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.
Related
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)) ;
I am trying out one of the matlab programming question.
Question:
Write a function called hulk that takes a row vector v as an input and
returns a matrix H whose first column consist of the elements of v,
whose second column consists of the squares of the elements of v, and
whose third column consists of the cubes of the elements v. For
example, if you call the function likes this, A = hulk(1:3) , then A
will be [ 1 1 1; 2 4 8; 3 9 27 ].
My Code:
function H = hulk(v)
H = [v; v.^2; v.^3];
size(H) = (n,3);
end
When I test my code using A = hulk(1:3), it throws an error on console.
Your function made an error for argument(s) 0
Am I doing something incorrect? Have I missed anything?
Remove the line size(H) = (n,3);
and add the line H = H';
Final code should be as follows
function H = hulk(v)
H = [v; v.^2; v.^3];
H = H';
end
Your code giving error in matlab editor on the size(H) = (n,3); line
That's why you should use the matlabeditor itself
For your future reference, you can very easily generalise this function in Matlab to allow the user to specify the number of cols in your output matrix. I also recommend that you make this function a bit more defensive by ensuring that you are working with column vectors even if your user submits a row vector.
function H = hulk(v, n)
%//Set default value for n to be 3 so it performs like your current function does when called with the same signature (i.e. only 1 argument)
if nargin < 2 %// nargin stands for "Number of ARGuments IN"
n = 3;
end if
%// Next force v to be a row vector using this trick (:)
%// Lastly use the very useful bsxfun function to perform the power calcs
H = bsxfun(#power, v(:), 1:n);
end
You could reduce the number of operations using cumprod. That way, each v.^k is computed as the previous v.^k times v:
function H = hulk(v, n)
H = cumprod(repmat(v,n,1),1);
The first input argument is the vector, and the second is the maximum exponent.
In an attempt to speed up for loops (or eliminate all together), I've been trying to pass matrices into functions. I have to use sine and cosine as well. However, when I attempt to find the integral of a matrix where the elements are composed of sines and cosines, it doesn't work and I can't seem to find a way to make it do so.
I have a matrix SI that is composed of sines and cosines with respect to a variable that I have defined using the Symbolic Math Toolbox. As such, it would actually be even better if I could just pass the SI matrix and receive a matrix of values that is the integral of the sine/cosine function at every location in this matrix. I would essentially get a square matrix back. I am not sure if I phrased that very well, but I have the following code below that I have started with.
I = [1 2; 3 4];
J = [5 6; 7 8];
syms o;
j = o*J;
SI = sin(I + j);
%SI(1,1) = sin(5*o + 1)
integral(#(o) o.*SI(1,1), 0,1);
Ideally, I would want to solve integral(#(o) o*SI,0,1) and get a matrix of values. What should I do here?
Given that A, B and C are all N x N matrices, for the moment, let's assume they're all 2 x 2 matrices to make the example I'm illustrating more succinct to understand. Let's also define o as a mathematical symbol based on your comments in your question above.
syms o;
A = [1 2; 3 4];
B = [5 6; 7 8];
C = [9 10; 11 12];
Let's also define your function f according to your comments:
f = o*sin(A + o*B + C)
We thus get:
f =
[ o*sin(5*o + 10), o*sin(6*o + 12)]
[ o*sin(7*o + 14), o*sin(8*o + 16)]
Remember, for each element in f, we take the corresponding elements in A, B and C and add them together. As such, for the first row and first column of each matrix, we have 1, 5 and 9. As such, A + o*B + C for the first row, first column equates to: 1 + 5*o + 9 = 5*o + 10.
