I'm new to MATLAB, and programming in general, and I am having difficulty accomplishing what I am sure is a very, very simple task:
I have a list of vectors v_i for i from 1 to n (n in some number), all of the same size k. I would like to create a vector v that is a "concatenation" (don't know if this is the correct terminology) of these vectors in increasing order: what I mean by this is that the first k entries of v are the k entries of v_1, the k+1 to 2k entries of v are the k entries of v_2 etc. etc. Thus v is a vector of length nk.
How should I create v?
To put this into context, here is function I've began writing (rpeakindex will just a vector, roughq would be the vector v I mentioned before):
function roughq = roughq(rpeakindex)
for i from 1 to size(rpeakindex) do
v_i = [rpeakindex(i)-30:1:rpeakindex(i)+90]
end
Any help is appreciated
Let's try two things.
First, for concatenating vectors there are a couple of methods here, but the simplest would be
h = horzcat(v_1, v_2);
The bigger problem is to enumerate all vectors with a "for" loop. If your v_n vectors are in a cell array, and they are in fact v{i}, then
h= [];
for j=1:n
h = horzcat(h, v{i});
end
Finally, if they only differ by name, then call them with
h=[];
for j=1:n
h= horzcat(h, eval(sprintf('v_%d',j));
end
Let the arrays (vectors) be:
v_1=1:10;
v_2=11:20;
v_3=21:30;
v_4=31:40;
and so on.
If they are few (e. g. 4), you can directly set then as input in the cat function:
v=cat(2,v_1,v_2,v_3,v_4)
or the horzcat function
v=horzcat(v_1,v_2,v_3,v_4)
otherwise you can use the eval function within a loop
v1=[];
for i=1:4
eval(['v1=[v1 v_' num2str(i) ']'])
end
Hope this helps.
Concatenating with horzcat is definitely an option, but since these vectors are being created in a function, it would be better to concatenate these vectors automatically in the function itself rather than write out horzcat(v1,v2,....vn) manually.
Given the function mentioned in the question, I would suggest something like this:
function v = roughq(rpeakindex)
v = zeros(121,length(rpeakindex)); %// create a 2D array of all zeros
for i = 1:size(rpeakindex)
v(:,i) = [rpeakindex(i)-30:1:rpeakindex(i)+90]; %// set result to ith column of v
end
v = v(:)'; %'//reshape v to be a single vector with the columns concatenated
end
Here's a simplified example of what's going on:
N = 3;
v = zeros(5,N);
for i = 1:N
v(:,i) = (1:5)*i;
end
v = v(:)';
Output:
v =
1 2 3 4 5 2 4 6 8 10 3 6 9 12 15
You may want to read up on MATLAB's colon operator to understand the v(:) syntax.
If you mean 2d matrix, you are using for holding vectors and each row hold vector v then you can simply use the reshape command in matlab like below:
V = [] ;
for i = 1:10
V(i,:) = randi (10,1 ,10) ;
end
V_reshpae = reshape (V, 1, numel(V)) ;
Related
Suppose I have a column vector of formulae like this
N =
4*k2 + 5*k3 + k1*x
7*k2 + 8*k3 + k1*y
and a column vector of symbolic variables like this
k =
k1
k2
k3
The formulae are linear with respect to k. I'd like to find a matrix M such that M*k equals N.
I can do this with N/k. However, that gives
[ (4*k2 + 5*k3 + k1*x)/k1, 0, 0]
[ (7*k2 + 8*k3 + k1*y)/k1, 0, 0]
which is correct, but not what I want. What I want is the matrix
x 4 5
y 7 8
which seems to me the simplest answer in that it involves no variables from k.
How do I convince Matlab to factor out the specified variables from a formula or a vector of formulae?
You can use coeffs, specifically the form
C = coeffs(p,vars) returns coefficients of the multivariate polynomial p with respect to the variables vars.
