Sorry for asking such a simple and silly question, but Matlab is really too hard to use for me. My question is just how to find the position the minimal value of a three dimensional array in Matlab.
For example, suppose I define a three dimensional array
m=zeros(2,2,2);
m(1,2,2)=-2;
The minimal value of m should be -2, located at (1,2,2). I can find the minimal value by
m0=min(min(min(m)));
But when I find its position by using
[x y z]=find(m==m0);
Instead of returning x=1, y=2 and z=2, it returns x=1, y=4 and z=1.
I appreciate if anyone would answer this question!
You can use min to find the minimum index of m and then convert it to x, y and z coordinates. There is no need to use find also.
min can be used with multiple output arguments to return the index of the minimum element. Here, I also use : to return every element of m as a column vector.
>> m=zeros(2,2,2);
>> m(1,2,2)=-2;
>> m(:)
ans =
0
0
0
0
0
0
-2
0
>> [~, ind] = min(m(:))
ind =
7
Now we have our index we need to convert it back into x, y and z coordinates. This can be done using ind2sub or manually by hand.
>> [x y z] = ind2sub(size(m), ind)
x =
1
y =
2
z =
2
You're correct. This is more complicated than it should be. The problem is that MATLAB is hardwired to work with matrices (i.e. arrays of rank 2), rather than arrays of general rank. Here's the solution:
m0 = min(m(:))
[x y z] = ind2sub(size(m), find(m(:) == m0))
Explanation:
If you type help find, you may notice that your original code was using the [rows, cols, vals] version of find, which is not what you expected.
Instead, min(m(:)) is a simplification of your min(min(min(m))). It automatically reshapes m into a rank one array (i.e. a vector).
The expression find(m(:) == m0) returns a single index for the minimum position in this reshaped vector. Finally, ind2sub converts this single index into a set of three indices, given the shape of m.
Related
How to find the approximate numbers in two matrix?
For example,
There are two matrixes
A=[1.567 1.679 1.366 ;
2.467 3.587 6.134 ;
3.497 5.877 9.465]
B=[3.134 5.100 7.555 ;
7.465 4.715 4.267 ;
2.347 4.111 4.503]
So in A matrix 2.467 is close to 2.347 in B matrix.
How can I find them by coding?
Create a meshgrid of all the values in A and all the values in B so we can compare them:
[Amesh,Bmesh] = meshgrid(A(:),B(:))
Now find the absolute value of the difference:
absdiff = abs(Amesh-Bmesh)
This is a matrix of the absolute difference between every value in A vs. every value in B. So the minimum value in this table is your closest match between a value in A and B. To programatically find that value:
[x,y] = find(absdiff == min(absdiff(:)))
x =
3
y =
2
This calculates the minimum value of that matrix, and then finds the x/y position of that value. In this case x is the index into matrix B and y is the index into matrix A.
>> A(y)
ans =
2.4670
>> B(x)
ans =
2.3470
I have inequalities with two unknown variables. So how could I assume one variable with different values and get the others?
For instance: -15<10*x+2*y<20.
How could I assume x=2, 3, and so on, and then find answer of (y) depending on the value of (x)?
I have been trying to apply the assume and find commands, but unfortunately, I could not. So I hope anyone could help me, please.
Looking forward to hearing from you.
I am new to Matlab, so I have been trying to apply solve, assume, and find commands
clear all;
clc;
syms x y real;
z=solve(-15<10*x+2*y,[x y])
b=solve(10*x+2*y<20,[x y])
yinterval = [ z,b]
I expect the output: to assume x=different numbers and then y= be a list of possible results depending on the value of x
Thanks,
For each value of x, technically there are infinite values of y that satisfy those equations, so for my solution, I assumed x and y were integer values. As well, it appears that you want to give the program a set of x values and have it calculate y values for each x value. Instead of using the solve command, we can simply use a couple of loops to find all satisfactory integer values of y for each value of x.
