DOUBLE cannot convert the input expression into a double array

Question

xueqi 2013-3-3

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/65703-double-cannot-convert-the-input-expression-into-a-double-array

This is the code. Could anyone run it and give me some suggestion about the error 'DOUBLE cannot convert the input expression into a double array' ? Cheers!

if true % DD =

    1.2000    0.6000    1.6000    0.5000    1.3000    1.4000  100.0000
    1.2000    0.6000    1.6000    0.5000    1.3000    1.6000  100.0000
    1.2000    0.6000    1.6000    0.5000    1.3000    1.8000  100.0000
    1.2000    0.8000    1.6000    0.7000    1.2000    1.4000  100.0000
    1.2000    0.6000    1.4000    0.7000    1.2000    1.4000  100.0000
    1.3000    0.6000    1.6000    0.7000    1.3000    1.4000  100.0000
    1.3000    0.6000    1.6000    0.7000    1.3000    1.4000  100.0000
    1.4000    0.4000    1.6000    0.6000    1.5000    1.4000  100.0000
    1.4000    1.6000    0.4000    0.8000    1.1000    1.2000  100.0000
    1.4000    1.8000    0.4000    0.8000    1.1000    1.2000  100.0000
    1.4000    2.0000    0.4000    0.8000    1.1000    1.2000  100.0000
    1.2000    2.2000    0.3000    0.8000    1.1000    1.2000  100.0000
    1.5000    1.6000    0.4000    0.6000    1.1000    1.3000  100.0000
    1.5000    1.3000    0.4000    0.6000    1.5000    1.3000  100.0000
    1.2000    2.0000    0.7000    0.8000    1.1000    1.2000  100.0000
    1.1000    2.0000    0.7000    0.8000    1.1000    1.2000  100.0000
    0.5000    1.3000    1.5000    1.4000    0.6000    0.8000  100.0000
    0.4000    1.5000    1.5000    1.4000    0.6000    0.8000  100.0000
    1.1000    0.5000    1.5000    0.4000    1.2000    1.5000  100.0000
    1.1000    0.5000    1.5000    0.4000    1.2000    1.7000  100.0000
    1.1000    0.5000    1.5000    0.6000    1.1000    1.3000  100.0000
    1.3000    0.3000    1.5000    0.5000    1.4000    1.3000  100.0000
    1.1000    0.5000    1.6000    0.4000    1.2000    1.5000  100.0000
    1.1000    0.5000    1.7000    0.4000    1.2000    1.5000  100.0000
    1.1000    0.6000    1.5000    0.4000    1.2000    1.5000  100.0000
    1.3000    0.5000    1.5000    0.4000    1.2000    1.5000  100.0000
    1.1000    0.5000    1.6000    0.4000    1.2000    1.5000  100.0000
    1.1000    0.8000    1.6000    0.7000    1.2000    1.5000  100.0000
    1.1000    0.5000    1.5000    0.4000    1.4000    1.7000  100.0000
    1.1000    0.6000    1.5000    0.4000    1.2000    1.7000  100.0000
    1.1000    0.5000    1.5000    0.4000    1.2000    1.7000  100.0000
    1.1000    0.5000    1.5000    0.4000    1.2000    1.7000  100.0000
    1.1000    0.5000    1.5000    0.8000    1.2000    1.7000  100.0000
    1.1000    0.5000    1.5000    0.4000    1.2000    1.7000  100.0000
    1.1000    0.9000    1.5000    0.4000    1.2000    1.7000  100.0000
    1.8000    0.5000    1.5000    0.4000    1.2000    1.7000  100.0000
    0.9000    0.5000    1.5000    0.8000    1.2000    1.3000  100.0000
    1.1000    0.6000    1.5000    0.8000    1.2000    1.7000  100.0000
    0.8000    0.5000    1.8000    0.8000    1.2000    1.7000  100.0000
rn=size(DD,1);
sub=[0.25,0.25,0.25,0.08]
p1=sub(1);
p2=sub(2);
p3=sub(3);
r=sub(4);
eutotal=zeros(rn,5);
eu=zeros(1,rn);
 for i=1:rn;
    d1=DD(i,1:3)-1;
d2=DD(i,4:6)-1;
edw=DD(i,7);
syms x;
 %f=@(x) 
 f=p1*(d1(1,1)-d2(1,1))*exp(-r*x*(d1(1,1)-d2(1,1)))...
    +p2*(d1(1,2)-d2(1,2))*exp(-r*x*(d1(1,2)-d2(1,2)))...
    +p3*(d1(1,3)-d2(1,3))*exp(-r*x*(d1(1,3)-d2(1,3)));
y=solve(f,x)
isempty(y);
if isempty(y)
   PF(1,1)=10;
else
PF=double(y);
end
 end
end

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Walter Roberson 2013-3-4

After the solve(), please show use "y", and show class() and size() of all the names that are mentioned in y (that is, the ones that would be affected by the subs())

xueqi 2013-3-4

It shows y = 125*log(z1) + 250*pi*k*i but doesnt give any information about z1,pi,k. I think it is becasue of that they are produced after the code is run... Can you have a look at the whole code I comment below?

