Undefined Function or Variable

I am trying to create a function at the end of my script. I believe in my function I defined the variable ‘u.’ But then later down in the function when I try to use ‘u’ I get an error saying that ‘u’ is not defined. I am attaching an image with my code and the error. Let me know if anyone has any suggestions. Any help is appreciated.

For some reason I am having trouble submitting multiple images but I have two ‘end’s at the end of my code. Also the error message is in line 272 ‘if u==1’

Answer 1

[Tommy]

length(o) and length(n) will both be 1, as o and n are scalars, so your code never enters your for loops. You should instead use just o and n in their place. i.e.

for W = 1:o-1

[Maddie Sanders]

Thank you so much for your help!!

[Maddie Sanders]

I have another question if you wouldn’t mind trying to answer it. I am having issues when I try to call my function. I get an error saying “Output argument “message” (and maybe others) not assigned during call to “MatlabProject>conclusion” “Error in v=conclusion(V2)” Also when I don’t write v= and just write conclusion(V2) nothing happens. I get no error but also no result.

[Tommy]

Absolutely!

When you use v=conclusion(V2), you are requesting output from your function conclusion, and so when your function doesn't provide that output, you get an error. When you just use conclusion(V2), you aren't requesting any output, and so it doesn't matter that the function doesn't provide any output.

When you define conclusion with

function message = conclusion(A)

you are offering message as the output. The error is saying that message is never defined within your function. You have

if u==0
    message=%something
elseif u==1
    y=%something
    message=%something
elseif u==2
    y=%something
    message=%something
end

Clearly message is being defined here, so therefore these lines are never reached.

u is a vector of length o-1 which contains possibly many 0s, 1s, and 2s. Consider these three if statements:

>> if ([0 1 2] == 0), disp('hi'), end
>> if ([0 1 2] == 1), disp('hi'), end
>> if ([0 1 2] == 2), disp('hi'), end

'hi' was never displayed, because each if statement evaluated to false. See here for an explanation.

Seems to me that you'll have to rethink your logic. How exactly does the output depend on u, which contains many different values?

Also note that if your if statements did evaluate to true and the lines which define message were reached, you would still get an error:

>> message = disp('hi')
Error using disp
Too many output arguments.

If you just want to assign the string to message, you don't need the disp().

Finally, in the future, copy and paste your code here rather than taking pictures, it's easier for everyone, probably including you :)

[Maddie Sanders]

Thank you so much for your help. It is very much appreciated. I was having issues with being able to ask questions on my computer so I had to do it on my phone which doesn't have access to my code, so I apologize for the inconvenience and thank you again for your help.

I do have one more question because my code isn't doing what I want it to.

V2 is a vector that is 51X1000. I want to use the conclusion function on this vector. In each row of the vector the values get larger. Ex: (1,1)=50 and (2,1)=60. Through my function I am trying to show this consistent relationship that as you go down in rows, the values get larger, but my code is getting these results so I know I am doing something wrong.

I am trying to code for all the columns in the vector V2 if row 1 is greater than row 2 display a 1 in vector u, if row 2>row 3, display a 1, and so on.

If row 1<row 2 display a 2, if row 2<row 3 display a 2...

I don't know if maybe in my u(W)=1/2/0 if I need to do u(W,K) or something like that.

I am just confused what my function is actually doing because when I display 'u' the values are nothing like what I predicted (which would be all 2s).

function [message] = conclusion(A)
%UNTITLED3 Summary of this function goes here
%   Detailed explanation goes here
[o,n]=size(A);
for W=1:o-1
      for K=1:n
        if abs(A(W,K))>abs(A(W+1,K))
            u(W)=1;
        elseif abs(A(W,K))<abs(A(W+1,K))
            u(W)=2;
        else
            u(W)=0;
        end
      end
end
if u==0
      message=disp('The results are inconconclusive.');
elseif u==1
      y=input('Is this graph showing multiple values of length of the distributed load or multiples values of the force per meter value of the distributed load? Say either length or force.');
      message=disp('As the value of the length of the distributed load increases, the internal load decreases.');
elseif u==2
      y=input('Is this graph showing multiple values of length of the distributed load or multiples values of the force per meter value of the distributed load? Say either length or force.');
      message=disp('As the value of the length of the distributed load increases, the internal load decreases.');
else
    message='The results are inconconclusive.'
    disp(num2str(u))
end
end

[Tommy]

Always happy to help! Don't sweat it.

"I don't know if maybe in my u(W)=1/2/0 if I need to do u(W,K) or something like that."

If you do want to test this for all columns, then you would need a second index. Otherwise, the results of each column are being overwritten by the results of the next column. Check if u contains the expected output for the very last column in A - I believe that should be the case...

Also, are any values within A negative?

If I am understanding correctly, the point of this function is to tell the user whether values in A increase or decrease as you move down rows? Can one column increase while another column decreases?

You could simplify the process by using diff and sign. Consider these examples:

>> A = [10 20 30;40 50 60;70 80 90]

A =

    10    20    30
    40    50    60
    70    80    90

>> sign(diff(A))

ans =

     1     1     1
     1     1     1

>> A = [70 80 90;40 50 60;10 20 30]

A =

    70    80    90
    40    50    60
    10    20    30

>> sign(diff(A))

ans =

    -1    -1    -1
    -1    -1    -1

>> A = [10 20 30;70 80 90;40 50 60]

A =

    10    20    30
    70    80    90
    40    50    60

>> sign(diff(A))

ans =

     1     1     1
    -1    -1    -1

>> A = [10 80 60;70 50 90;40 20 30]

A =

    10    80    60
    70    50    90
    40    20    30

>> sign(diff(A))

ans =

     1    -1     1
    -1    -1    -1

Then it seems to me that you'd want to check if every row is increasing across every column. You might then use all. Am I on the right track?