To start, we need to make a results matrix to store each x,y pair that satisfies the equations you've given. This is called pre-allocation, as we're pre-allocating the space needed to store our answers. Using the equations, we can deduce that there will be 17 satisfactory y values per x. So, our first two lines of code will be initializing the desired x-values and the results matrix:
xVec = 1:5; %x-vector, change this to whatever x-values you want to test
results = zeros(length(xVec)*14, 2); %results matrix
Note: If you decide to iterate x or y by a value different than +1 (more on that later), you'll need to come up with a different method of creating this results matrix. You could also just not pre-allocate the results matrix, but your code will run slower as the size of the results matrix will be changing on each loop.
Next are the loops. Admittedly, this is not the most elegant solution, but it'll get the job done. First, we need an index to keep up with where we are in our results matrix. This is pretty easy, we'll just call it index and start at 1 (since MATLAB indexes from 1 in matrices. Remember that!):
index = 1; %index for results matrix
Next, we need to loop through each value in our x-vector. Simply use a for loop:
for x = xVec
...
For each value of x, there is a minimum value of y. This value can be solved for in
-15 < 10*x + 2*y --> -14 = 10*x + 2*y_min
So, simply solving for y gives us our next line of code:
y = -7 - 5*x; %solving for y
Note: each time we iterate x in our for loop, a new starting value of y will be calculated.
Finally, we need to loop through values of y that still satisfy the inequalities given. This is performed through use of a while loop:
while 10*x + 2*y > -15 && 10*x + 2*y < 20
...
Note: && is the 'and' statement while using loops. You can't use a single equation for this (i.e. you can't say something like -15 < x < 20, you have to split them up using &&).
Since we solved for the first value of y, we can go ahead and record the current x and y values in our results matrix:
results(index, :) = [x, y]; %storing current x- and y-values
Then, we need to iterate y, as otherwise we'd be stuck in this while-loop forever.
y = y + 1;
Note: You can iterate this y-value with whatever amount you want. I chose to iterate by 1 each time, as I assumed you wanted to find integer values. Just change the +1 to whatever value you want.
Finally, we iterate our index, so that the next pair of x,y values that satisfy our equations don't overwrite our previous solutions.
index = index + 1;
All that's left is to close our loops and run! As I said, this isn't the most efficient solution, so I wouldn't use this for large amounts of x- and y-values. As well, like with iterating the y-values, the x-values can have any 'step-size' you want. As it's coded currently, it jumps +1 between each x, but changing the xVec input to any vector will still work (ex. xVec = 1:0.1:5; iterates the x-value by +0.1 each step instead of +1).
Here's the code all together, sans comments (since I wrote the comments while making the above code snippets):
xVec = 1:5;
results = zeros(length(xVec)*14, 2);
index = 1;
for x = xVec
y = -7 - 5*x;
while 10*x + 2*y > -15 && 10*x + 2*y < 20
results(index, :) = [x, y];
y = y + 1;
index = index + 1;
end
end
Let me know if you have any questions!
I would like to generate an array which contains all ordered samples of length k taken from a set of n elements {a_1,...,a_n}, that is all the k-tuples (x_1,...,x_k) where each x_j can be any of the a_i (repetition of elements is allowed), and whose total number is n^k.
Is there a built-in function in Matlab to obtain it?
I have tried to write a code that iteratively uses the datasample function, but I couldn't get what desired so far.
An alternative way to get all the tuples is based on k-base integer representation.
If you take the k-base representation of all integers from 0 to n^k - 1, it gives you all possible set of k indexes, knowing that these indexes start at 0.
Now, implementing this idea is quite straightforward. You can use dec2base if k is lower than 10:
X = A(dec2base(0:(n^k-1), k)-'0'+1));
For k between 10 and 36, you can still use dec2base but you must take care of letters as there is a gap in ordinal codes between '9' and 'A':
X = A(dec2base(0:(n^k-1), k)-'0'+1));
X(X>=17) = X(X>=17)-7;
Above 36, you must use a custom made code for retrieving the representation of the integer, like this one. But IMO you may not need this as 2^36 is quite huge.
What you are looking for is ndgrid: it generates the grid elements in any dimension.