Now if you want to integrate, just use the int command. This will find the exact integral, provided that the integral can be solvable in closed form. int also can handle matrices so it will integrate each element in the matrix. You can call it like so:
out = int(f,a,b);
This will integrate f for each element from the lower bound a to the upper bound b. As such, supposing our limits were from 0 to 1 as you said. Therefore:
out = int(f,0,1);
We thus get:
out =
[ sin(15)/25 - sin(10)/25 - cos(15)/5, sin(18)/36 - sin(12)/36 - cos(18)/6]
[ sin(21)/49 - sin(14)/49 - cos(21)/7, sin(24)/64 - sin(16)/64 - cos(24)/8]
Bear in mind that out is defined in the symbolic math toolbox. If you want the actual numerical values, you need to cast the answer to double. Therefore:
finalOut = double(out);
We thus get:
finalOut =
0.1997 -0.1160
0.0751 -0.0627
Obviously, this can generalize for any size M x N matrices, so long as they all share the same dimensions.
Caveat
sin, cos, tan and the other related functions have their units in radians. If you wish for the degrees equivalent, append a d at the end of the function (i.e. sind, cosd, tand, etc.)
I believe this is the answer you're after. Good luck!
Is given:
stationary mass ms=1;
Eta-constant eta=0.45;
variable number of repetitions, e.g. N=5;
omega OM=sqrt(ks/ms);
angular frequency om=eta*OM;
time period T=2*pi/om;
upper bound TTT=1.5;
variable for creating function t=0:0.001:TTT;
I made a function like that:
kt=zeros(size(t));
for j=1:2*N+1
n= j-(N+1);
if n==0
k(j)=ks/2;
else
k(j)=i/pi/n;
end
kt=kt+k(j)*exp(i*n*om*t);
end
It’s a Sawtooth wave and there is my problem. From the complex vector kt with value 1x1501 double I have to make the Hermitean matrix for variable N. This means that N can be 5, can be 50, 100, etc. The matrix should look like (picture):
Where k1 is k for N=1, k0 is k for N=0 or k-1 is k for N=-1. Size of matrix is 2*N+1 and 2*N+1.
Thank you for your help and responding!
That's a Toeplitz matrix, you can use the toeplitz command to generate the matrix above. In the general case, this would have been written as:
H = toeplitz(kt(N:end), kt(1:N + 1))
where the first N values in kt correspond to k-N, ... k-1, and the last N + 1 values are k0, ... kN. However, since H is Hermitian, this can be simplified to:
H = toeplitz(kt(N:end));
Try this code:
k=[1 2+i 3+i 4+i 5+i];
N=7;
M=diag(k(1)*ones(N,1));
for j=1:length(k)-1
M=M+diag(k(j+1)*ones(N-j,1),j)+diag(conj(k(j+1))*ones(N-j,1),-j)
end;
Here N should be equal or greater than the length of k array
Just a simple question today. If I have an m*n matrix and I want to cycle through every value in it and apply a probability based function.
Basically, if the probability is p, then each value in the matrix has p chance of having the function applied to it.
I have the loop and the function itself all worked out, but I haven't found how to actually apply the probability itself.
Any advice would be greatly appreciated! Thanks in advance.
Here's your data matrix:
>> X = reshape(1:9, 3, 3);
and you want to (possibly) apply the following function to every element (note how I've vectorized it, so that it can take a matrix as an argument)
>> f = #(x) x.^2;
You want to apply the function with probability p
>> p = 0.25;
So generate some random numbers between 0 and 1, and see which ones are less than p
>> idx = rand(3,3) < p;
And now apply the function to the relevant indexes
>> X(idx) = f(X(idx));
Here's your result:
>> X
X =
1 16 7
2 5 64
3 6 81
The trick is that you can generate the random numbers first, and then apply the other formulas.
For example:
R = rand(m,n) < p
Now each value of R(row,col) corresponds to the outcome that you need to process your original matrix(row,col).
So I suggest applying your function to every cell and then setting the values to a default value based on some probability. So lets assume M is the result of applying to function to everycell:
default = NaN % Or 0 or whatever
p = 0.8;
M(rand(size(M)) > p) = default;
I think you might have to reshape m after this... not sure
M = reshape(M, m, n);