Since the first input needs to be a polynomial, you need to pass each component of N:
coeffs(N(1), k)
coeffs(N(2), k)
Or use a loop and store all results in a symbolic array:
result = sym('result', [numel(N) numel(k)]); % create symbolic array
for m = 1:numel(N)
result(m,:) = coeffs(N(m), k);
end
In your example, this gives
result =
[ 5, 4, x]
[ 8, 7, y]
Based on #LuisMendo's answer, I used coeffs. But there are a couple of problems with coeffs. The first is that its result doesn't include any coefficients that are 0. The second is that it doesn't seem to guarantee that the coefficients are ordered the same way as the variables in its second argument. I came up with the following function to replace coeffs.
Luckily coeffs returns a second result that lists the variables associated with each item in the first result. (It's more complicated if the formula is not linear.)
function m = factorFormula(f, v )
% Pre: f is a 1x1 sym representing a
% linear function of the variables in v.
% Pre: v is a column vector of variables
% Post: m is a row vector such that m*v equals f
% and the formulas in m do not contain the
% variables in v
[cx,tx] = coeffs(f,v)
n = size(v,1)
m = sym(zeros(1,n))
for i = 1:n
j = find(tx==v(i))
if size(j,2) == 1
m(i) = cx(j)
end
end
end
This only works for one formula, but it can be extended to a vector using the loop in #LuisMendo's answer or this equivalent expression in #Sanchises comment there.
cell2sym(arrayfun( #(f)factorFormula(f,k),N,'UniformOutput',false ) )
I hope there is a better answer than this.
I have a matrix:
1|2|3|4
4|5|6|7
7|8|9|10
10|11|12|13
I want to multiply the indices of this matrix with indices of another matrix of different size:
7|8|9
9|10|10
10|11|11
for these two matrices I have used the following for loops:
for x=1:4
for y=1:4
for m=1:3
for n=1:3
c=(m*x+n*y);
end
end
end
end
Is there any way to rewrite the above code without using loops? If the indices of each element can be generated in the above matrices, I think it can be done. Please help
mx = m'*x;
mx = mx(:);
ny = n'*y;
ny = ny(:);
mxe = repmat(mx, [length(ny), 1]);
nye = repmat(ny, [length(mx), 1]);
c = mxe+nye;
This will result in c containing all the values that get put in during that loop you have there (note that in your loop, value gets assigned and overwritten).
This question already has answers here:
Element-wise array replication in Matlab
(7 answers)
Closed 6 years ago.
This is a basic program but since I'm new to MATLAB, I'm not able to figure out the solution.
I have a column vector "Time" in which I want to print value "1" in first 147 cells, followed by "2" in 148 to 2*147 cells and so on. For that, I have written the following script:
Trial>> c=1;
Trial>> k=0;
Trial>> for i = c:146+c
Time(i,1)=1+k;
c=i;
k=k+1;
end
I know I need to iterate the loop over "Time(i,1)=1+k;" before it executes the next statement. I tried using break but that's not supposed to work. Can anyone suggest me the solution to get the desired results?(It was quite simple in C with just the use of curly braces.)
I am sure you don't want to run c=i; in every iteration.
My code should work for you:
x = 10; % Replace 10 by the max number you need in your array.
k = 1;
for i = 1 : x * 147
Time(i, 1) = k;
if rem(i, 147) == 0
k = k + 1;
end
end
This is the prime example of a piece of code that should be vectorized can help you understand vectorization. Your code can be written like this:
n = 147;
reps = 10; %% Replace this by the maximum number you want your matrix to have
Time = reshape(bsxfun(#plus, zeros(n,1), 0:reps), 1, []);
Explanation:
Let A be a column vector (1 column, n rows), and B be a row vector (1 row, m columns.
What bsxfun(#plus, A, B) will do here is to add all elements in A with all elements in B, like this:
A(1)+B(1) A(1)+B(2) A(1)+B(3) ... A(1)+B(m)
A(2)+B(1) A(2)+B(2) ............. A(2)+B(m)
............................................
