I am trying to find the value of 3 variables that satisfy a condition, namely that their sum is less than or equal to 1. My approach is to use ndgrid to sweep over all combinations of the variables and define a matrix I that contains a 1 if the condition is satisfied. My code is below
ss=0.25;
[pp1,pp2,pp3] = ndgrid(0:ss:1,0:ss:1,0:ss:1);
I = pp1+pp2+pp3<=1
My question is, how do I generate a list of all valid variable combinations? I wish to have a 3 x n vector p that contains all n valid values for pp1,pp2, and pp3.
I've found a solution which I'll post as an answer. It is simply
p = [pp1(I) pp2(I) pp3(I)]
If anyone has a better solution I would appreciate any comments.
Related
I am currently in the process of writing a custom function to compute the RREF of a given m x n matrix. Since I am a complete newbie to MATLAB, I thought it would be a good idea to sample the built-in rref() function.
While examining the part of code that found "the value and index of largest element in the remainder" of the leading column, I had that:
[p,k] = max(abs(A(i:m,j)))
where m is the number of rows of the matrix, and i=j=1.
I understand that max(abs(A(i:m,j))) gives you the value of the largest element in the leading column - a single scalar answer. However, I cannot understand why it manages to assign two values to [p,k], with kbeing the index number for p. could someone please be kind enough to help?
k is the position in your vector where the maximum value is.
For instance, assume we use the vector [1,2,5,2,1]. There the max value is 5. This value is at the third position in the vector. So [p,k] = max([1,2,5,2,1]);will return p=5 and k=3.
The function will assing values depending on how you call it.
p = max(...
will assign only p
[p,k] = max(...
will assign p and k.
Might sound too simple to you but I need some help in regrad to do all folowings in one shot instead of defining redundant variables i.e. tmp_x, tmp_y:
X= sparse(numel(find(G==0)),2);
[tmp_x, temp_y] = ind2sub(size(G), find(G == 0));
X(:)=[tmp_x, tmp_y];
(More info: G is a sparse matrix)
I tried:
X(:)=ind2sub(size(G), find(G == 0));
but that threw an error.
How can I achieve this without defining tmp_x, tmp_y?
A couple of comments with your code:
numel(find(G == 0)) is probably one of the worst ways to determine how many entries that are zero in your matrix. I would personally do numel(G) - nnz(G). numel(G) determines how many elements are in G and nnz(G) determines how many non-zero values are in G. Subtracting these both would give you the total number of elements that are zero.
What you are doing is first declaring X to be sparse... then when you're doing the final assignment in the last line to X, it reconverts the matrix to double. As such, the first statement is totally redundant.
If I understand what you are doing, you want to find the row and column locations of what is zero in G and place these into a N x 2 matrix. Currently with what MATLAB has available, this cannot be done without intermediate variables. The functions that you'd typically use (find, ind2sub, etc.) require intermediate variables if you want to capture the row and column locations. Using one output variable will give you the column locations only.
You don't have a choice but to use intermediate variables. However, if you want to make this more efficient, you don't even need to use ind2sub. Just use find directly:
[I,J] = find(~G);
X = [I,J];
I need to generate m unique random numbers in range 1 to n. Currently what I have implemented is:
round(rand(1,m)*(n-1)+1)
However, some numbers are repeated in the array. How can I get only unique numbers?
You can use randperm.
From the description:
p = randperm(n,k) returns a row vector containing k unique integers
selected randomly from 1 to n inclusive.
Thus, randperm(6,3)
might be the vector
[4 2 5]
Update
The two argument version of randperm only appeared in R2011b, so if you are using an earlier version of MATLAB then you will see that error. In this case, use:
A = randperm(n);
A = A(1:m);
As pointed out above, in Matlab versions older than R2011b randperm only accepts one input argument. In that case the easiest approach, assuming you have the Statistics Toolbx, is to use randsample:
randsample(n,m)
The randperm approach described by #Stewie appears to be the way to go in most cases. However if you can only use Matlab with 1 input argument and n is really large, it may not be feasible to use randperm on all numbers and select the first few. In this case here is what you can do:
Generate an integer between 1 and n
Generate an integer between 1 and n-1, this is the choice out of the available integers.
Repeat until you have m numbers
This can be done with randi and could even be vectorized by just drawing a lot of random numbers at each step until the unique amount is correct.
Use Shuffle, from the MATLAB File Exchange.
Index = Shuffle(n, 'index', m);
This can be done by sorting a random vector of floats:
[i,i]=sort(rand(1,range));
output=i(1:m);
i have two matrices
r=10,000x2
q=10,000x2
i have to find out those rows of q which are one value or both values(as it is a two column matrix) different then r and allocate them in another matrix, right now i am trying this.i cannot use isequal because i want to know those rows
which are not equal this code gives me the individual elements not the complete rows different
can anyone help please
if r(:,:)~=q(:,:)
IN= find(registeredPts(:,:)~=q(:,:))
end
You can probably do this using ismember. Is this what you want? Here you get the values from q in rows that are different from r.
q=[1,2;3,4;5,6]
r=[1,2;3,5;5,6]
x = q(sum(ismember(q,r),2) < 2,:)
x =
3 4
What this do:
ismember creates an array with 1's in the positions where q == r, and 0 in the remaining positions. sum(.., 2) takes the column sum of each of these rows. If the sum is less than 2, that row is included in the new array.
Update
If the values might differ some due to floating point arithmetic, check out ismemberf from the file exchange. I haven't tested it myself, but it looks good.
I just started matlab and need to finish this program really fast, so I don't have time to go through all the tutorials.
can someone familiar with it please explain what the following statement is doing.
[Y,I]=max(AS,[],2);
The [] between AS and 2 is what's mostly confusing me. And is the max value getting assigned to both Y and I ?
According to the reference manual,
C = max(A,[],dim) returns the largest elements along the dimension of A specified by scalar dim. For example, max(A,[],1) produces the maximum values along the first dimension (the rows) of A.
[C,I] = max(...) finds the indices of the maximum values of A, and returns them in output vector I. If there are several identical maximum values, the index of the first one found is returned.
I think [] is there just to distinguish itself from max(A,B).
C = max(A,[],dim) returns the largest elements along the dimension of A specified by scalar dim. For example, max(A,[],1) produces the maximum values along the first dimension (the rows) of A.
Also, the [C, I] = max(...) form gives you the maximum values in C, and their indices (i.e. locations) in I.
Why don't you try an example, like this? Type it into MATLAB and see what you get. It should make things much easier to see.
m = [[1;6;2] [5;8;0] [9;3;5]]
max(m,[],2)
AS is matrix.
This will return the largest elements of AS in its 2nd dimension (i.e. its columns)
This function is taking AS and producing the maximum value along the second dimension of AS. It returns the max value 'Y' and the index of it 'I'.
note the apparent wrinkle in the matlab convention; there are a number of builtin functions which have signature like:
xs = sum(x,dim)
which works 'along' the dimension dim. max and min are the oddbal exceptions:
xm = max(x,dim); %this is probably a silent semantical error!
xm = max(x,[],dim); %this is probably what you want
I sometimes wish matlab had a binary max and a collapsing max, instead of shoving them into the same function...