Consider a general vector which represent some non-linear function
for example:
x = [1 2 3 4 5 6 7 8 9 10];
f = [-1 6 8 7 5 2 0.1 -2 -3];
Is there a method in matlab that can find the solutions of f(x)=0? with some given accuracy
If you think about it, when you have a random distribution f, finding zeros can only be done with linear interpolation between the data points:
For your example, I would define a function myFunc as:
function y = myFunc(val)
x = [1 2 3 4 5 6 7 8 9 10];
f = [-1 6 8 7 5 2 0.1 -2 -3 3];
P = griddedInterpolant (x, f, 'linear', 'linear');
y = P(val);
end
and apply a root searching algorithm via something like fzero:
val = 0;
x = [1 2 3 4 5 6 7 8 9 10];
x = [-inf x inf]; % Look outside boundary too
fun = #myFunc;
sol = zeros(1, numel(x)-1);
cnt = 0;
for i = 1:length(x)-1 % fzero stops at the 1st zero hence the loop over each interval
bound = [x(i) x(i+1)];
try
z = fzero(fun, bound);
cnt = cnt+1;
sol(cnt) = z;
catch
% No answer within the boundary
end
end
sol(cnt+1:end) = [];
Maybe you can try interp1 in arrayfun like below (linear interpolation was adopted)
x0 = arrayfun(#(k) interp1(f(k:k+1),x(k:k+1),0),find(sign(f(1:end-1).*f(2:end))<0));
such that
x0 =
1.1429 7.0476 9.5000
DATA
x = [1 2 3 4 5 6 7 8 9 10];
f = [-1 6 8 7 5 2 0.1 -2 -3 3];
I've made a function that does it, but feel it is something quite "regular" that matlab must have built in answers...so if someone has any write it down and I will accept it as an answer.
function sol = find_zeros(x,f)
f_vec = round(f*10^2)/10^2;
ind=find(diff(sign(f_vec))~=0);
K = length(ind);
if (K>0)
sol = zeros(1,K);
for k=1:K
if (f_vec(ind(k))<f_vec(ind(k)+1))
df = f_vec(ind(k)):0.01:f_vec(ind(k)+1);
else
df = flip(f_vec(ind(k)+1):0.01:f_vec(ind(k)));
end
dx = linspace(x(ind(k)),x(ind(k)+1),length(df));
j = find(df==0);
sol(k) = dx(j);
end
else
sol=[];
end
sol=unique(sol);
end
Related
I am a bit new in using Matlab, and I have a question about defining a multivariable function for vector input.
If the function is a single function, say f(t), I know how to make it for vector input. The general way is to use arrayfun after defining a f(t). How about for multivariable function, say f(x,y)? What I want to do is to get two inputs, say [1 2 3] for x and [4 5 6 7] for y (dimension may be different, but both of them are either column vector or row vector) so that I can calculate to give
[f(1,4),f(1,5),f(1,6),f(1,7);
f(2,4),f(2,5),f(2,6),f(2,7);
f(3,4),f(3,5),f(3,6),f(3,7)]
The difficulty is that the vector input for x and y may not be in the same dimension.
I understand it may be difficult to illustrate if I do not have an example of f(x,y). For my use of f(x,y), it may be very complicated to display f(x,y). For simplicity, treat f(x,y) to be x^2+y, and once defined, you cannot change it to x.^2+y for vector inputs.
Here is a set of suggestions using ndgrid:
testfun = #(x,y) x^2+y; % non-vectorized form
x = 1:3;
y = 4:7;
[X,Y] = ndgrid(x,y);
% if the function can be vectorized (fastest!):
testfun_vec = #(x,y) x.^2+y; % vectorized form
A = testfun_vec(X,Y);
% or without ndgrid (also super fast):
B = bsxfun(testfun_vec,x.',y); % use the transpose to take all combinations
% if not, or if it's not bivariate operation (slowest):
C = arrayfun(testfun,X(:),Y(:));
C = reshape(C,length(x),length(y));
% and if you want a loop:
D = zeros(length(x),length(y));
for k = 1:length(X(:))
D(k) = testfun(X(k),Y(k));
end
Which will output for all cases (A,B,C and D):
5 6 7 8
8 9 10 11
13 14 15 16
As mentioned already, if you can vectorize your function - this is the best solution, and if it has only two inputs bsxfun is also a good solution. Otherwise if you have small data and want to keep your code compact use arrayfun, if you are dealing with large arrays use an un-nested for loop.