A(n)+B(1) A(n)+B(2) .............. A(n)+B(m)
Now, for the two vectors we have: zeros(n,1), and 0:reps, this will give us;
0+0 0+1 0+2 0+reps
0+0 0+1 0+2 0+reps
% n rows of this
So, what we need to do now is place each column underneath each other, so that you will have the column with zeros first, then the row with ones, ... and finally the one with reps (147 in your case).
This can be achieved by reshaping the matrix:
reshape(bsxfun(#plus, zeros(n,1), 0:reps), [], 1);
^ ^ ^ ^
| | | Number of rows in the new matrix. When [] is used, the appropriate value will be chosen by Matlab
| | Number of rows in the new matrix
| matrix to reshape
reshape command
Another approach is using kron:
kron(ones(reps+1, 1) * 0:(n-1)
For the record, a review of your code:
You should always preallocate memory for matrices that are created inside loops. In this case you know it will become a matrix of dimensions ((reps+1)*n-by-1). This means you should do Time = zeros((reps+1)*n, 1);. This will speed up your code a lot.
You shouldn't use i and j as variable names in Matlab, as they denote the imaginary unit (sqrt(-1)). You can for instance do: for ii = 1:(n*147) instead.
You don't want c=i inside the loop, when the loop is supposed to go from c to c + 146. That doesn't make much sense.
You can use repmat,
x = 10; % Sequence length (or what ever it can be called)
M = repmat(1:x,147,1); % Replicate array 1:x for 147 columns
M = M(:); % Reshape the matrix so that is becomes a column vector.
I can assume that this is a task to practice for loops, but this will work.
An alternative solution may be to do
n = 147;
reps = 10;
a = ceil( (1:(n*reps)) / n);
You first construct an array with the length you want. Then you divide, and round of upwards. 1 to 147 will then become 1.
So I have the following matrices:
A = [1 2 3; 4 5 6];
B = [0.5 2 3];
I'm writing a function in MATLAB that will allow me to multiply a vector and a matrix by element as long as the number of elements in the vector matches the number of columns. In A there are 3 columns:
1 2 3
4 5 6
B also has 3 elements so this should work. I'm trying to produce the following output based on A and B:
0.5 4 9
2 10 18
My code is below. Does anyone know what I'm doing wrong?
function C = lab11(mat, vec)
C = zeros(2,3);
[a, b] = size(mat);
[c, d] = size(vec);
for i = 1:a
for k = 1:b
for j = 1
C(i,k) = C(i,k) + A(i,j) * B(j,k);
end
end
end
end
MATLAB already has functionality to do this in the bsxfun function. bsxfun will take two matrices and duplicate singleton dimensions until the matrices are the same size, then perform a binary operation on the two matrices. So, for your example, you would simply do the following:
C = bsxfun(#times,mat,vec);
Referencing MrAzzaman, bsxfun is the way to go with this. However, judging from your function name, this looks like it's homework, and so let's stick with what you have originally. As such, you need to only write two for loops. You would use the second for loop to index into both the vector and the columns of the matrix at the same time. The outer most for loop would access the rows of the matrix. In addition, you are referencing A and B, which are variables that don't exist in your code. You are also initializing the output matrix C to be 2 x 3 always. You want this to be the same size as mat. I also removed your checking of the length of the vector because you weren't doing anything with the result.
As such:
function C = lab11(mat, vec)
[a, b] = size(mat);
C = zeros(a,b);
for i = 1:a
for k = 1:b
C(i,k) = mat(i,k) * vec(k);
end
end
end
Take special note at what I did. The outer-most for loop accesses the rows of mat, while the inner-most loop accesses the columns of mat as well as the elements of vec. Bear in mind that the number of columns of mat need to be the same as the number of elements in vec. You should probably check for this in your code.
