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How to resolve Matrix dimensions error?

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gp
gp 2020-12-10
评论： gp 2020-12-12
Hi, May I know how to resolve this error? Thank You!
Error part Matrix dimensions must agree.
Error in MathCW (line 15) v = [ones(21,1) I2 I2.^2].\PN0;

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Paul Hoffrichter
Paul Hoffrichter 2020-12-10
I made the code more readable to me, and adjusted the dimensions which are in the annotations.
% Increment of current
I=0:0.5:(0.5*20); % 1x21
I2=transpose(I); % Power measurements 21x1
Po=I.*I*100; % 1x21
for experiment = 1:10 % Noise with sd 0
noise0=randn(1,21); % 1x21
PN0=Po+(0.*noise0); % 1x21
scatter(I,Po); hold on
scatter(I,PN0,'filled');hold on;
hold off
x = polyfit(I,PN0,2); % 1x3
A = [ones(21,1) I2 I2.^2]; % 21x3
PN0tr = PN0'; % 21x1
xx = A .\ PN0tr; % 21x3 .\ 21x1
% I has a 0 in it, so you are dividing by 0 - not good
% A .\ PN0tr is the matrix with elements: PN0tr(i,j) / A(i,j)
y = polyval(x,I); % 1x21
errDiff = Po-y; % ( 1x21 - 1x21 ) .^ 2
E =(errDiff).^2;
Em=mean(E);
disp( ['Em = ' num2str(Em) ])
end
Now the output is:
Em = 6.0104e-25
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Paul Hoffrichter
Paul Hoffrichter 2020-12-11
I and other experts were starting to help you in your other question. Can you tell me why it was deleted, or should I contact Mathworks directly?
gp
gp 2020-12-12
I have no idea why it was deleted. I just noticed that issues. I will now post a new question and hope you can help me on this. Thank you

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更多回答（3 个）

AVB
AVB 2020-12-10
Next time please make sure you copy the code block in your question using the 'Insert a line of code' option.
There were two issues.
  1. The matrix [ones(21,1) I2 I2.^2] is of 21x3 size hence PNO should have compatible size which means it should be of size 21x1
  2. The first argument in the polyval function should be polynomial coefficients (in descending powers) of an nth-degree polynomial. See polyval
Below is your updated code:
% Increment of current
I=0:0.5:(0.5*20);
I2=transpose(I);
% Power measurements
Po=I.*I*100;
for experiment = 1:10
% Noise with sd 0
noise0=randn(1,21);
PN0=Po+(0.*noise0);
scatter(I,Po);
hold on
scatter(I,PN0,'filled');
hold off
x = polyfit(I,PN0,2);
v = [ones(21,1) I2 I2.^2].\PN0';
y = polyval(x,I);
% Error
E =(Po-y).^2;
Em=mean(E);
end
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AVB
AVB 2020-12-10
did you change the polyval function arguments? should be ......
y = polyval(x,I);
If you want to keep your y as is as below,
y = polyval(I,x);
then do transpose on y while computing E
E =(Po-y').^2;
gp
gp 2020-12-10
Yup, I did changed and transpose it also

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Paul Hoffrichter
Paul Hoffrichter 2020-12-10
E =(Po-y).^2; % ( 1x21 - 1x3 ) .^ 2
is this what you want:
E =(Po-y').^2; % ( 1x21 -3x1 ) .^ 2
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gp
gp 2020-12-10
Is it used to get the same dimension size?
Paul Hoffrichter
Paul Hoffrichter 2020-12-10
编辑：Paul Hoffrichter 2020-12-10
Yes. However, I have already fixed a core problem, so now the dimensions are the same. But the above dimensions are interesting. I am pretty sure this would not have worked 10 years ago. Here is a simple example to illustrate some newer matrix/vector operations.
>> t % 1x4
t =
10 20 30 40
>> u % 7x1
u =
0
2
4
6
8
9
10
>> t - u
ans =
10 20 30 40
8 18 28 38
6 16 26 36
4 14 24 34
2 12 22 32
1 11 21 31
0 10 20 30
If u was also a row vector, then its dimensions would have to equal t's dimensions. For example:
s = % 4x1
5 5 5 5
>> t - s
ans =
5 15 25 35

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Paul Hoffrichter
Paul Hoffrichter 2020-12-10
y = polyval(I,x); % Error
Did you mean
y = polyval(x,I);

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