Error using load Must be a string scalar or character vector
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thanks a lot for your answer I now understand my mistake but after this when I am trying to write the rain flow script in order to reduce a spectrum of varying stress there is this error which is showing"Error using load Must be a string scalar or character vector.I am also attaching the script thanks in advance for your help.
5 Comments
Harish Ramachandran
on 20 Dec 2017
Edited: Harish Ramachandran
on 20 Dec 2017
load is primarily used for adding variables from a file into a work space. I am not quite sure what you are trying to achieve by the command -
s1=load(i);
@Prateek Srivastava: What do you expect these to do?:
s1=load(i);
s2=load(i+1);
s3=load(i+2);
s4=load(i+3);
Inventing syntaxes is a very inefficient way of writing code. You would be much better off reading the documentation and using syntaxes that are actually supported by MATLAB.
Prateek Srivastava
on 20 Dec 2017
Harish Ramachandran
on 20 Dec 2017
Is the signal in the .txt file? If so can you please attach that as well?
Jan
on 21 Dec 2017
@Prateek: The code
abs(s2-s1)>=abs(s2-s3),abs(s3-s4)>=abs(s2-s3)
% ^
does not make sense. The comma means, that both parts are evaluated, but independent from each other. A simplified code:
x = 1;
y = 2;
if x == 1, y == 7
Now the condition of the IF command is "x == 1". Afterwards y is compared to 7, but this does not concern the IF in any way. Maybe you want:
if abs(s2-s1)>=abs(s2-s3) & abs(s3-s4)>=abs(s2-s3)
% or
if abs(s2-s1)>=abs(s2-s3) | abs(s3-s4)>=abs(s2-s3)
But
if abs(s2-s1)>=abs(s2-s3) , abs(s3-s4)>=abs(s2-s3)
is equivalent to:
if abs(s2-s1)>=abs(s2-s3)
Answers (2)
I am trying to extract the first four values from my signal
The symbol "load" was not defined as a variable. Then the command load() is used, but this cannot be meant. I cannot know which variable is meant, perhaps B or t? Where do you expect "load" to be defined?
Other hints:
if abs(s2-s1)>=abs(s2-s3),abs(s3-s4)>=abs(s2-s3);
What is the purpose of the part behind the comma? Perhaps you want:
if abs(s2-s1)>=abs(s2-s3) & abs(s3-s4)>=abs(s2-s3)
"sign" is a built-in function. Shadowing it by a local variable can cause serious confusion if you want to use the function later on. Avoid to use the names of important Matlab functions as local variables. See:
clear('sign')
sign(2)
sign = rand(1, 10);
sign(2)
5 Comments
Prateek Srivastava
on 20 Dec 2017
Jan
on 20 Dec 2017
@Prateek Srivastava: Seriously? Did you read my answer? Let me repeat the essential part:
"load" is not a defined variable. Then Matlab use it as command load(), which expects a file name as first input. And such file names are provided as string or char vector, as the error message tells you.
Think twice: What do you expect the line "s1 = load(i)" to do? Do you want to copy the i.th element of the array "load" to the variable s1? Then why do you assume that "load" is an existing variable? It is not.
Use the debugger to check your code. Type this in the command window:
dbstop if error
or set a breakpoint in the failing line. Then run the code again until it stops at the error. Now check the variables, which are used in this line:
which load
You will see, that "load" is the function to load MAT files. Then "load(i)" must fail.
Prateek Srivastava
on 20 Dec 2017
Edited: Jan
on 21 Dec 2017
Prateek Srivastava
on 21 Dec 2017
Jan
on 21 Dec 2017
@Prateek Srivastava: There is no variable called "load" in your posted code. The error message means, that there is no such variable at all. In consequence, you cannot obtain the i.th element of this variable, because this variable does not exist at all.
Maybe you want the data of A or B?
Harish Ramachandran
on 21 Dec 2017
You imported the data and stored the stress values etc. inside a struct A. s1 should get the value from the structure instead of using load command.
A=importdata('rain.txt');
This will create a struct A with A.data containing your required values.
>> A.data
ans =
1 -10
2 10
3 0
4 30
5 -20
6 0
7 -10
8 20
9 0
If you want to get the first stress value - access it by indexing through 'A.data'
s1 = A.data(1,2)
s1 =
-10
>> s2 = A.data(2,2)
s2 =
10
and so on... Hope this helps.
7 Comments
Prateek Srivastava
on 21 Dec 2017
Edited: Jan
on 21 Dec 2017
Harish Ramachandran
on 21 Dec 2017
Edited: Harish Ramachandran
on 22 Dec 2017
- Firstly, there is no stress_PV in the script you provided. Are you referring to sign_PV?
- If so, then sign_PV is not a table but a column vector which contains 9 elements. You want to access the element not via a 'load' command but using vector indexing.
>> sign_PV
sign_PV =
-10
10
0
30
-20
0
-10
20
0
while i<=(length(sign_PV))
s1=sign_PV(i);
s2=sign_PV(i+1);
s3=sign_PV(i+2);
s4=sign_PV(i+3);
...
And then proceed with the computation you were talking about while taking care of the index limits and boundary conditions.
Walter Roberson
on 21 Dec 2017
We would usually call that a column vector (a vector in the form of a column) rather than a row vector (which would be a vector in the form of a row)
Harish Ramachandran
on 22 Dec 2017
Oops my bad. I've corrected that mistake.
Prateek Srivastava
on 22 Dec 2017
Walter Roberson
on 22 Dec 2017
Edited: Walter Roberson
on 22 Dec 2017
Your code does not assign any value to i before it uses i
Harish Ramachandran
on 28 Dec 2017
Use a 'for' loop, initialize the value of i and then proceed to assignment. Or you could initialize and update the value of i accordingly in the 'while' loop.
Thank you Walter Roberson!
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