This question already has an answer here:
sum(Array) says index exceeds matrix dimensions [duplicate]
(1 answer)
Closed 3 years ago.
I am trying to find the minimum value of a function of two variables, and then find the value of the variables.
My method is to iterate the function through several values of the variables and then use the min function to find the lowest value.
minval = -10;
maxval = 10;
n = 1;
for x = minval:maxval
for y = minval:maxval
f(n) = abs(x-1)+abs(y-1)+abs(x-3)+abs(y-5)+abs(x-8)+abs(y-3);
n=n+1;
end
end
f(n) = abs(x-1)+abs(y-1)+abs(x-3)+abs(y-5)+abs(x-8)+abs(y-3);
fmin = min(f)
The problem is with the last line:
fmin = min(f)
I am getting the error
??? Index exceeds matrix dimensions.
Error in ==> Lab2 at 65
fmin = min(f)
Why is this? Any help is greatly appreciated.
Don't define a variable named min. Try this:
which min
What does it tell you?
Note that functions in MATLAB can be overloaded by creating variables with the same name. When you do so, you prevent the function from being accessed by MATLAB. This is RARELY a good idea, so don't do it. The solution is
clear min
So you will delete that variable you have created. Of course, if there was something important in that variable, stuff it somewhere else first.
It does indeed look like you have declared a variable called min and so Matlab now treats it like a variable and not a function so it thinks you are trying to index the variable min with the vector f.
But just a comment on your code, leaving off whatever f(442) is you could achieve the same thing in a much more matlabesque fashion without loops like this:
minval = -10;
maxval = 10;
X = minval:maxval;
Y = X;
[xx, yy] = meshgrid(X, Y);
F = abs(xx-1) + abs(yy-1) + abs(xx-3) + abs(yy-5) +abs(xx-8) + abs(yy-3);
Your f is now equivalent to F(:)' (without the final value...), prove it to yourself like this: sum(f(1:end-1) == F(:)')
F as a matrix probably makes more sense than f as a flat vector anyway and you can find the min of F like this: min(F(:))
This code runs perfectly when I plug it into my version of Matlab.
If the error is occuring on line 65, then there must be something else going on in your program. Try and turn this part of your program into a function, so that it won't be impacted by everything else that you're working on.
Related
It says the index in position 1 of the diff function must be a positive integer or a logical value which is it so why am I getting this error? I'm trying to implement the basic Euler method in MATLAB
y=zeros(1,6);
h=0;
x(1)= 0;
y(1)= 0;
i=1;
diff(y,x)= x+y
while h<=1
y(i+1)=y(i) + h*f(x(i))
h=h+0.2;
i=i+1;
end
Edit: Changed it to the code below but it still raises the same error in the line y(i+1)=...
y=zeros(1,6);
x=zeros(1,6);
h=0;
i=1;
g=x+y;
while h<=1
y(i+1)=y(i) + h*g(x(i),y(i));
h=h+0.2;
i=i+1;
end
Approach: I would recommend defining an anonymous function
diffh =#(x,y) x + y; % define this prior to use
to use later inside the loop.
Then changing one line
y(ii+1)=y(ii) + h*diffh(x(ii),y(ii));
should work. I've added the "h" to the end as a convention to remind me this is an anonymous function (see note at end).
% MATLAB R2019a
y = zeros(1,6);
x = zeros(1,6);
h=0;
ii=1;
diffh =#(x,y) x + y;
while h <= 1
y(ii+1)=y(ii) + h*diffh(x(ii),y(ii));
x(ii+1) = x(ii)+h;
h=h+0.2;
ii=ii+1;
end
Side note: I've also changed the index i to ii by convention (though MATLAB doesn't require this). Unless you overwrite their values, both i and j default as the sqrt(-1). You can absolutely use them as indices without issue (provided you don't later need complex or imaginary numbers). To ensure this never becomes an issue, many people just use ii and jj as a convention to preserve the default values.
Note that diff is a MATLAB function itself.
Using i as an index is mostly not a good idea in Matlab, because i is also imaginary number. Perhaps another name could solve the problem.
I'm writing a piece of code for submission through an online grader, as showcased below. B is some given array filled any/all integers 1 through K, and I want to extract the corresponding logical indices of matrix X and perform some operations on those elements, to be put into a return array:
for i = 1:K
A = X(B == i, :);
returnArr(i, :) = sum(A) / length(A);
end
This did not pass the grader at all, and so I looked to change my approach, instead indexing array X indirectly via first using the "find" function, as below:
for i = 1:K
C = find(B == i);
returnArr(i,:) = sum(X(C,:)) / length(C);
end
To my surprise, this code passed the grader without any issues. I know there are a plethora of variations between graders, and one might handle a particular function differently than another, but from a MATLAB functionality/coding perspective, what am I missing in terms of discrepancies between the two approaches? Thanks!
I think the problem is that:
length(C) == sum(B == i)
while
length(A) == max([sum(B == i) , size(X , 2)])
In other words, to obtain the same result of the second example with the first one, you should modify it like this:
A = X(B == i , :);
returnArr(i, :) = sum(A) / size(A,1);
The function length returns the length of largest array dimension
I am a complete beginner at Octave!
