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if i have y4='0.2+6' is there a way to make p=0.2+6 but keeping the numbers like that and not equaling it to 6.2.

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Elliott Cameron 2020-4-26

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关闭： MATLAB Answer Bot 2021-8-20

if i have y4='0.2+6' is there a way to make p=0.2+6 but keeping the numbers like that and not equaling it to 6.2.

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darova 2020-4-26

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Can you explain more? The question is unclear

Elliott Cameron 2020-4-27

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so y4='0.2+6' with the function inside ''. However i want to have p=0.2+6 without the '' but also i do not want p=6.2, i want p=0.2+6 staying like that and not adding together

darova 2020-4-27

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I think it's too much for MATLAB. There is no such option

Tommy 2020-4-27

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What is the ultimate use?

Elliott Cameron 2020-4-27

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to multiply something by (0.2) + 6, but not 6.2

Ameer Hamza 2020-4-27

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Elliot, what about Walter's answer here: https://www.mathworks.com/matlabcentral/answers/520415-hello-if-i-have-two-equations-1-2x-2-and-2-3x-2-4x-7-is-there-a-way-to-remove-the-x-in-a-new-equat. Didn't it pointed you to the right direction?

Tommy 2020-4-28

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Can you explain a bit further? Seems to me that multiplying something by (0.2) + 6 is the same thing as multiplying by 6.2. Are you hoping for the two terms to remain separate when displayed to the user?

Elliott Cameron 2020-4-28

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multiplying something first by 0.2 and then adding 6 is not the same as multiplying by 6,2? no walters answer still had the equation within '....'

Rik 2020-4-28

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This only makes sense in a linear concatenation of char arrays. What exact situation are you trying to use this in? I don't see how this will likely result in valid Matlab syntax. I also don't understand how this would be valid mathematics.

Elliott Cameron 2020-4-28

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this is my demand equation demand=-0.2*x + 20 and later on i am using this find (RandData(1,:).*(-0.2)+20>RandData where this is multiplied by the slope of the line and then the intercect is added. If it were to be multiplied by 19.8 then the answer is different

Rik 2020-4-28

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Why not leave it as a function?

demand_eq=@(x) -0.2*x+20;
result=demand_eq(RandData(1,:));
result=result>RandData;

This is much easier to understand than this.

demand_eq='(-0.2)+20';
result=zeros(1,size(RandData,2));
for n=1:numel(result)
    result(n)=eval(sprintf('%.2f*%s',RandData(1,n),demand_eq))
end
result=result>RandData;

You see the train wreck that results in? You even need eval to get anything at all. You lose any option for vectorization and optimization. You also can't use other Matlab functions that accept functions as inputs.

Tommy 2020-4-28

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Ah I did not interpret 'multiply something by (0.2) + 6' as 'multiply something by (0.2) and then add 6'. I agree with Rik. I don't think you should store this as a character vector in the first place. If for whatever reason you have to start with a character vector, I would first parse it to pull out the numbers and then go from there.

Steven Lord 2020-4-28

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If the function to be evaluated is always going to be a polynomial, another option would be to store it as a vector of coefficients and use polyval to evalute it.

p = [-0.2, 20];
x = 0:0.25:5;
y = polyval(p, x)

One potential benefit this has over the anonymous function is that you could manipulate or query the coefficients by indexing into p.

David Goodmanson 2020-4-28

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Hi Elliott,

could you just do something like

x = 6.2
xintdec = [floor(x) x-floor(x)]
   xintdec =  0.2000    6.0000
y = xintdec*rand  
   y =  0.1265    3.7942

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if i have y4='0.2+6' is there a way to make p=0.2+6 but keeping the numbers like that and not equaling it to 6.2.

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