2 4.9 4.6 4.3 4.0];
h = 6.625e-34;
m = 9.109e-31;
e = 1.602e-19;
lambda = h ./ sqrt(2 * m * e * U * 1000);
R = 65e-3;
ri = [0.7 0.75 0.8 0.85 0.9 0.9 1.05 1.05 1.1 1.1 1.1 1.15 1.15 1.10 1.10 1.10 1.10 1.15 1.10 1.15 1.2; ...
      0.75 0.75 0.85 0.95 1 1 1.05 1.1 1.1 1.1 1.1 1.15 1.2 1.2 1.25 1.25 1.25 1.3 1.3 1.3 1.35; ...
      0.7 0.75 0.75 0.8 0.85 0.85 0.9 0.95 1 1 1 1 1.05 1.05 1.05 1.1 1.1 1.15 1.15 1.15 1.2];
ro = [1.2 1.3 1.35 1.4 1.4 1.45 1.5 1.55 1.55 1.6 1.6 1.67 1.67 1.75 1.75 1.8 1.9 1.9 2 2.05 2.1; ...
     1.25 1.3 1.4 1.4 1.45 1.45 1.5 1.55 1.6 1.65 1.65 1.7 1.7 1.75 1.85 1.9 1.9 1.95 2 2.05 2.1; ...
     1.2 1.25 1.3 1.35 1.4 1.45 1.5 1.5 1.55 1.6 1.6 1.65 1.7 1.7 1.75 1.8 1.85 1.9 1.95 2 2.05];
di = 200 * R * lambda ./ ri;   % Auto-expand, >= R2016b
do = 200 * R * lambda ./ ro;   % Auto-expand, >= R2016b
dinner = mean(di, 1)
douter = mean(do, 1)