I am struggling to figure out the minimum value from the array mse_array. I keep getting an error saying: "subscript indices must be either positive integers less than 2^31 or logicals"
Could someone help me? Here's my code:
function randomLines(data1, data2)
mse_array = []
A = rand(2,10)*100
for i = columns(A)
max = A (1, i)
min = A (2, i)
m = (max-min)/0.16
p = [m min];
array_function_values = polyval(p,data2);
current_mse = mseFunction(data1,array_function_values);
mse_array(end + 1) = current_mse
endfor
min_value = min(mse_array)
endfunction
Q: How to find a minimum from an array in Octave?
octave has built-in self-describing documentation to use once in doubts
A: Always re-read both help min and doc min
octave-3.2.4.exe:1> help min
`min' is a function from the file C:\Octave\3.2.4_gcc-4.4.0\libexec\octave\3.2.4\oct\i686-pc-mingw32\max.oct
-- Loadable Function: min (X)
-- Loadable Function: min (X, Y)
-- Loadable Function: min (X, Y, DIM)
-- Loadable Function: [W, IW] = min (X)
For a vector argument, return the minimum value. For a matrix
argument, return the minimum value from each column, as a row
vector, or over the dimension DIM if defined. For two matrices
(or a matrix and scalar), return the pair-wise minimum. Thus,
min (min (X))
returns the smallest element of X, and
min (2:5, pi)
=> 2.0000 3.0000 3.1416 3.1416
compares each element of the range `2:5' with `pi', and returns a
row vector of the minimum values.
For complex arguments, the magnitude of the elements are used for
comparison.
If called with one input and two output arguments, `min' also
returns the first index of the minimum value(s). Thus,
[x, ix] = min ([1, 3, 0, 2, 0])
=> x = 0
ix = 3
See also: max, cummin, cummax
Re-defining (thus rendering un-usable) elements as reserved words just increases troubles. Try to use your own naming convention, that will not harm the rest of the system -- i.e. anArrayMinimumVALUE or aSmallestVALUE, rather than "killing" the Octave function min() by creating a variable named by accident as "also" min
There are several things wrong with your code. Here are a few to get you started.
The error is probably occurring at this line:
min_value = min(mse_array)
You're using mse_array as an index into min, but the values in mse_array are likely to valid array indices as the error states. (I'm guessing at this purely based on the name of the variable, mse_array, which appear to be from some form of mean squared error calculation.) It's important to check the line number and everything else in error messages.
Another issue in that your for loop only evaluates once. The columns function will return the scalar number of columns of A. Thus your your loop declaration is equivalent to:
for i = 10
In other words, the code will only look at the last column of A. You probably want to use:
for i = 1:columns(A)
Lastly, it's a bad idea to overwrite the built-in min and max functions in Octave with your own variable names. If you overwrite the built-in min function, you won't be able to use it anymore directly until you call clear min (alternatively, you can call builtin, but you should avoid that). Instead, just choose better variable names.
I am trying to produce something like this in MATLAB with function handle
f=#(x,y)(x(1)*x(2)+y);
c=[2 3 4;5 9 2];
h=[5 1 2];
f(c,h)
The answer should be:
15 11 12
But when I write this code below, it just builds a number not an array.
f=#(x)(x(1)*x(2))
f(c)
answer:
10
Can someone explain me where I went wrong?
I do not know what you expected here. The cause of the problem is quite clear.
a = 1;
b = 2;
c = [3 4];
d = a*b+c;
is a scalar + vector operation which always returns
ans = [a*b+c(1), a*b+c(2)];
however scalar*scalar which was the second case always returns a scalar. What you do is that you multiply the first matrix element of x (or c) with the second element. That is to say element c(1,1)*c(2,1) since matlab works columnwise. If you looks at your values you would probably notice the answer is incorrect as well, if you are trying to do what I think you are. You could try this instead,
f=#(x,y)(x(1,:).*x(2,:)+y);
c=[2 3 4;5 9 2];
h=[5 1 2];
f(c,h)
which multipies the elements on the first row of x with the same column on the second row and then adds y. An anonymous function takes a number of inputs and perform a defined operation, the same as ordinary functions or ordinary codes. You can see them as functions that does not require a call to another m file. The main differences (except that ordinary functions gives more freedom), are how they are handled by matlab and not in the syntax.
I want to store data coming from for-loops in an array. How can I do that?
sample output:
for x=1:100
for y=1:100
Diff(x,y) = B(x,y)-C(x,y);
if (Diff(x,y) ~= 0)
% I want to store these values of coordinates in array
% and find x-max,x-min,y-max,y-min
fprintf('(%d,%d)\n',x,y);
end
end
end
Can anybody please tell me how can i do that. Thanks
Marry
So you want lists of the x and y (or row and column) coordinates at which B and C are different. I assume B and C are matrices. First, you should vectorize your code to get rid of the loops, and second, use the find() function:
Diff = B - C; % vectorized, loops over indices automatically
[list_x, list_y] = find(Diff~=0);
% finds the row and column indices at which Diff~=0 is true
Or, even shorter,
[list_x, list_y] = find(B~=C);
Remember that the first index in matlab is the row of the matrix, and the second index is the column; if you tried to visualize your matrices B or C or Diff by using imagesc, say, what you're calling the X coordinate would actually be displayed in the vertical direction, and what you're calling the Y coordinate would be displayed in the horizontal direction. To be a little more clear, you could say instead
[list_rows, list_cols] = find(B~=C);
To then find the maximum and minimum, use
maxrow = max(list_rows);
minrow = min(list_rows);
and likewise for list_cols.
If B(x,y) and C(x,y) are functions that accept matrix input, then instead of the double-for loop you can do
[x,y] = meshgrid(1:100);
Diff = B(x,y)-C(x,y);
mins = min(Diff);
maxs = max(Diff);
min_x = mins(1); min_y = mins(2);
max_x = maxs(1); max_y = maxs(2);
If B and C are just matrices holding data, then you can do
Diff = B-C;
But really, I need more detail before I can answer this completely.
So: are B and C functions, matrices? You want to find min_x, max_x, but in the example you give that's just 1 and 100, respectively, so...what do you mean?