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Answer 1

Walter Roberson 2013-3-4

1
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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/65703-double-cannot-convert-the-input-expression-into-a-double-array#answer_77317

在 MATLAB Online 中打开

Okay, I think I understand what is going on.

Sometimes MuPAD (the symbolic engine) creates expressions along the lines of

... * z1 .... where z1 = ...

That is, it extracts a complicated sub-expression and gives it a name, and then at the end says "where" and gives a definition for the sub-expression.

Unfortunately when this happens, the MATLAB interface drops the clause defining the subexpression, leaving a z* variable that did not exist in the original expression. This is, of course, a bug. And it seems to happen more in more recent versions (including in R2012a)

If you start up a MuPAD notepad with the "mupad" command, and solve() the expression inside it, you will see MuPAD's fuller answer.

In the case of the "f" you show above, general solution is

125*ln(RootOf(7*z^14-2*z^5-7,z))

where RootOf(7*z^14-2*z^5-7,z) means all of the values of "z" such that 7*z^14-2*z^5-7 comes out as 0 (the "roots" of the expression).

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Walter Roberson 2013-3-6

在 MATLAB Online 中打开

If you want the analytic solution then you need to use solve(), through MATLAB (which will call MuPAD for you), or more directly by calling to MuPAD, or by buying MAPLE and installing it as an alternative symbolic toolbox.

When you use solve() from MATLAB and it calls MuPAD for you, MATLAB does not correctly retrieve the full solution: this is a bug in the MATLAB interface to MuPAD. I do not know of any way to correct that problem, but if you open a technical support case they might be able to tell you a way.

If you use a MuPAD notebook, you can get the full solution there. I believe there are ways within MuPAD to export solutions in forms that MATLAB can read. This route would not usually be convenient.

There is a possibility that if you were to use

feval(symengine, 'solve', exp(-(2*x)/125)/20 + (7*exp(-(7*x)/125))/40 - (7*exp((7*x)/125))/40, x)

that you might get the complete solution. I make no promise about that.

What are you going to do with the explicit solution? You take it and promptly apply double() to it, which converts it into a numeric solution instead of a formula. Why not just go for a numeric solution anyhow?

The general form of the explicit solution is:

RootOf(exp(z) * p3 * d1(1,3) - exp(z) * p3 * d2(1,3) + p1 * exp(z * (d1(1,1) - d2(1,1)) / (d1(1,3) - d2(1,3))) * d1(1,1) - p1 * exp(z * (d1(1,1) - d2(1,1)) / (d1(1,3) - d2(1,3))) * d2(1,1) + p2 * exp(z * (d1(1,2) - d2(1,2)) / (d1(1,3) - d2(1,3))) * d1(1,2) - p2 * exp(z * (d1(1,2) - d2(1,2)) / (d1(1,3) - d2(1,3))) * d2(1,2), z) / r / (-d1(1,3) + d2(1,3));

RootOf( expression, z ) means the values of z that make the expression 0 -- the roots of the expression. There is no known closed form solution for the roots of expressions in that particular form, so unless you wish to do some kind of further symbolic analysis on the expression, you are going to have to convert to numeric form anyhow, so you might as well have used numeric form to start with.

xueqi 2013-3-7

This is the answer. Thanks!

xueqi 2013-3-7

And it makes my program much much faster!

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Answer 2

Azzi Abdelmalek 2013-3-3

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/65703-double-cannot-convert-the-input-expression-into-a-double-array#answer_77214

编辑：Azzi Abdelmalek 2013-3-3

在 MATLAB Online 中打开

Try

subs(y)

Example

y=solve('x+1');
whos y
r=subs(y)
whos r

Also

if isempty(y)==1 is the same as if isempty(y)

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xueqi 2013-3-4

f =

exp(-(2*x)/125)/20 + (7*exp(-(7*x)/125))/40 - (7*exp((7*x)/125))/40

xueqi 2013-3-4

编辑：xueqi 2013-3-4

OK. Sorry to be chaos. But when I run this code in my office pc, there is no problem at all. And also this error is not always showing in my own laptop. I just have no idea what is happening... I don.t know it is because I am using 2010 in my own laptop but 2012 in my office pc, or this error just shows randomly!

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DOUBLE cannot convert the input expression into a double array

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