In the case k is fixed at the moment of coding, get all indexes of all elements a this way:
[X_1, ..., X_k] = ndgrid(1:n);
Then build the matrix X from vector A:
X = [A(X_1(:)), ..., A(X_k(:))];
If k is a parameter, my advice would be to look at the code of ndgrid and adapt it in a new function so that the output is a matrix of values instead of storing them in varargout.
What about this solution, I don't know if it's as fast as yours, but do you think is correct?
function Y = ordsampwithrep(X,K)
%ordsampwithrep Ordered samples with replacement
% Generates an array Y containing in its rows all ordered samples with
% replacement of length K with elements of vector X
X = X(:);
nX = length(X);
Y = zeros(nX^K,K);
Y(1,:) = datasample(X,K)';
k = 2;
while k < nX^K +1
temprow = datasample(X,K)';
%checknew = find (temprow == Y(1:k-1,:));
if not(ismember(temprow,Y(1:k-1,:),'rows'))
Y(k,:) = temprow;
k = k+1;
end
end
end
Suppose in MATLAB I have a real matrix A which is n x m and a binary matrix B of the same size. The latter matrix defines the optimization set (all indices for which the element of B equals one): over this set I would like to find the maximal element of A. How can I do this?
The first idea I had is that I consider C = A.*B and look for the maximal element of C. This works fine for all matrices A which have at least one positive element, however it does not work for matrices with all negative elements.
You can do
C = A(B==1);
to give you an array of just the values of A corresponding to a value of 1 in B. And
max( C )
will give you the maximum value of A where B is 1
With this method you don't run into a problem when all values of A are negative as the zeros don't appear in C.
Obviously you can condense this to
desiredValue = max(A(B(:)==1));
I am using the colon operator to make sure that the result of A(B(:)==1) is a column vector - if B is all ones I am not sure if Matlab would return a vector or a nxm matrix (and I can't confirm right now).
update to get the index of the value, you can do:
f = find(B==1);
[m mi] = max(A(f));
maxIndex = f(mi);
And to get that back to the 2D elements:
[i j] = ind2sub(size(A), maxIndex);
I've been searching the net for a couple of mornings and found nothing, hope you can help.
I have an anonymous function like this
f = #(x,y) [sin(2*pi*x).*cos(2*pi*y), cos(2*pi*x).*sin(2*pi*y)];
that needs to be evaluated on an array of points, something like
x = 0:0.1:1;
y = 0:0.1:1;
w = f(x',y');
Now, in the above example everything works fine, the result w is a 11x2 matrix with in each row the correct value f(x(i), y(i)).
The problem comes when I change my function to have constant values:
f = #(x,y) [0, 1];
Now, even with array inputs like before, I only get out a 1x2 array like w = [0,1];
while of course I want to have the same structure as before, i.e. a 11x2 matrix.
I have no idea why Matlab is doing this...
EDIT 1
Sorry, I thought it was pretty clear from what I wrote in the original question, but I see some of you asking, so here is a clarification: what I want is to have again a 11x2 matrix, since I am feeding the function with arrays with 11 elements.
This means I expect to have an output exactly like in the first example, just with changed values in it: a matrix with 11 rows and 2 columns, with only values 0 in the first column and only values 1 in the second, since for all x(i) and y(i) the answer should be the vector [0,1].
It means I expect to have:
w = [0 1
0 1
0 1
...
0 1]
seems pretty natural to me...
You are defining a function f = #(x,y) [0, 1]; which has the input parameters x,y and the output [0,1]. What else do you expect to happen?
Update:
This should match your description:
g=#(x,y)[zeros(size(x)),ones(size(y))]
g(x',y')
Defining an anonymous function f as
f = #(x,y) [0,1];
naturally returns [0,1] for any inputs x and y regardless of the length of those vectors.
This behavior puzzled me also until I realized that I expected f(a,b) to loop over a and b as if I had written
for inc = 1:length(a)
f(a(inc), b(inc))
end
However, f(a,b) does not loop over the length of its inputs, so it merely returns [0,1] regardless of the length of a and b.
The desired behavior can be obtained by defining f as
g=#(x,y)[zeros(size(x)),ones(size(y))]
as Daniel stated in his answer.