Here is the code using for loops and inline functions:
x = [1 2 3];
y = [4 5 6 7];
f = #(x,y) x^2 +y;
A = zeros(length(x), length(y));
for m = 1:length(x)
for n = 1:length(y)
A(m, n) = f(x(m), y(n));
end
end
disp(A);
Result:
A =
5 6 7 8
8 9 10 11
13 14 15 16
>> x = [1 2 3];
>> y = [4 5 6 7];
>> outValue = foo(x, y);
>> outValue
outValue =
5 6 7 8
8 9 10 11
13 14 15 16
Make this function:
function out = foo(x, y)
for i = 1 : length(x)
for j = 1 : length(y)
out(i, j) = x(i)^2 + y(j);
end
end
I have a 3x3 Matrix and want to save the indices and values into a new 9x3 matrix. For example A = [1 2 3 ; 4 5 6 ; 7 8 9] so that I will get a matrix x = [1 1 1; 1 2 2; 1 3 3; 2 1 4; 2 2 5; ...] With my code I only be able to store the last values x = [3 3 9].
A = [1 2 3 ; 4 5 6 ; 7 8 9];
x=[];
for i = 1:size(A)
for j = 1:size(A)
x =[i j A(i,j)]
end
end
Thanks for your help
Vectorized approach
Here's one way to do it that avoids loops:
A = [1 2 3 ; 4 5 6 ; 7 8 9];
[ii, jj] = ndgrid(1:size(A,1), 1:size(A,2)); % row and column indices
vv = A.'; % values. Transpose because column changes first in the result, then row
x = [jj(:) ii(:) vv(:)]; % result
Using your code
You're only missing concatenation with previous x:
A = [1 2 3 ; 4 5 6 ; 7 8 9];
x = [];
for i = 1:size(A)
for j = 1:size(A)
x = [x; i j A(i,j)]; % concatenate new row to previous x
end
end
Two additional suggestions:
Don't use i and j as variable names, because that shadows the imaginary unit.
Preallocate x instead of having it grow in each iteration, to increase speed.
The modified code is:
A = [1 2 3 ; 4 5 6 ; 7 8 9];
x = NaN(numel(A),3); % preallocate
n = 0;
for ii = 1:size(A)
for jj = 1:size(A)
n = n + 1; % update row counter
x(n,:) = [ii jj A(ii,jj)]; % fill row n
end
end
I developed a solution that works much faster. Here is the code:
% Generate subscripts from linear index
[i, j] = ind2sub(size(A),1:numel(A));
% Just concatenate subscripts and values
x = [i' j' A(:)];
Try it out and let me know ;)
MATLAB:
I am trying to do basis expansion of a huge matrix(1000x15).
For example,
X =
x1 x2
1 4
2 5
3 6
I want to build a new matrix.
Y =
x1 x2 x1*x1 x1*x2 x2*x2
1 4 1 4 16
2 5 4 10 25
3 6 9 18 36
Could any one please suggest a easier way to do this
% your input
A = [1 4; 2 5; 3 6];
% generate pairs
[p,q] = meshgrid(1:size(A,2), 1:size(A,2));
% only retain unique pairs
ii = tril(p) > 0;
% perform element wise multiplication
res = [A A(:,p(ii)) .* A(:,q(ii))];
Using the one-liner from this answer to get the 2-combinations of the indices, you can generate the matrix without the interleaved ordering with
function Y = columnCombo(Y)
comb = nchoosek(1:size(Y,2),2);
Y = [Y , Y.^2 , Y(:,comb(:,1)).*Y(:,comb(:,2))];
end
For the interleaved ordering, I came up with this, possibly sub-optimal, solution:
function Y = columnCombo(Y)
[m,n] = size(Y);
comb = nchoosek(1:n,2);
Y = [Y,zeros(m,n + size(comb,1))];
col = n+1;
for k = 1:n-1
Y(:,col) = Y(:,k).*Y(:,k) ;
ms = comb(comb(:,1)==k,:) ;
ncol = size(ms,1) ;
Y(:,col+(1:ncol)) = Y(:,ms(:,1)).*Y(:,ms(:,2)) ;
col = col + ncol + 1 ;
end
Y(:,end) = Y(:,n).^2;
end
This problem is a succession of my previous problem:
1) Extract submatrices, 2) vectorize and then 3) put back
Now, I have two patients, named Ann and Ben.
Indeed the matrices A and B are data for Ann and the matrix C is data for Ben:
Now, I need to design a matrix M such that y = M*x where
y = [a11, a21, a12, a22, b11, b21, b12, b22]' which is a vector, resulting from concatenation of the top-left sub-matrices, Ann and Ben;
x = [2, 5, 4, 6, 7, 9, 6, 2, 9, 3, 4, 2]' which is a vector, resulting from concatenation of sub-matrices A, B and C.
Here, the M is a 8 by 12 matrix that
a11 = 2 + 7, a21 = 5 + 9, .., a22 = 6 + 2 and b11 = 9, ..b22 = 2.