If you don't like using the bsxfun approach, one alternative is to take the vector vec and make a matrix out of this that is the same size as mat by stacking the vector vec on top of itself for as many times as we have rows in mat. After this, you can do element-by-element multiplication. You can do this stacking by using repmat which repeats a vector or matrices a given number of times in any dimension(s) you want. As such, your function would be simplified to:
function C = lab11(mat, vec)
rows = size(mat, 1);
vec_mat = repmat(vec, rows, 1);
C = mat .* vec_mat;
end
However, I would personally go with the bsxfun route. bsxfun basically does what the repmat paradigm does under the hood. Internally, it ensures that both of your inputs have the same size. If it doesn't, it replicates the smaller array / matrix until it is the same size as the larger array / matrix, then applies an element-by-element operation to the corresponding elements in both variables. bsxfun stands for Binary Singleton EXpansion FUNction, which is a fancy way of saying exactly what I just talked about.
Therefore, your function is further simplified to:
function C = lab11(mat, vec)
C = bsxfun(#times, mat, vec);
end
Good luck!
I have a non-fixed dimensional matrix M, from which I want to access a single element.
The element's indices are contained in a vector J.
So for example:
M = rand(6,4,8,2);
J = [5 2 7 1];
output = M(5,2,7,1)
This time M has 4 dimensions, but this is not known in advance. This is dependent on the setup of the algorithm I'm writing. It could likewise be that
M = rand(6,4);
J = [3 1];
output = M(3,1)
so I can't simply use
output=M(J(1),J(2))
I was thinking of using sub2ind, but this also needs its variables comma separated..
#gnovice
this works, but I intend to use this kind of element extraction from the matrix M quite a lot. So if I have to create a temporary variable cellJ every time I access M, wouldn't this tremendously slow down the computation??
I could also write a separate function
function x= getM(M,J)
x=M(J(1),J(2));
% M doesn't change in this function, so no mem copy needed = passed by reference
end
and adapt this for different configurations of the algorithm. This is of course a speed vs flexibility consideration which I hadn't included in my question..
BUT: this is only available for getting the element, for setting there is no other way than actually using the indices (and preferably the linear index). I still think sub2ind is an option. The final result I had intended was something like:
function idx = getLinearIdx(J, size_M)
idx = ...
end
RESULTS:
function lin_idx = Lidx_ml( J, M )%#eml
%LIDX_ML converts an array of indices J for a multidimensional array M to
%linear indices, directly useable on M
%
% INPUT
% J NxP matrix containing P sets of N indices
% M A example matrix, with same size as on which the indices in J
% will be applicable.
%
% OUTPUT
% lin_idx Px1 array of linear indices
%
% method 1
%lin_idx = zeros(size(J,2),1);
%for ii = 1:size(J,2)
% cellJ = num2cell(J(:,ii));
% lin_idx(ii) = sub2ind(size(M),cellJ{:});
%end
% method 2
sizeM = size(M);
J(2:end,:) = J(2:end,:)-1;
lin_idx = cumprod([1 sizeM(1:end-1)])*J;
end
method 2 is 20 (small number of index sets (=P) to convert) to 80 (large number of index sets (=P)) times faster than method 1. easy choice
For the general case where J can be any length (which I assume always matches the number of dimensions in M), there are a couple options you have:
You can place each entry of J in a cell of a cell array using the num2cell function, then create a comma-separated list from this cell array using the colon operator:
cellJ = num2cell(J);
output = M(cellJ{:});
You can sidestep the sub2ind function and compute the linear index yourself with a little bit of math:
sizeM = size(M);
index = cumprod([1 sizeM(1:end-1)]) * (J(:) - [0; ones(numel(J)-1, 1)]);
output = M(index);
Here is a version of gnovices option 2) which allows to process a whole matrix of subscripts, where each row contains one subscript. E.g for 3 subscripts:
J = [5 2 7 1
1 5 2 7
4 3 9 2];
sizeM = size(M);
idx = cumprod([1 sizeX(1:end-1)])*(J - [zeros(size(J,1),1) ones(size(J,1),size(J,2)-1)]).';