I design the M manually by:
M=zeros(8,12)
M(1,1)=1; M(1,5)=1; % compute a11
M(2,2)=1; M(2,6)=1; % compute a21
M(3,3)=1; M(3,7)=1; % compute a12
M(4,4)=1; M(4,8)=1; % compute a22
M(5,9)=1; % for setting b11 = 9, C(1,1)
M(6,10)=1; % for setting b21 = 3, C(2,1)
M(7,11)=1; % for setting b12 = 4, C(1,2)
M(8,12)=1 % for setting b22 = 2, C(2,2)
Obviously, in general for M(i,j), i means the 8 linear-index position of vector y and j means linear-index position of vector x.
However, I largely simplified my original problem that I want to construct this M automatically.
Thanks in advance for giving me a hand.
Here you have my solution. I have essentially build the matrix M automatically (from the proper indexes) as you suggested.
A = [2 4 8;
5 6 3;
10 3 6];
B = [7 6 3;
9 2 9;
10 2 3];
C = [9 4 7;
3 2 5;
10 3 4];
% All matrices in the same array
concat = cat(3, A, B, C);
concat_sub = concat(1:2,1:2,:);
x = concat_sub(:);
n = numel(x)/3; %Number of elements in each subset
M2 = zeros(12,8); %Transpose of the M matrix (it could be implemented directly over M but that was my first approach)
% The indexes you need
idx1 = 1:13:12*n; % Indeces for A
idx2 = 5:13:12*2*n; % Indices for B and C
M2([idx1 idx2]) = 1;
M = M2';
y = M*x
I have taken advantage of the shape that the matrix M shold take:
You can index into things and extract what you want without multiplication. For your example:
A = [2 4 8; 5 6 3; 10 3 6];
B = [7 6 3; 9 2 9; 10 2 3];
C = [9 4 7;3 2 5; 10 3 4];
idx = logical([1 1 0;1 1 0; 0 0 0]);
Ai = A(idx);
Bi = B(idx);
Ci = C(idx);
output = [Ai; Bi; Ci];
y = [Ai + Bi; Ci]; % desired y vector
This shows each step individually but they can be done in 2 lines. Define the index and then apply it.
idx = logical([1 1 0;1 1 0;0 0 0]);
output = [A(idx); B(idx); C(idx)];
y = [Ai + Bi; Ci]; % desired y vector
Also you can use linear indexing with idx = [1 2 4 5]' This will produce the same subvector for each of A B C. Either way works.
idx = [1 2 4 5]';
or alternatively
idx = [1;2;4;5];
output = [A(idx); B(idx); C(idx)];
y = [Ai + Bi; Ci]; % desired y vector
Either way works. Check out doc sub2ind for some examples of indexing from MathWorks.
I want to create a matrix of the following form
Y = [1 x x.^2 x.^3 x.^4 x.^5 ... x.^100]
Let x be a column vector.
or even some more variants such as
Y = [1 x1 x2 x3 (x1).^2 (x2).^2 (x3).^2 (x1.x2) (x2.x3) (x3.x1)]
Let x1,x2 and x3 be column vectors
Let us consider the first one. I tried using something like
Y = [1 : x : x.^100]
But this also didn't work because it means take Y = [1 x 2.*x 3.*x ... x.^100] ?
(ie all values between 1 to x.^100 with difference x)
So, this also cannot be used to generate such a matrix.
Please consider x = [1; 2; 3; 4];
and suggest a way to generate this matrix
Y = [1 1 1 1 1;
1 2 4 8 16;
1 3 9 27 81;
1 4 16 64 256];
without manually having to write
Y = [ones(size(x,1)) x x.^2 x.^3 x.^4]
Use this bsxfun technique -
N = 5; %// Number of columns needed in output
x = [1; 2; 3; 4]; %// or [1:4]'
Y = bsxfun(#power,x,[0:N-1])
Output -
Y =
1 1 1 1 1
1 2 4 8 16
1 3 9 27 81
1 4 16 64 256
If you have x = [1 2; 3 4; 5 6] and you want Y = [1 1 1 2 4; 1 3 9 4 16; 1 5 25 6 36] i.e. Y = [ 1 x1 x1.^2 x2 x2.^2 ] for column vectors x1, x2 ..., you can use this one-liner -
[ones(size(x,1),1) reshape(bsxfun(#power,permute(x,[1 3 2]),1:2),size(x,1),[])]
Using an adapted Version of the code found in Matlabs vander()-Function (which is also to be found in the polyfit-function) one can get a significant speedup compared to Divakars nice and short solution if you use something like this:
N = 5;
x = [1:4]';
V(:,n+1) = ones(length(x),1);
for j = n:-1:1
V(:,j) = x.*V(:,j+1);
end
V = V(:,end:-1:1);
It is about twice as fast for the example given and it gets about 20 times as fast if i set N=50 and x = [1:40]'. Although I state that is not easy to compare the times, just as an option if speed is an issue, you might have a look at this solution.
in octave, broadcasting allows to write
N=5;
x = [1; 2; 3; 4];
y = x.^(0:N-1)
output -
y =
1 1 1 1 1
1 2 4 8 16
1 3 9 27 81
1 4 16 64 256